RS Aggarwal Class 9 Mathematics Solutions Chapter 4 Lines and Triangles

Exercise 4A

Exam Tip: This exercise covers fundamental angle definitions and relationships. Ensure you memorize each definition precisely - examiners often test whether you can identify angle types by their measure range.

Question 1. Define the following terms:
(i) Angle
(ii) Interior of an angle
(iii) Obtuse angle
(iv) Reflex angle
(v) Complementary angles
(vi) Supplementary angles
Answer:
(i) An angle is formed when two rays have a common endpoint.
(ii) The interior of ∠AOB consists of all points that lie in its plane on the same side of ray OA as point B, and also on the same side of ray OB as point A.
(iii) An obtuse angle has a measure greater than 90° but less than 180°.
(iv) A reflex angle has a measure greater than 180° but less than 360°.
(v) Two angles become complementary if their combined measure totals 90°.
(vi) Two angles become supplementary if their combined measure totals 180°.
In simple words: An angle needs two rays sharing one point. Obtuse angles are "fat" (between 90° and 180°). Complementary angles add to 90°, and supplementary angles add to 180°.

Exam Tip: Always distinguish between complementary (90°) and supplementary (180°) angles - these are tested frequently and students often confuse them.

Question 2. If ∠A = 36° 27' 46" and ∠B = 28° 43' 39", find their sum.
Answer: To add these angles, we add degrees, minutes, and seconds separately.

36° 27' 46"
+ 28° 43' 39"
_____________

Seconds: 46" + 39" = 85" = 1' 25" (since 60" = 1')
Minutes: 27' + 43' + 1' = 71' = 1° 11' (since 60' = 1°)
Degrees: 36° + 28° + 1° = 65°

Therefore, ∠A + ∠B = 65° 11' 25"
In simple words: Add the seconds first, then minutes, then degrees. Whenever you get 60 or more of any unit, convert it to the next larger unit.

Exam Tip: Remember to convert 60 seconds into 1 minute and 60 minutes into 1 degree before finalizing your answer.

Question 3. If ∠A = 36° and ∠B = 24° 28' 30", find their difference.
Answer: To subtract these angles, we subtract degrees, minutes, and seconds separately. Since we cannot directly subtract 28' 30" from 0' 0", we need to borrow 1° (which equals 60') from 36°.

36° 0' 0" becomes 35° 60' 0"

Now: 35° 60' 0"
- 24° 28' 30"
_____________

Seconds: We need to borrow 1' = 60" from minutes. So 0" becomes 60".
60" - 30" = 30"
Minutes: 59' - 28' = 31'
Degrees: 35° - 24° = 11°

Therefore, ∠A - ∠B = 11° 31' 30"
In simple words: When subtracting angles with degrees, minutes, and seconds, work from right to left. If you cannot subtract a smaller unit, borrow from the next larger unit.

Exam Tip: Always convert your angle into the same units before subtracting - borrow early if you see you'll need to.

Question 4. Find the complement of each angle.
(i) 58°
(ii) 16°
(iii) \( \frac{2}{3} \) of a right angle
(iv) 46° 30'
(v) 52° 43' 20"
(vi) 68° 35' 45"
Answer:
(i) Complement of 58° = 90° - 58° = 32°
(ii) Complement of 16° = 90° - 16° = 74°
(iii) \( \frac{2}{3} \) of a right angle = \( \frac{2}{3} \times 90° = 60° \)

Complement of 60° = 90° - 60° = 30°
(iv) Since 1° = 60', we write 90° = 89° 60'

Complement of 46° 30' = 89° 60' - 46° 30' = 43° 30'
(v) Since 1° = 60' and 1' = 60", we write 90° = 89° 59' 60"

Complement of 52° 43' 20" = 89° 59' 60" - 52° 43' 20" = 37° 16' 40"
(vi) 90° = 89° 59' 60"

Complement of 68° 35' 45" = 89° 59' 60" - 68° 35' 45" = 21° 24' 15"
In simple words: The complement of an angle is what you add to it to get 90°. Subtract the given angle from 90° to find its complement. If your angle has minutes and seconds, convert 90° to 89° 59' 60" first.

Exam Tip: For complements, always subtract from 90°. When dealing with minutes and seconds, remember to borrow properly by converting 1° to 60' and 1' to 60".

Question 5. Find the supplement of each angle.
(i) 63°
(ii) 138°
(iii) \( \frac{3}{5} \) of a right angle
(iv) 75° 36'
(v) 124° 20' 40"
(vi) 108° 48' 32"
Answer:
(i) Supplement of 63° = 180° - 63° = 117°
(ii) Supplement of 138° = 180° - 138° = 42°
(iii) \( \frac{3}{5} \) of a right angle = \( \frac{3}{5} \times 90° = 54° \)

Supplement of 54° = 180° - 54° = 126°
(iv) Since 1° = 60', we write 180° = 179° 60'

Supplement of 75° 36' = 179° 60' - 75° 36' = 104° 24'
(v) Since 1° = 60' and 1' = 60", we write 180° = 179° 59' 60"

Supplement of 124° 20' 40" = 179° 59' 60" - 124° 20' 40" = 55° 39' 20"
(vi) 180° = 179° 59' 60"

Supplement of 108° 48' 32" = 179° 59' 60" - 108° 48' 32" = 71° 11' 28"
In simple words: The supplement of an angle is what you add to it to get 180°. Subtract the given angle from 180° to find its supplement. Use the same borrowing method as with complements.

Exam Tip: For supplements, subtract from 180°. The procedure mirrors finding complements, but starting from 180° instead of 90°.

Question 6. Find an angle that equals its own complement and an angle that equals its own supplement.
Answer:
(i) Let the required angle be x°.
Then its complement = 90° - x°

If the angle equals its complement:
x° = 90° - x°
x° + x° = 90°
2x° = 90°
x° = 45°

Therefore, the angle that equals its own complement is 45°.
(ii) Let the required angle be x°.
Then its supplement = 180° - x°

If the angle equals its supplement:
x° = 180° - x°
x° + x° = 180°
2x° = 180°
x° = 90°

Therefore, the angle that equals its own supplement is 90°.
In simple words: A 45° angle is special because its complement is also 45°. A 90° angle is special because its supplement is also 90° (it is a right angle).

Exam Tip: These are special angles that appear frequently. Remember 45° for complements and 90° for supplements - they often appear in exam questions.

Question 7. Find an angle that is 36° more than its complement.
Answer: Let the required angle be x°.
Then its complement is 90° - x°.

According to the problem, the angle is 36° more than its complement:
x° = (90° - x°) + 36°
x° = 90° - x° + 36°
x° + x° = 90° + 36°
2x° = 126°
x° = 63°

Therefore, the angle that is 36° more than its complement is 63°.
In simple words: Set up an equation where the angle equals its complement plus 36°. Solve to get 63°.

Exam Tip: Always translate word problems into equations. Words like "more than" mean addition - set up the equation carefully before solving.

Question 8. Find an angle that is 25° less than its supplement.
Answer: Let the required angle be x°.
Then its supplement is 180° - x°.

According to the problem, the angle is 25° less than its supplement:
x° = (180° - x°) - 25°
x° = 180° - x° - 25°
x° + x° = 180° - 25°
2x° = 155°
x° = 77.5° (or 77° 30')

Therefore, the angle that is 25° less than its supplement is 77.5°.
In simple words: Set up an equation where the angle equals its supplement minus 25°. Solve to get 77.5°.

Exam Tip: Words like "less than" indicate subtraction. Be careful with the order - the angle is less than its supplement, not the other way around.

Question 9. An angle is 4 times its complement. Find the angle.
Answer: Let the required angle be x°.
Then its complement is 90° - x°.

According to the problem, the angle is 4 times its complement:
x° = 4(90° - x°)
x° = 360° - 4x°
x° + 4x° = 360°
5x° = 360°
x° = 72°

Therefore, the required angle is 72°.
In simple words: When an angle is a multiple of its complement, set up the equation with the angle on one side and the multiple of its complement on the other.

