CBSE Class 10 Mathematics Some Applications of Trigonometry VBQs Set 02

Question. When a point is observed, the angle formed by the line of sight with the horizontal where the point being viewed is above the horizontal level is known as
(a) angle of elevation
(b) angle of depression
(c) angle of a triangle
(d) right angle
Answer: (a) angle of elevation

Question. When we raise our head to look at an object, the angle formed by the line of sight with the horizontal is known as
(a) acute angle
(b) angle of elevation
(c) angle of depression
(d) right angle
Answer: (b) angle of elevation

Question. When we lower our head to look at an object, the angle formed by the line of sight horizontal is known as
(a) obtuse angle
(b) angle of elevation
(c) angle of depression
(d) acute angle
Answer: (c) angle of depression

Question. When the length of the shadow of a pole of height 10m is equal to 10m, then the elevation of source of light is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer: (b) 45°

Question. A man from the top of a 100m high tower sees a car moving towards the tower at an angle of depression of 30°. After some time, the angle of depression becomes 60°. The distance travelled by the car during this time is
(a) \( 100\sqrt{3} \) m
(b) \( \frac{100\sqrt{3}}{3} \) m
(c) \( \frac{200\sqrt{3}}{3} \) m
(d) \( 200\sqrt{3} \) m
Answer: (c) \( \frac{200\sqrt{3}}{3} \) m

Question. A man of height 6m observes the top of a tower and the foot of the tower at angles of 45° and 30° of elevation and depression respectively. The height of tower is
(a) \( 6\sqrt{3} \) m
(b) 12 m
(c) \( 6(\sqrt{3} - 1) \) m
(d) \( 6(\sqrt{3} + 1) \) m
Answer: (d) \( 6(\sqrt{3} + 1) \) m

Question. A man observes a car from the top of 50m high tower that is moving towards the tower with uniform speed of 20m/s. If the angle of depression of the car at that instant is 45°, then time taken by car to reach the tower is
(a) 2 sec
(b) 2.5 sec
(c) 4 sec
(d) 5 sec
Answer: (b) 2.5 sec

Question. From the top of a cliff 25m high, the angle of elevation of the top of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is
(a) 40m
(b) 50m
(c) 60m
(d) 75m
Answer: (b) 50m

Question. Position of four ships A, B, C and D in the sea are as follows: Ship C is due north of D. Ships A, B and C are in a straight line and ∠CBD = 60°. Distance between ships B and D is 100 km. Distance between ships C and D is same as the distance between ships C and B. If ∠CAD = 30°, find the distance between ships A and D.
(a) 100 km
(b) \( 50\sqrt{3} \) km
(c) \( 100\sqrt{3} \) km
(d) 200 km
Answer: (c) \( 100\sqrt{3} \) km

Question. The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of 100 m from its base is 45°. If the angle of elevation of the top of the complete pillar at the same point is to be 60°, then the height of their complete pillar is to be increased by
(a) \( 50\sqrt{2} \) m
(b) 100 m
(c) \( 100\sqrt{3} \) m
(d) \( 100(\sqrt{3} - 1) \) m
Answer: (d) \( 100(\sqrt{3} - 1) \) m

Question. If a pole 6 m high casts a shadow \( 2\sqrt{3} \) m along the ground, then the Sun’s elevation is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer: (a) 60°

Question. The angle of elevation of the top of a 15 m high tower at a point 15 m away from the base of the tower is
(a) 30°
(b) 60°
(c) 45°
(d) 75°
Answer: (c) 45°

ASSERTION AND REASON QUESTIONS

Question. Assertion (A): The length of the shadow of a vertical tower is \(\sqrt{3}\) times the height of tower. So, the angle of elevation of the Sun at this instant is 45°.
Reason (R): The value of \(\tan 45^\circ\) is 1.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.

Question. At some time of the day, the length of the shadow of a tower is equal to its height. Then the Sun’s altitude at that time is
(a) \( 30^\circ \)
(b) \( 60^\circ \)
(c) \( 90^\circ \)
(d) \( 45^\circ \)
Answer: (d) \( 45^\circ \)

Question. A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that here is no slack in the string, the angle of elevation of the kite at the ground is
(a) \( 45^\circ \)
(b) \( 30^\circ \)
(c) \( 60^\circ \)
(d) \( 90^\circ \)
Answer: (b) \( 30^\circ \)

Question. Length of the shadow of a person is \( x \) when the angle of elevation of the Sun is \( 45^\circ \). If the length of the shadow increases by \( (\sqrt{3} - 1)x \), then the angle of elevation of the Sun should become
(a) \( 60^\circ \)
(b) \( 45^\circ \)
(c) \( 30^\circ \)
(d) \( 20^\circ \)
Answer: (c) \( 30^\circ \)

Question. When the length of the shadow of a pole of height 10 m is equal to 10 m, then the elevation of source of light is
(a) \( 30^\circ \)
(b) \( 45^\circ \)
(c) \( 60^\circ \)
(d) \( 90^\circ \)
Answer: (b) \( 45^\circ \)

Question. The angles of elevation of two cars, from the car to the top of a building are \( 45^\circ \) and \( 30^\circ \). If the cars are on the same side of the building and are 50 m apart, what is the height of the building?
(a) \( 25(\sqrt{3} - 1) \) m
(b) \( 25(\sqrt{3} + 1) \) m
(c) \( 50(\sqrt{3} - 1) \) m
(d) \( 50(\sqrt{3} + 1) \) m
Answer: (b) \( 25(\sqrt{3} + 1) \) m

Question. From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression \( 30^\circ \) and \( 45^\circ \) respectively. Find the distance between the cars. [Use \( \sqrt{3} = 1.732 \)]
Answer: Let the tower be \( AB \) of height 100 m. Let cars be at \( C \) and \( D \).
In right \(\Delta ABC\), \( \tan 30^\circ = \frac{100}{BC} \implies BC = 100\sqrt{3} \) m.
In right \(\Delta ABD\), \( \tan 45^\circ = \frac{100}{BD} \implies BD = 100 \) m.
Distance between cars \( = BC + BD = 100\sqrt{3} + 100 = 100(\sqrt{3} + 1) = 100(1.732 + 1) = 273.2 \) m.