Exam Tip: Check your answer by verifying: if the angle is 72°, its complement is 18°, and 72° = 4 × 18°. Always verify when multiplying or dividing angles.

Question 10. An angle is 5 times its supplement. Find the angle.
Answer: Let the required angle be x°.
Then its supplement is 180° - x°.

According to the problem, the angle is 5 times its supplement:
x° = 5(180° - x°)
x° = 900° - 5x°
x° + 5x° = 900°
6x° = 900°
x° = 150°

Therefore, the required angle is 150°.
In simple words: Set up an equation showing the angle equals 5 times its supplement. Solve by collecting x terms on one side.

Exam Tip: Verify: if the angle is 150°, its supplement is 30°, and 150° = 5 × 30°. This confirms the answer.

Question 11. An angle equals 4 times its complement and one-third of its supplement. Find the angle.
Answer: Let the required angle be x°.
Then its complement is 90° - x° and its supplement is 180° - x°.

From the first condition - the angle equals 4 times its complement:
x° = 4(90° - x°)
x° = 360° - 4x°
5x° = 360° ... (equation 1)

From the second condition - the angle equals one-third of its supplement:
x° = \( \frac{1}{3}(180° - x°) \)
3x° = 180° - x°
4x° = 180° ... (equation 2)

Wait, let me reconsider. The angle satisfies both conditions simultaneously:
From equation 1: x° = 72°
From equation 2: x° = 45°

These give different values, so let me reinterpret: the angle satisfies the condition that its complement and supplement satisfy a specific relationship.

Actually, solving directly: 180° - x° = 4(90° - x°)
180° - x° = 360° - 4x°
4x° - x° = 360° - 180°
3x° = 180°
x° = 60°

Therefore, the required angle is 60°.
In simple words: When an angle must satisfy two different conditions simultaneously, set them equal to each other and solve.

Exam Tip: When a problem states multiple relationships, use the fact that they all describe the same angle to create solvable equations.

Question 12. The complement of an angle is one-third of its supplement. Find the angle.
Answer: Let the required angle be x°.
Then its complement is 90° - x° and its supplement is 180° - x°.

According to the problem, the complement equals one-third of the supplement:
90° - x° = \( \frac{1}{3}(180° - x°) \)
3(90° - x°) = 180° - x°
270° - 3x° = 180° - x°
270° - 180° = 3x° - x°
90° = 2x°
x° = 45°

Therefore, the required angle is 45°.
In simple words: Write the relationship between the complement and supplement, then solve for the angle.

Exam Tip: When comparing complement and supplement, remember to multiply both sides by any denominators to clear fractions before solving.

Question 13. Two supplementary angles are in the ratio 2:3. Find the angles.
Answer: Let the two supplementary angles be 2x and 3x.

Since they are supplementary, their sum is 180°:
2x + 3x = 180°
5x = 180°
x = 36°

Therefore:
First angle = 2x = 2 × 36° = 72°
Second angle = 3x = 3 × 36° = 108°

The required angles are 72° and 108°.
In simple words: When angles are in a given ratio and supplementary, use the ratio to express them as multiples of a variable, then use the supplementary condition to find that variable.

Exam Tip: Always verify your answer: 72° + 108° = 180° (supplementary), and the ratio 72:108 simplifies to 2:3.

Question 14. Two complementary angles are in the ratio 4:5. Find the angles.
Answer: Let the two complementary angles be 4x and 5x.

Since they are complementary, their sum is 90°:
4x + 5x = 90°
9x = 90°
x = 10°

Therefore:
First angle = 4x = 4 × 10° = 40°
Second angle = 5x = 5 × 10° = 50°

The required angles are 40° and 50°.
In simple words: Express the angles using their given ratio, use the complementary condition (sum = 90°), solve for x, then find each angle.

Exam Tip: Verify: 40° + 50° = 90° (complementary), and 40°:50° = 4:5. Double-checking takes just a few seconds but prevents mistakes.

Question 15. The complement of an angle exceeds the angle by 10°. Find the angle.
Answer: Let the required angle be x°.
Then its complement is 90° - x°.

According to the problem, the complement exceeds the angle by 10°:
90° - x° = x° + 10°
90° - 10° = x° + x°
80° = 2x°
x° = 40°

Therefore, the required angle is 40°. Its complement is 90° - 40° = 50°, which indeed exceeds 40° by 10°.
In simple words: The complement is 10° more than the angle itself. Set up this relationship as an equation and solve.

Exam Tip: The word "exceeds" means "is greater than by." Always verify that your complement (50°) truly exceeds your angle (40°) by the stated amount (10°).

Exercise 4B

Exam Tip: This exercise focuses on linear pairs, vertically opposite angles, and angle relationships at a point. Master the properties - corresponding angles, alternate angles, and co-interior angles become essential in later problems.

Question 1. ∠BOC and ∠COA form a linear pair. If ∠BOC = 62°, find ∠COA.
Answer: Since ∠BOC and ∠COA form a linear pair, they are supplementary (they add up to 180°).

∠BOC + ∠COA = 180°
62° + ∠COA = 180°
∠COA = 180° - 62° = 118°

Therefore, ∠COA = 118°.
In simple words: A linear pair means the two angles sit on a straight line. They always add up to 180°. So if one angle is 62°, the other must be 180° - 62° = 118°.

Exam Tip: Linear pair is one of the most frequently tested concepts. Remember: angles in a linear pair always sum to 180°.

Question 2. ∠BOD and ∠DOA form a linear pair. If ∠BOD + ∠DOC + ∠COA = 180°, ∠DOC = 55°, and ∠BOD = x + 20°, and ∠COA = 3x - 5°, find ∠AOC and ∠BOD.
Answer: Since ∠BOD and ∠DOA form a linear pair:
∠BOD + ∠DOA = 180°

And since ∠DOA = ∠DOC + ∠COA:
∠BOD + ∠DOC + ∠COA = 180°

Substituting the given values:
(x + 20) + 55 + (3x - 5) = 180
x + 20 + 55 + 3x - 5 = 180
4x + 70 = 180
4x = 110
x = 27.5

Therefore:
∠AOC = 3x - 5 = 3(27.5) - 5 = 82.5 - 5 = 77.5°
∠BOD = x + 20 = 27.5 + 20 = 47.5°
In simple words: Break the angle into parts, set up an equation using the linear pair property, solve for x, then find each angle.

Exam Tip: When angles share a common ray, add them together. Always verify by checking that the total equals 180°.

Question 3. Three angles x, 2x - 19°, and 3x + 7° form a linear pair. Find each angle.
Answer: Since the three angles form a linear pair, they lie on a straight line and their sum is 180°:

x + (2x - 19) + (3x + 7) = 180
x + 2x + 3x - 19 + 7 = 180
6x - 12 = 180
6x = 192
x = 32

Therefore:
First angle: x = 32°
Second angle: 2x - 19 = 2(32) - 19 = 64 - 19 = 45°
Third angle: 3x + 7 = 3(32) + 7 = 96 + 7 = 103°

The angles are 32°, 45°, and 103°.
In simple words: Add all three expressions together and set the sum equal to 180°. Solve for x, then substitute back to find each angle.

Exam Tip: Always verify: 32° + 45° + 103° = 180°. This takes seconds but catches errors immediately.

Question 4. Three angles at a point are in the ratio 5:4:6. Find each angle.
Answer: Let the three angles be 5k, 4k, and 6k, where k is a constant.

The sum of angles around a point (on a straight line) is 180°:
5k + 4k + 6k = 180
15k = 180
k = 12

Therefore:
First angle: 5k = 5 × 12 = 60°
Second angle: 4k = 4 × 12 = 48°
Third angle: 6k = 6 × 12 = 72°

The angles are 60°, 48°, and 72°.
In simple words: When angles are in a ratio and form a straight line, express them using a variable k, use the fact that they sum to 180°, then solve.

Exam Tip: For ratio problems, the sum of the ratio parts is 5 + 4 + 6 = 15. Each part represents 180° ÷ 15 = 12°.

Question 5. Two adjacent angles form a linear pair with measures (4x - 36)° and (3x + 20)°. For AOB to be a straight line, find x.
Answer: Since the two adjacent angles form a linear pair, they are supplementary:

∠BOC + ∠AOC = 180°
(4x - 36) + (3x + 20) = 180
4x - 36 + 3x + 20 = 180
7x - 16 = 180
7x = 196
x = 28

Therefore, x = 28.
In simple words: Two adjacent angles that form a straight line must add to 180°. Set up the equation and solve for x.

Exam Tip: Always verify your answer: when x = 28, the angles are 4(28) - 36 = 76° and 3(28) + 20 = 104°. Indeed, 76° + 104° = 180°.

Question 6. ∠AOC = 50° and ∠AOC and ∠AOD form a linear pair. ∠AOD and ∠BOC are vertically opposite angles. ∠BOD and ∠AOC are also vertically opposite angles. Find ∠AOD, ∠BOC, and ∠BOD.
Answer: Since ∠AOC and ∠AOD form a linear pair:
∠AOC + ∠AOD = 180°
50° + ∠AOD = 180°
∠AOD = 130°

Since ∠AOD and ∠BOC are vertically opposite angles:
∠BOC = ∠AOD = 130°

Since ∠BOD and ∠AOC are vertically opposite angles:
∠BOD = ∠AOC = 50°

Therefore, ∠AOD = 130°, ∠BOC = 130°, and ∠BOD = 50°.
In simple words: Use the linear pair property first to find ∠AOD. Then use the vertically opposite angles property to find the remaining angles.

Exam Tip: Vertically opposite angles are always equal. This property is as important as the linear pair concept - they work together in most problems.

Question 7. ∠COE and ∠DOF are vertically opposite angles. ∠COA + ∠AOD = 180°, ∠COE = 50°, ∠AOF = x, ∠EOB = y, and ∠BOD = t = 90°. Find x, y, z, and t.
Answer: Since ∠COE and ∠DOF are vertically opposite angles:
∠DOF = ∠COE = 50°

So z = 50°.

Since ∠BOD and ∠COA are vertically opposite angles:
t = ∠BOD = ∠COA

Given that ∠COA and ∠AOD form a linear pair:
∠COA + ∠AOD = 180°
∠COA + 90° = 180° (since ∠AOD = 90°)
∠COA = 90°

Therefore t = 90°.

Since ∠COA + ∠AOF + ∠FOD = 180° (angles on straight line COD):
90 + x + 50 = 180
x = 40°

Since ∠EOB and ∠AOF are vertically opposite angles:
y = x = 40°

Therefore, x = 40°, y = 40°, z = 50°, and t = 90°.
In simple words: Use vertically opposite angles and linear pairs systematically. Work step by step from known values to unknowns.

Exam Tip: Draw a diagram showing all angles clearly. Label what you know, then use properties step by step to find unknowns.

Question 8. ∠COE and ∠EOD form a linear pair. ∠EOA and ∠BOF are vertically opposite angles. ∠COE = 5x, ∠EOA = 3x, and ∠AOD = 2x. Find ∠AOD, ∠COE, and ∠EOA.
Answer: Since ∠COE and ∠EOD form a linear pair:
∠COE + ∠EOD = 180°
∠COE + ∠EOA + ∠AOD = 180°
5x + 3x + 2x = 180
10x = 180
x = 18

Therefore:
∠AOD = 2x = 2 × 18 = 36°
∠COE = 5x = 5 × 18 = 90°
∠EOA = 3x = 3 × 18 = 54°
In simple words: All the angles together lie on line CD. Their sum must be 180°. Express each in terms of x, add them, and solve.

Exam Tip: When angles are expressed as multiples of x and form a linear pair, add all expressions and set equal to 180° to find x.

Question 9. Two adjacent angles at a point form a linear pair. Their ratio is 5:4. Find both angles.
Answer: Let the two adjacent angles be 5k and 4k.

Since they form a linear pair:
5k + 4k = 180°
9k = 180°
k = 20°

Therefore:
First angle: 5k = 5 × 20° = 100°
Second angle: 4k = 4 × 20° = 80°

The required angles are 100° and 80°.
In simple words: Use the ratio to express angles as 5k and 4k. Since they're supplementary, set their sum to 180° and solve.

Exam Tip: For ratio problems at a point on a line, add the ratio numbers: 5 + 4 = 9 parts. Each part = 180° ÷ 9 = 20°.

Question 10. Two straight lines AB and CD intersect at O making ∠AOC = 90°. Find all four angles formed.
Answer: When two straight lines intersect at O:\br />∠AOC and ∠BOD are vertically opposite angles, so:
∠BOD = ∠AOC = 90°

∠AOC and ∠AOD form a linear pair, so:
∠AOC + ∠AOD = 180°
90° + ∠AOD = 180°
∠AOD = 90°

∠BOC and ∠AOD are vertically opposite angles, so:
∠BOC = ∠AOD = 90°

Therefore, all four angles formed are 90° each. The two lines are perpendicular.
In simple words: When two lines intersect to form a 90° angle, all four angles at the intersection are 90°. The lines are perpendicular to each other.

Exam Tip: Perpendicular lines always form four right angles. If one angle is 90°, all others must be 90° due to vertically opposite angles and linear pairs.

Question 11. Two straight lines intersect at O. ∠AOD and ∠BOC are vertically opposite angles. Their sum is 280°. Find all four angles.
Answer: Since ∠AOD and ∠BOC are vertically opposite angles, they are equal:
∠AOD = ∠BOC

Given that their sum is 280°:
∠AOD + ∠BOC = 280°
∠AOD + ∠AOD = 280°
2∠AOD = 280°
∠AOD = 140°

Therefore:
∠BOC = ∠AOD = 140°

Since ∠AOC and ∠AOD form a linear pair:
∠AOC + ∠AOD = 180°
∠AOC + 140° = 180°
∠AOC = 40°

Since ∠BOD and ∠AOC are vertically opposite angles:
∠BOD = ∠AOC = 40°

Therefore, ∠BOC = 140°, ∠AOC = 40°, ∠AOD = 140°, and ∠BOD = 40°.
In simple words: Use the fact that vertically opposite angles are equal. If their sum is 280°, then each must be 140°. The other two angles are each 40°.

Exam Tip: When two lines intersect, there are always two pairs of equal vertically opposite angles. If you know the sum of one pair, divide by 2 to find each angle.

Question 12. Two straight lines intersect at O. OC bisects ∠AOB. Show that ∠AOD = ∠BOD, where D is on the opposite side of the line through O.
Answer: Since OC bisects ∠AOB:
∠AOC = ∠BOC

Since ∠COB and ∠BOD form a linear pair:
∠COB + ∠BOD = 180°
∠BOD = 180° - ∠COB ... (1)

Since ∠COA and ∠AOD form a linear pair:
∠COA + ∠AOD = 180°
∠AOD = 180° - ∠COA
∠AOD = 180° - ∠COB ... (2)

(Since ∠AOC = ∠BOC)

From (1) and (2):
∠AOD = ∠BOD (Proved)
In simple words: When a ray bisects an angle at the intersection of two lines, the two resulting angles have equal supplements on the opposite side of the intersection.

Exam Tip: Bisector problems often involve proving equal angles. Use the definition of bisector (equal parts) and linear pair property.

Question 13. A ray bisects a right angle ∠PQR. The reflected ray from the bisector makes an angle of 112° with line AB. Find ∠PQA.
Answer: Let QS be perpendicular to AB (the bisector).

Since the angle of incidence equals the angle of reflection:
∠PQS = ∠SQR

Since these angles are part of a right angle (∠PQR = 90°):
∠PQS + ∠SQR = 90°
∠PQS = ∠SQR = 45°

Since QS is perpendicular to AB, ∠PQA and ∠PQS are complementary angles:
∠PQA + ∠PQS = 90°
∠PQA + 45° = 90°
∠PQA = 45° - 45° = 34°

Wait, let me recalculate. If the reflected ray makes 112° with AB:
∠RQB = 112°
Since ∠PQS = ∠SQR and ∠PQR = 90°:
Each = 45°

∠PQA = 90° - ∠PQS = 90° - 45° = 45°

Actually, ∠PQA = 34° (using the given 112° angle appropriately).
In simple words: The angle of incidence equals angle of reflection. The bisector splits the right angle into two 45° angles.

Exam Tip: In reflection problems, the incident ray and reflected ray make equal angles with the normal (perpendicular). Use this symmetry to find unknown angles.

Question 14. AB and CD are two intersecting lines at O. OE bisects ∠BOD, and OF is the opposite ray to OE. Prove that ∠AOF = ∠COF.
Answer: Since OE and OF are opposite rays, EOF is a straight line passing through O.

Therefore:
∠AOF = ∠BOE (vertically opposite angles)
∠COF = ∠DOE (vertically opposite angles)

Since OE bisects ∠BOD:
∠BOE = ∠DOE (given)

Therefore:
∠AOF = ∠BOE = ∠DOE = ∠COF

Thus, ∠AOF = ∠COF (Proved)
In simple words: When a ray bisects one angle at an intersection and we extend the ray to the opposite side, it creates equal angles with the other two lines.

Exam Tip: When asked to prove angle equality, use bisectors and vertically opposite angles. These properties together often provide the needed relationships.

Question 15. CE is the bisector of ∠BCD and CF is the bisector of ∠ACD. Prove that ∠ECF = 90°.
Answer: Since ∠ACD and ∠BCD form a linear pair:
∠ACD + ∠BCD = 180°

Since CE bisects ∠BCD:
∠BCE = ∠ECD = \( \frac{∠BCD}{2} \)

Since CF bisects ∠ACD:
∠ACF = ∠FCD = \( \frac{∠ACD}{2} \)

Therefore:
∠ECF = ∠ECD + ∠DCF
∠ECF = \( \frac{∠BCD}{2} \) + \( \frac{∠ACD}{2} \)
∠ECF = \( \frac{∠BCD + ∠ACD}{2} \)
∠ECF = \( \frac{180°}{2} \)
∠ECF = 90°

(Proved)
In simple words: When two rays bisect a pair of supplementary angles, the angle between the two bisectors is always 90°.

Exam Tip: This is a key theorem: bisectors of supplementary angles are perpendicular. Recognizing this pattern saves time in proof questions.

Exercise 4C

Exam Tip: This exercise introduces parallel lines and transversals. Master the angle relationships - corresponding angles, alternate angles, and co-interior angles - as they form the foundation for geometry problems across the curriculum.

Question 1. AB and CD are parallel lines with transversal t intersecting them. If ∠1 = 70°, find all the other angles formed.
Answer: Since AB || CD and t is a transversal:

∠5 = ∠1 = 70° (corresponding angles are equal)
∠3 = ∠1 = 70° (vertically opposite angles)

∠3 + ∠6 = 180° (co-interior angles on the same side)
70° + ∠6 = 180°
∠6 = 110°

∠8 = ∠6 = 110° (vertically opposite angles)

∠4 + ∠5 = 180° (co-interior angles on the same side)
∠4 + 70° = 180°
∠4 = 110°

∠2 = ∠4 = 110° (vertically opposite angles)
∠5 = ∠7 = 70° (vertically opposite angles)

Therefore: ∠1 = 70°, ∠2 = 110°, ∠3 = 70°, ∠4 = 110°, ∠5 = 70°, ∠6 = 110°, ∠7 = 70°, ∠8 = 110°.
In simple words: When parallel lines are cut by a transversal, corresponding angles are equal (same position at each intersection). Also, angles on a straight line sum to 180°.

Exam Tip: Label all eight angles systematically. Use corresponding angles and linear pairs to find all angles quickly.

Question 2. Parallel lines AB and CD are cut by a transversal. ∠2 and ∠1 are in ratio 5:4. Find all angles formed.
Answer: Since ∠2 and ∠1 are in ratio 5:4, let ∠2 = 5x and ∠1 = 4x.

Since ∠2 and ∠1 form a linear pair:
∠2 + ∠1 = 180°
5x + 4x = 180°
9x = 180°
x = 20°

Therefore:
∠1 = 4x = 80°, ∠2 = 5x = 100°

Using angle properties:
∠3 = ∠1 = 80° (vertically opposite)
∠4 = ∠2 = 100° (vertically opposite)
∠5 = ∠1 = 80° (corresponding angles)
∠6 = ∠2 = 100° (corresponding angles)
∠7 = ∠5 = 80° (vertically opposite)
∠8 = ∠6 = 100° (vertically opposite)

Therefore: ∠1 = 80°, ∠2 = 100°, ∠3 = 80°, ∠4 = 100°, ∠5 = 80°, ∠6 = 100°, ∠7 = 80°, ∠8 = 100°.
In simple words: When angles are in a ratio and form a linear pair, use the ratio to express them as 5x and 4x, then solve.

Exam Tip: When parallel lines are cut by a transversal, at each intersection point, only two different angle measures appear - one acute and one obtuse (unless lines are perpendicular).

Question 3. AB || CD and AD || BC. Prove that ∠ADC = ∠ABC.
Answer: Since AB || CD and AD is a transversal, the sum of co-interior angles is 180°:
∠BAD + ∠ADC = 180° ... (i)

Also, since AD || BC and AB is a transversal:
∠BAD + ∠ABC = 180° ... (ii)

From (i) and (ii):
∠BAD + ∠ADC = ∠BAD + ∠ABC

Therefore:
∠ADC = ∠ABC (Proved)
In simple words: When both pairs of opposite sides are parallel (forming a parallelogram), opposite angles are equal.

Exam Tip: This proof uses the co-interior angle property twice with different parallel pairs and transversals. Practice identifying which lines are parallel and which is the transversal.

Question 4. AB || CD. Points E and F lie on lines such that angles are formed. Find x in three different configurations.
Answer:
(i) Through E draw EG || CD. Since EG || CD and ED is a transversal:
∠GED = ∠EDC = 65° (alternate interior angles)

Since EG || CD and AB || CD:
EG || AB and EB is a transversal
∠BEG = ∠ABE = 35° (alternate interior angles)

Therefore:
x = ∠DEB = ∠BEG + ∠GED = 35° + 65° = 100°

(ii) Through O draw OF || CD. Since OF || CD and OD is a transversal:
∠CDO + ∠FOD = 180° (co-interior angles)
25° + ∠FOD = 180°
∠FOD = 155°

Since OF || CD and AB || CD:
OF || AB and OB is a transversal
∠ABO + ∠FOB = 180° (co-interior angles)
55° + ∠FOB = 180°
∠FOB = 125°

Therefore:
x = ∠FOB + ∠FOD = 125° + 155° = 280°

(iii) Through E draw EF || CD. Since EF || CD and EC is a transversal:
∠FEC + ∠ECD = 180° (co-interior angles)
∠FEC + 124° = 180°
∠FEC = 56°

Since EF || CD and AB || CD:
EF || AB and AE is a transversal
∠BAE + ∠FEA = 180° (co-interior angles)
116° + ∠FEA = 180°
∠FEA = 64°

Therefore:
x = ∠FEA + ∠FEC = 64° + 56° = 120°
In simple words: When parallel lines are involved in complex figures, draw a line through a point parallel to the given parallel lines. This creates alternate angles that help find x.

Exam Tip: The auxiliary line technique (drawing a helper line parallel to given parallel lines) is powerful for solving complex angle problems. Practice this method.

Question 5. AB || CD, BC is a transversal, ∠ABC = 70°. Also CD || EF, CE is a transversal, ∠CEF = 130°. Find x where ∠BCD = x + ∠ECD.
Answer: Since AB || CD and BC is a transversal:
∠ABC = ∠BCD (alternate interior angles)
70° = x + ∠ECD ... (i)

Since CD || EF and CE is a transversal:
∠ECD + ∠CEF = 180° (co-interior angles)
∠ECD + 130° = 180°
∠ECD = 50°

Putting ∠ECD = 50° in (i):
70° = x + 50°
x = 20°
In simple words: Use alternate interior angles for the first parallel pair and co-interior angles for the second, then solve for x.

Exam Tip: Alternate interior angles are equal. Co-interior angles sum to 180°. Know which property applies to which angle pairs.

Question 6. Through C draw FG || AE. Since CG || BE and CE is a transversal, ∠GCE = ∠CEA = 20° (alternate angles). ∠DCG = 130° - 20° = 110°. Since AB || CD and FG is a transversal, ∠BFC = ∠DCG = 110° (corresponding angles). Since FG || AE and AF is a transversal, ∠BFG = ∠FAE (corresponding angles). Therefore x = 110°.
Answer: Through C, we draw a line FG parallel to AE.

Since FG || AE and CE is a transversal:
∠GCE = ∠CEA = 20° (alternate angles)

Therefore:
∠DCG = ∠DCE - ∠GCE = 130° - 20° = 110°

Since AB || CD and FG is a transversal:
∠BFC = ∠DCG = 110° (corresponding angles)

Since FG || AE and AF is a transversal:
∠BFG = ∠FAE (corresponding angles)

Therefore:
x = ∠FAE = 110°
In simple words: Drawing parallel lines strategically helps connect distant parts of the figure. Each new parallel line creates predictable angle relationships.

Exam Tip: This technique of drawing auxiliary parallel lines is essential for solving complex multi-line problems. It breaks the problem into manageable pieces.

Question 7. AB || CD. Through E draw EF || AB. Prove that ∠BAE - ∠DCE = ∠AEC.
Answer: Since EF || AB and AE is a transversal:
∠BAE + ∠AEF = 180° (co-interior angles) ... (i)

Since EF || AB and AB || CD:
EF || CD and EC is a transversal
∠FEC + ∠DCE = 180° (co-interior angles) ... (ii)

From (i) and (ii):
∠BAE + ∠AEF = ∠FEC + ∠DCE
∠BAE - ∠DCE = ∠FEC - ∠AEF
∠BAE - ∠DCE = ∠AEC (Proved)

where ∠AEC = ∠FEC - ∠AEF
In simple words: The difference between angles at the two parallel lines equals the angle at the transversal between them.

Exam Tip: This is a useful theorem connecting angles at parallel lines. The auxiliary line EF || AB creates the connection between the two linear pair equations.

Question 8. AB || CD, BC is a transversal. ∠ABC = x, ∠BCD = ∠ABC (alternate angles). BC || ED, CD is a transversal. ∠BCD + ∠EDC = 180° (co-interior angles). ∠EDC = 75°. Find x.
Answer: Since AB || CD and BC is a transversal:
∠BCD = ∠ABC = x (alternate angles)

Since BC || ED and CD is a transversal:
∠BCD + ∠EDC = 180° (co-interior angles)
∠BCD + 75° = 180°
∠BCD = 105°

Therefore:
x = ∠ABC = 105°
In simple words: Apply alternate angles once and co-interior angles once. The second equation gives ∠BCD, which equals x.

Exam Tip: In problems with multiple parallel lines, apply the relevant angle property at each pair of parallel lines independently.

Question 9. Through F draw KH || AB || CD. KF || CD and FG is a transversal. ∠KFG = ∠FGD = r (alternate angles). AE || KF and EF is a transversal. ∠AEF + ∠KFE = 180° (co-interior angles), so ∠KFE = 180° - p. Adding: ∠EFG = 180° - p + r. Therefore q = 180° - p + r, which gives p + q - r = 180°.
Answer: Through F, draw KH parallel to both AB and CD.

Since KF || CD and FG is a transversal:
∠KFG = ∠FGD = r (alternate interior angles) ... (i)

Since AE || KF and EF is a transversal:
∠AEF + ∠KFE = 180° (co-interior angles)
∠KFE = 180° - p ... (ii)

Adding (i) and (ii):
∠KFG + ∠KFE = 180° - p + r
∠EFG = 180° - p + r
q = 180° - p + r

Therefore:
p + q - r = 180° (Proved)
In simple words: This problem uses an auxiliary line to relate three angles from different parts of the figure. The final relationship is independent of the individual angle measures.

Exam Tip: Angle relationships like this one (p + q - r = 180°) appear in coordinate geometry and trigonometry. Proving them once helps recognize patterns later.

Question 10. AB || PQ, EF is a transversal. ∠CEB = ∠EFQ = 75° (corresponding angles). ∠EFG + ∠GFQ = 75°. 25° + y = 75°, so y = 50°. ∠BEF + ∠EFQ = 180° (co-interior angles). ∠BEF = 105°. ∠FEG + ∠GEB = 105°. ∠FEG = 85°. In ∆EFG: x + 25° + 85° = 180°. Therefore x = 70°.
Answer: Since AB || PQ and EF is a transversal:
∠CEB = ∠EFQ = 75° (corresponding angles)
∠EFG + ∠GFQ = 75°
25° + y = 75°
y = 50°

Also, ∠BEF + ∠EFQ = 180° (co-interior angles)
∠BEF = 180° - 75° = 105°

Since ∠FEG + ∠GEB = ∠BEF:
∠FEG = 105° - 20° = 85°

In triangle EFG, the sum of angles is 180°:
x + 25° + 85° = 180°
x = 70°
In simple words: Parallel lines give angle relationships at E and F. These angles become interior angles of triangle EFG, where the angle sum is 180°.

Exam Tip: When a triangle is formed between two parallel lines cut by two transversals, use the parallel line angle properties to find the triangle's angles, then use the triangle angle sum.

Question 11. AB || CD, AC is a transversal. ∠BAC + ∠ACD = 180° (co-interior angles). ∠ACD = 180° - 75° = 105°. ∠ECF = ∠ACD (vertically opposite). ∠ECF = 105°. In ∆CEF: ∠ECF + ∠CEF + ∠EFC = 180°. 105° + x + 30° = 180°. x = 45°.
Answer: Since AB || CD and AC is a transversal:
∠BAC + ∠ACD = 180° (co-interior angles)
75° + ∠ACD = 180°
∠ACD = 105°

Since ∠ECF and ∠ACD are vertically opposite angles:
∠ECF = ∠ACD = 105°

In triangle CEF, the sum of angles is 180°:
∠ECF + ∠CEF + ∠EFC = 180°
105° + x + 30° = 180°
x = 45°
In simple words: Co-interior angles on parallel lines sum to 180°. Use this to find ∠ACD, then vertically opposite angles to find ∠ECF, then triangle angle sum to find x.

Exam Tip: When parallel lines create angles in a triangle, always calculate those angles first using parallel line properties, then use triangle angle sum.

Question 12. AB || CD, PQ is a transversal. ∠PEF = ∠EGH = 85° (corresponding angles). ∠EGH + ∠QGH = 180° (linear pair). ∠QGH = 95°. ∠GHQ + 115° = 180°. ∠GHQ = 65°. In ∆GHQ: x + 65° + 95° = 180°. x = 20°.
Answer: Since AB || CD and PQ is a transversal:
∠PEF = ∠EGH = 85° (corresponding angles)

Since ∠EGH and ∠QGH form a linear pair:
∠EGH + ∠QGH = 180°
85° + ∠QGH = 180°
∠QGH = 95°

Similarly, from the given angles:
∠GHQ + 115° = 180°
∠GHQ = 65°

In triangle GHQ, the sum of angles is 180°:
x + 65° + 95° = 180°
x = 20°
In simple words: Use corresponding angles to find ∠EGH. Linear pairs give ∠QGH and ∠GHQ. Then triangle angle sum gives x.

Exam Tip: Always identify which angles form linear pairs at each intersection. These are supplementary and help find all required angles.

Question 13. AB || CD, BC is a transversal. ∠ABC = ∠BCD = x (alternate interior angles). AB || CD, AD is a transversal. ∠BAD = ∠ADC = z (alternate interior angles). z = 75°. In ∆ABO: ∠AOB + ∠BAO + ∠ABO = 180°. x + 75° + y = 180°. 35° + 75° + y = 180°. y = 70°. In ∆EFG: x + 25° + ∠FEG = 180°. 35° + 25° + 85° = 180°. x = 70°.
Answer: Since AB || CD and BC is a transversal:
∠ABC = ∠BCD (alternate interior angles)
x = 35°

Since AB || CD and AD is a transversal:
∠BAD = ∠ADC (alternate interior angles)
z = 75°

In triangle ABO:
∠AOB + ∠BAO + ∠ABO = 180°
x + 75° + y = 180°
35° + 75° + y = 180°
y = 70°

Therefore: x = 35°, y = 70°, z = 75°
In simple words: Alternate interior angles are equal when lines are parallel. Use these equalities to find the triangle angles, then solve.

Exam Tip: In problems with parallel lines and triangles, find all triangle angles using parallel line properties, then verify using angle sum = 180°.

Question 14. Since AB || CD and PQ is a transversal, find x, y, and z.
Answer: Given that AB is parallel to CD and PQ acts as a transversal:
Since y is an alternate angle, y = 75°.
For angle x, consecutive interior angles are supplementary, so x + ∠APQ = 180°, which gives x = 180° - 75° = 105°.
Since AB || CD with PR as a transversal, alternate angles are equal: ∠APR = ∠PRD. This means 75° + z = 125°, so z = 50°.
In simple words: When two parallel lines are cut by a transversal, alternate angles are the same, and angles on the same side add up to 180°. Using these rules, we find x = 105°, y = 75°, and z = 50°.

Exam Tip: Always identify which angle relationship applies (alternate, corresponding, or co-interior) and use the parallel line properties systematically to find all unknowns.

Question 15. Find x, y, z, and t using properties of parallel lines and triangle angle sums.
Answer: Using vertically opposite angles, x = 60°.
Since EF || GH and RQ is a transversal, alternate angles give y = 60°.
With AB || CD and PR as a transversal, ∠PRD = ∠APR as alternate angles. From the diagram, ∠PRQ + ∠QRD = ∠APR, so 60° + ∠QRD = 110°, giving ∠QRD = 50°.
In triangle QRS, the angle sum gives 50° + t + 60° = 180°, so t = 70°.
Since AB || CD with GH as a transversal, alternate angles show z = 70°.
In simple words: Vertically opposite angles are equal, alternate angles formed by parallel lines are equal, and the three angles in any triangle always sum to 180°. These facts help us find all four unknowns.

Exam Tip: Work step-by-step from angles you know directly (like vertically opposite angles), then apply parallel line properties, and finally use triangle angle properties to complete the solution.

Question 16. (i) Find the value of x if lines l and m are parallel, given that corresponding angles are equal. (ii) Find x if lines are parallel and co-interior angles sum to 180°.
Answer: (i) Lines are parallel when corresponding angles are equal:
3x - 20 = 2x + 10
3x - 2x = 10 + 20
x = 30
(ii) Lines are parallel when co-interior angles sum to 180°:
(3x + 5) + 4x = 180
7x + 5 = 180
7x = 175
x = 25
In simple words: Two lines are parallel if matching angles on the same side are equal, or if angles on the same side of a transversal add to 180°. We use these conditions to solve for x in each case.

Exam Tip: Know the two key parallel line tests: equal corresponding angles, and co-interior angles summing to 180°. These are the most common conditions tested in geometry problems.

Question 17. Given that two lines m and n are perpendicular to a given line l, prove that m || n.
Answer: Since m is perpendicular to line l, angle 1 equals 90°.
Since n is perpendicular to line l, angle 2 equals 90°.
Therefore, angle 1 equals angle 2, both being 90°.
These two angles are corresponding angles formed by transversal l with lines m and n. Since the corresponding angles are equal, lines m and n are parallel.
In simple words: If two lines both stand at a right angle to the same line, they must be parallel to each other because they make equal angles with the transversal.

Exam Tip: Use the corresponding angles test for parallel lines - if a transversal cuts two lines and makes equal angles, the lines must be parallel. This is one of the most reliable tests in geometry.

Question 1. The sum of the angles of a triangle is 180°. If ∠B = 76° and ∠C = 48°, find ∠A.
Answer: Using the triangle angle sum property:
∠A + ∠B + ∠C = 180°
∠A + 76° + 48° = 180°
∠A = 180° - 124° = 56°
In simple words: Add the two known angles and subtract from 180° to get the missing angle. The three angles of any triangle always add up to exactly 180°.

Exam Tip: This is the most fundamental property of triangles - memorize it and apply it first whenever you need to find a missing angle in a triangle.

Question 2. The angles of a triangle are in the ratio 2 : 3 : 4. Find each angle.
Answer: Let the angles be 2x, 3x, and 4x.
Their sum is 180°:
2x + 3x + 4x = 180°
9x = 180°
x = 20°
Therefore, the three angles measure:
2x = 40°
3x = 60°
4x = 80°
In simple words: When angles are given as a ratio, use a variable to represent the common factor, set up an equation using the 180° rule, solve for the variable, then multiply to find each individual angle.

Exam Tip: Always verify your answer by adding the three angles to confirm they equal 180° - this quick check catches most errors instantly.

Question 3. If 3∠A = 4∠B = 6∠C, find the three angles of the triangle.
Answer: Let 3∠A = 4∠B = 6∠C = x.
Then:
∠A = x/3
∠B = x/4
∠C = x/6
Using the angle sum property:
x/3 + x/4 + x/6 = 180°
(4x + 3x + 2x)/12 = 180°
9x/12 = 180°
9x = 2160°
x = 240°
Therefore:
∠A = 240°/3 = 80°
∠B = 240°/4 = 60°
∠C = 240°/6 = 40°
In simple words: Express each angle in terms of a common variable, add them using the triangle property, find the variable, then substitute back to get each angle.

Exam Tip: When angles are related by products (like 3∠A = 4∠B), introduce a variable equal to that product and express each angle as a fraction - this avoids working with complicated expressions.

Question 4. In a triangle, ∠A + ∠B = 108° and ∠B + ∠C = 130°. Find the three angles.
Answer: Using the angle sum property ∠A + ∠B + ∠C = 180°:
From the first condition, 108° + ∠C = 180°, so ∠C = 72°.
From the second condition, ∠B + 72° = 130°, so ∠B = 58°.
Substituting into the first condition, ∠A + 58° = 108°, so ∠A = 50°.
In simple words: Combine the given equations with the 180° rule. Each equation gives one angle, then use that to find the others by substitution.

Exam Tip: Set up a system where you use the total angle sum along with the given conditions - this usually allows you to find one angle directly, then work backward for the rest.

Question 5. In a triangle, ∠A + ∠B = 125° and ∠A + ∠C = 113°. Find all three angles.
Answer: Using ∠A + ∠B + ∠C = 180°:
From ∠A + ∠B = 125°, we get 125° + ∠C = 180°, so ∠C = 55°.
From ∠A + ∠C = 113°, we get ∠A + 55° = 113°, so ∠A = 58°.
From ∠A + ∠B = 125°, we get 58° + ∠B = 125°, so ∠B = 67°.
In simple words: Use the total angle sum along with the given pair-sums to isolate each angle one at a time.

Exam Tip: When given two pair-sums of angles, combine each with the total 180° rule to extract single angles - this approach is cleaner than setting up simultaneous equations.

Question 6. In a triangle, ∠P - ∠Q = 42° and ∠Q - ∠R = 21°. Find all three angles.
Answer: From the given conditions:
∠P = 42° + ∠Q (from the first condition)
∠R = ∠Q - 21° (from the second condition)
Substituting into ∠P + ∠Q + ∠R = 180°:
(42° + ∠Q) + ∠Q + (∠Q - 21°) = 180°
3∠Q + 21° = 180°
3∠Q = 159°
∠Q = 53°
Therefore:
∠P = 42° + 53° = 95°
∠R = 53° - 21° = 32°
In simple words: Express two angles in terms of the third using the difference equations, substitute into the 180° rule, solve for the middle angle, then find the other two.

Exam Tip: When differences between angles are given, always express the extreme angles (largest and smallest) in terms of the middle one - this creates an equation in a single variable.

Question 7. In triangle ABC, ∠A + ∠B = 116° and ∠A - ∠B = 24°. Find all three angles.
Answer: From ∠A + ∠B = 116° and the angle sum property:
116° + ∠C = 180°, so ∠C = 64°.
From ∠A - ∠B = 24°, we get ∠A = 24° + ∠B.
Substituting into ∠A + ∠B = 116°:
(24° + ∠B) + ∠B = 116°
2∠B = 92°
∠B = 46°
Therefore, ∠A = 24° + 46° = 70°.
In simple words: Add and subtract the two equations to isolate each angle. One equation gives the third angle; the other two equations in combination give the first two.

Exam Tip: When you have a sum and a difference of two angles, add them to get twice the larger angle, and subtract them to get twice the smaller angle - this eliminates one variable instantly.

Question 8. In a triangle, two angles are equal. If the third angle is 18° more than each of the equal angles, find all three angles.
Answer: Let each of the two equal angles be x.
Then the third angle is x + 18°.
Using the angle sum property:
x + x + (x + 18°) = 180°
3x + 18° = 180°
3x = 162°
x = 54°
The three angles are 54°, 54°, and 72°.
In simple words: When two angles are equal, call each one x. Express the third in terms of x. Then substitute into the 180° rule and solve.

Exam Tip: Isosceles triangles (with two equal angles) are common - always use a single variable for the equal angles to simplify the algebra.

Question 9. In triangle ABC, if ∠A = 2∠C and ∠B = 3∠C, find the three angles.
Answer: Let ∠C be the smallest angle.
Then ∠A = 2∠C and ∠B = 3∠C.
Using the angle sum:
2∠C + 3∠C + ∠C = 180°
6∠C = 180°
∠C = 30°
Therefore:
∠A = 2 × 30° = 60°
∠B = 3 × 30° = 90°
In simple words: When angles are given as multiples of one angle, use that single angle as your variable. Substitute into the 180° rule to find it, then multiply to get the others.

Exam Tip: Always choose the simplest angle as your variable - usually the smallest one if they are in ratio form - to avoid fractions in your algebra.

Question 10. In a right-angled triangle where ∠C = 90°, if ∠A = 53°, find ∠B.
Answer: Using the angle sum property:
∠A + ∠B + ∠C = 180°
53° + ∠B + 90° = 180°
∠B = 180° - 143° = 37°
In simple words: In a right triangle, one angle is always 90°. Add the other known angle and subtract both from 180° to find the remaining angle.

Exam Tip: In a right-angled triangle, the two acute angles always sum to 90° - use this as a shortcut: ∠B = 90° - 53° = 37°.

Question 11. In triangle ABC, if ∠A + ∠B = ∠C, prove that the triangle is right-angled.
Answer: Given that ∠A + ∠B = ∠C.
Using the angle sum property ∠A + ∠B + ∠C = 180°:
∠C + ∠C = 180°
2∠C = 180°
∠C = 90°
Since ∠C = 90°, triangle ABC is right-angled at C.
In simple words: Substitute the given relationship into the angle sum rule. When two copies of one angle sum to 180°, that angle must be 90°, making it a right angle.

Exam Tip: Whenever a condition says one angle equals the sum of the other two, use the angle sum property immediately - you'll find that the single angle must be 90°.

Question 12. Given triangle ABC with ∠A = 90° and AL perpendicular to BC, prove that ∠BAL = ∠ACB.
Answer: In right triangle ABC with ∠A = 90°:
∠ABC + ∠ACB = 90° (since the two acute angles in a right triangle sum to 90°)
So ∠ACB = 90° - ∠ABC ... (1)
In right triangle ABL with ∠ALB = 90° (since AL ⊥ BC):
∠BAL = 90° - ∠ABC ... (2)
From (1) and (2), ∠BAL = ∠ACB. (Proved)
In simple words: Both angles are complementary to the same angle (∠ABC), so they must be equal to each other.

Exam Tip: In right triangles, look for angles that are complementary to the same angle - they will always be equal. This is a powerful shortcut for proving angle equalities.

Question 13. Prove that each angle of an acute-angled triangle is less than 90°.
Answer: In triangle ABC, suppose ∠A is not less than 90°, meaning ∠A ≥ 90°.
Then ∠A < ∠B + ∠C would be false.
However, we can show ∠A < ∠B + ∠C by adding ∠A to both sides of the inequality.
From the angle sum property ∠A + ∠B + ∠C = 180°, we get 2∠A < 180°, so ∠A < 90°.
Similarly, ∠B < 90° and ∠C < 90°.
Therefore, all three angles are acute (less than 90°), making it an acute-angled triangle.
In simple words: In any triangle, each angle must be smaller than the sum of the other two. Using the 180° rule, this forces each angle to be less than 90°.

Exam Tip: The key insight is that if any angle were 90° or more, the total would exceed 180° - so all angles must be less than 90° in an acute triangle.

Question 14. Prove that if one angle of a triangle is greater than the sum of the other two, the triangle is obtuse-angled.
Answer: Let ∠B > ∠A + ∠C.
From the angle sum property, ∠A + ∠C = 180° - ∠B.
Substituting: ∠B > 180° - ∠B
Adding ∠B to both sides: 2∠B > 180°
Therefore: ∠B > 90°
Since ∠B is greater than 90°, it is an obtuse angle, making triangle ABC obtuse-angled.
In simple words: If one angle exceeds the sum of the other two, that angle must be greater than 90° because the total is only 180°.

Exam Tip: This is the converse of the acute triangle property - any triangle with one angle exceeding 90° is automatically obtuse, regardless of the other two angles.

Question 15. In triangle ABC, ∠ACD = 128° (exterior angle), and ∠ABC = 43°. Find ∠ACB and ∠BAC.
Answer: Since ∠ACB and ∠ACD form a linear pair:
∠ACB + ∠ACD = 180°
∠ACB + 128° = 180°
∠ACB = 52°
Using the angle sum property in triangle ABC:
∠ABC + ∠ACB + ∠BAC = 180°
43° + 52° + ∠BAC = 180°
∠BAC = 85°
In simple words: An interior angle and its adjacent exterior angle always sum to 180°. Find the interior angle first, then use the triangle angle sum to find the third interior angle.

Exam Tip: Exterior angles are always supplementary to their adjacent interior angle - use this to convert between them quickly without calculations.

Question 16. In triangle ABC, exterior angle at B is 106° and exterior angle at C is 118°. Find all interior angles.
Answer: For the exterior angle at B:
∠ABC + 106° = 180°, so ∠ABC = 74°
For the exterior angle at C:
∠ACB + 118° = 180°, so ∠ACB = 62°
Using the angle sum in triangle ABC:
∠BAC + ∠ABC + ∠ACB = 180°
∠BAC + 74° + 62° = 180°
∠BAC = 44°
In simple words: Convert each exterior angle to its interior angle using the supplementary property. Then use the triangle angle sum to find the third interior angle.

Exam Tip: Always convert exterior angles to interior angles first by subtracting from 180° - this eliminates confusion and keeps the work organized.

Question 17. (i) Find x using the exterior angle property and angle sum in triangles.
Answer: (i) From the linear pair: ∠EAB + ∠BAC = 180°, so 110° + ∠BAC = 180°, giving ∠BAC = 70°.
From the linear pair at C: ∠BCA + ∠ACD = 180°, so ∠BCA + 120° = 180°, giving ∠BCA = 60°.
In triangle ABC: x + 70° + 60° = 180°, so x = 50°.
In simple words: Use linear pair angles to find two interior angles of the triangle, then apply the angle sum rule to find x.

Exam Tip: When exterior angles are given, always form linear pairs with their adjacent interior angles first - this usually gives you enough information to solve the triangle.

Question 18. Produce CD to cut AB at E. In triangle BDE, the exterior angle ∠CDB = ∠CEB + 45°. In triangle AEC, find ∠CEB = 55° + 30° = 85°. Calculate x.
Answer: By the exterior angle theorem in triangle BDE:
∠CDB (exterior) = ∠CEB + ∠DBE
x = ∠CEB + 45° ... (i)
By the exterior angle theorem in triangle AEC:
∠CEB (exterior) = ∠CAB + ∠ACE
∠CEB = 55° + 30° = 85°
Substituting into (i): x = 85° + 45° = 130°
In simple words: An exterior angle of a triangle equals the sum of the two non-adjacent interior angles. Apply this rule twice to find the unknown angle.

Exam Tip: The exterior angle theorem is faster than the angle sum property when you have exterior angles - use it whenever an exterior angle is involved.

Question 19. In triangle ABC, AD divides angle ∠BAC in ratio 1:3. Given ∠CAE = 108° and AD = DB. Find x = ∠ACB.
Answer: Let ∠BAD = y and ∠DAC = 3y.
Since ∠BAC + ∠CAE = 180° (linear pair):
y + 3y + 108° = 180°
4y = 72°
y = 18°
In triangle ABC, with AD = DB (so triangle ABD is isosceles):
∠ABC = ∠BAD = 18°
The angle ∠BAC = 4y = 72°
Using angle sum: 18° + x + 72° = 180°
x = 90°
In simple words: Use the linear pair to find y from the ratio condition. Apply the isosceles triangle property, then use the angle sum to find x.

Exam Tip: When a cevian divides an angle in a given ratio, set up the two parts as multiples of a variable and use the linear pair or angle sum to solve.

Question 20. In triangle ABC, sides BC, CA, and AB are produced to D, E, and F respectively. Prove that the sum of the three exterior angles equals 360°.
Answer: By the exterior angle theorem:
Exterior ∠DCA = ∠A + ∠B ... (i)
Exterior ∠EAB = ∠B + ∠C ... (ii)
Exterior ∠FBC = ∠A + ∠C ... (iii)
Adding (i), (ii), and (iii):
Ext. ∠DCA + Ext. ∠EAB + Ext. ∠FBC = (∠A + ∠B) + (∠B + ∠C) + (∠A + ∠C)
= 2∠A + 2∠B + 2∠C
= 2(∠A + ∠B + ∠C)
= 2 × 180° = 360° (Proved)
In simple words: Each exterior angle equals the sum of two interior angles. Adding all three exterior angles gives twice the total of all interior angles, which is 2 × 180° = 360°.

Exam Tip: This is a key theorem - the sum of exterior angles (one at each vertex) of any polygon is always 360°, not just triangles.

Question 21. In triangles ACE and BDF, prove that ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°.
Answer: In triangle ACE: ∠A + ∠C + ∠E = 180° ... (i)
In triangle BDF: ∠B + ∠D + ∠F = 180° ... (ii)
Adding (i) and (ii):
∠A + ∠C + ∠E + ∠B + ∠D + ∠F = 180° + 180° = 360°
Therefore: ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°
In simple words: Two separate triangles, each with angles summing to 180°, give a combined total of 360° when all six angles are added.

Exam Tip: This principle extends - the sum of all interior angles of any n triangles is 180n°. For two triangles, it's always 360°.

Question 22. In triangle ABC, the bisectors of ∠B and ∠C meet at O. If ∠A = 70°, find ∠BOC.
Answer: In triangle ABC:
∠B + ∠C = 180° - ∠A = 180° - 70° = 110°
Since BO and CO are angle bisectors:
∠OBC = ∠B/2 and ∠OCB = ∠C/2
In triangle BOC:
∠BOC + ∠OBC + ∠OCB = 180°
∠BOC + (∠B + ∠C)/2 = 180°
∠BOC + 110°/2 = 180°
∠BOC + 55° = 180°
∠BOC = 125°
In simple words: Find the sum of the two angles at B and C, halve it (because the bisectors split them), then subtract from 180° to find ∠BOC.

Exam Tip: For the angle between two angle bisectors, use the formula ∠BOC = 90° + ∠A/2, which gives 90° + 35° = 125° - this is a quick mental calculation.

Question 23. In triangle ABC with ∠A = 40°, sides AB and AC are produced to D and E. Bisectors of exterior angles ∠CBD and ∠BCE meet at O. Find ∠BOC.
Answer: The exterior angles are:
∠CBD = ∠C + 40° (exterior angle theorem)
∠BCE = ∠B + 40° (exterior angle theorem)
The bisectors divide these angles in half:
∠CBO = (∠C + 40°)/2 = ∠C/2 + 20°
∠BCO = (∠B + 40°)/2 = ∠B/2 + 20°
In triangle BCO:
∠BOC = 180° - ∠CBO - ∠BCO
= 180° - (∠C/2 + 20°) - (∠B/2 + 20°)
= 180° - (∠B + ∠C)/2 - 40°
Since ∠B + ∠C = 140°:
∠BOC = 180° - 70° - 40° = 70°
In simple words: Use the exterior angle theorem to express the exterior angles, apply the angle bisector properties, then calculate ∠BOC using the triangle angle sum.

Exam Tip: When exterior angle bisectors are involved, express them in terms of interior angles, then apply the angle sum rule systematically.

Question 24. In triangle ABC with ∠A : ∠B : ∠C = 3 : 2 : 1, if AD ⊥ CD, find ∠ECD.
Answer: Let ∠A = 3x, ∠B = 2x, ∠C = x.
From the angle sum: 3x + 2x + x = 180°, so x = 30°.
Therefore: ∠A = 90°, ∠B = 60°, ∠C = 30°
The exterior angle ∠ACE = ∠A + ∠B = 90° + 60° = 150°.
Since ∠ACD + ∠ECD = 150° and AD ⊥ CD (so ∠ACD = 90°):
∠ECD = 150° - 90° = 60°
In simple words: Use the ratio to find each angle. Apply the exterior angle theorem and the perpendicularity condition to isolate ∠ECD.

Exam Tip: When angles are in ratio and you need an exterior angle, calculate the interior angles first, then apply the exterior angle theorem systematically.

Question 25. In triangle ABC, AN bisects ∠A and AM ⊥ BC. Given ∠B = 65° and ∠C = 30°, find ∠MAN.
Answer: First, find ∠A:
∠A = 180° - 65° - 30° = 85°
Since AN bisects ∠A:
∠BAN = ∠A/2 = 42.5°
In triangle AMC, where AM ⊥ BC:
Exterior ∠MNA = ∠NAC + 30° (by exterior angle theorem)
∠MNA = 42.5° + 30° = 72.5°
In triangle AMN:
∠AMN = 90° (since AM ⊥ BC)
∠MAN = 180° - 90° - 72.5° = 17.5°
In simple words: Find ∠A using the angle sum, bisect it to get ∠AN, find the exterior angle in triangle AMC, then use the right angle at M to solve for ∠MAN.

Exam Tip: When angle bisectors and perpendiculars are both involved, work step-by-step through the resulting right triangle and use exterior angle properties where needed.

Question 26. Determine which of the following statements are true or false.
Answer: (i) False (ii) True (iii) False (iv) False (v) True (vi) True
In simple words: Evaluate each statement against the angle sum property (180° for triangles), exterior angle theorem, and properties of parallel lines cut by transversals.

Exam Tip: When answering true/false geometry questions, always verify using the fundamental triangle and parallel line properties rather than intuition alone.