Question. From the top of a tower 100 m high, the angle of depression of the top and bottom of a pole are observed to be \( 45^\circ \) and \( 60^\circ \). Find the height of the pole.
Answer: Let height of pole be \( h \).
In right triangle for the bottom of pole: \( \tan 60^\circ = \frac{100}{x} \implies x = \frac{100}{\sqrt{3}} \).
In right triangle for the top of pole: \( \tan 45^\circ = \frac{100 - h}{x} \implies 1 = \frac{100 - h}{x} \implies x = 100 - h \).
\( 100 - h = \frac{100}{\sqrt{3}} \implies h = 100 - \frac{100}{\sqrt{3}} = 100 \left(1 - \frac{1}{\sqrt{3}}\right) = 100 \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) = \frac{100(3 - \sqrt{3})}{3} \) m.

Question. As observed from the top of light house, 50 m above sea level the angle of depression of ship, sailing directly towards it, changes from \( 30^\circ \) to \( 45^\circ \). Determine the distance travelled by ship.
Answer: Let the initial distance be \( d_1 \) and final distance be \( d_2 \).
\( \tan 30^\circ = \frac{50}{d_1} \implies d_1 = 50\sqrt{3} \) m.
\( \tan 45^\circ = \frac{50}{d_2} \implies d_2 = 50 \) m.
Distance travelled \( = d_1 - d_2 = 50\sqrt{3} - 50 = 50(\sqrt{3} - 1) \) m.

Question. A portion of 60 m long tree is broken by tornado and the top struck up the ground making an angle of \( 30^\circ \) with the ground level. Find the height of the point where the tree is broken.
Answer: Let height at which it is broken be \( h \). Length of broken part \( = 60 - h \).
\( \sin 30^\circ = \frac{h}{60 - h} \)
\( \implies \) \( \frac{1}{2} = \frac{h}{60 - h} \)
\( \implies \) \( 60 - h = 2h \)
\( \implies \) \( 3h = 60 \)
\( \implies \) \( h = 20 \) m.

Question. The height of a tower is half the height of the flag staff at its top. The angle of elevation of the top of the tower as seen from a distance of 10 metres from its foot is \( 30^\circ \). Find the angle of elevation of the top of the flagstaff from the same point.
Answer: Let height of tower be \( h \). Flagstaff height \( = 2h \).
\( \tan 30^\circ = \frac{h}{10} \implies h = \frac{10}{\sqrt{3}} \).
Total height \( = h + 2h = 3h = 3 \left(\frac{10}{\sqrt{3}}\right) = 10\sqrt{3} \).
Angle of elevation \( \alpha \) of the top of flagstaff: \( \tan \alpha = \frac{3h}{10} = \frac{10\sqrt{3}}{10} = \sqrt{3} \)
\( \implies \) \( \alpha = 60^\circ \).

Question. The length of a string between a kite and a point on the roof of a building 10 m high is 180 m. If the string makes an angle \( \theta \) with the level ground such that \( \tan \theta = \frac{4}{3} \), how high is the kite from the ground?
Answer: \( \tan \theta = \frac{4}{3} \implies \sin \theta = \frac{4}{5} \).
Let height of kite above building be \( h \). \( \sin \theta = \frac{h}{180} \)
\( \implies \) \( \frac{4}{5} = \frac{h}{180} \)
\( \implies \) \( h = 144 \) m.
Total height from ground \( = h + 10 = 144 + 10 = 154 \) m.

Question. Two lamp-posts are of equal height. A boy measured the elevation of the top of each lamp-post from the mid-point of the line-segment joining the feet of lamp-post as \( 30^\circ \). After walking 15m towards one of them, he measured the elevation of the top of nearest lamp-post at the point where he stands as \( 60^\circ \). Determine the height of each lamp-post and the distance between them.
Answer: Let height be \( h \) and half distance be \( x \). \( \tan 30^\circ = \frac{h}{x} \implies x = h\sqrt{3} \).
Distance from nearest post \( = x - 15 \).
\( \tan 60^\circ = \frac{h}{x - 15} \)
\( \implies \) \( \sqrt{3} = \frac{h}{h\sqrt{3} - 15} \)
\( \implies \) \( 3h - 15\sqrt{3} = h \)
\( \implies \) \( 2h = 15\sqrt{3} \implies h = 7.5\sqrt{3} \) m.
Distance \( 2x = 2h\sqrt{3} = 2(7.5\sqrt{3})\sqrt{3} = 45 \) m.

Question. An aeroplane is observed at the same time by two anti-aircraft guns 6000 m distant apart to be at elevations of \( 30^\circ \) and \( 45^\circ \) respectively. Find height of the plane and its horizontal distance from the nearer gun.
Answer: Let height be \( h \). Distances are \( x \) and \( 6000 - x \).
\( \tan 45^\circ = \frac{h}{x} \implies x = h \).
\( \tan 30^\circ = \frac{h}{6000 - h} \)
\( \implies \) \( \frac{1}{\sqrt{3}} = \frac{h}{6000 - h} \)
\( \implies \) \( 6000 - h = h\sqrt{3} \)
\( \implies \) \( h(\sqrt{3} + 1) = 6000 \implies h = \frac{6000}{\sqrt{3} + 1} = 3000(\sqrt{3} - 1) \) m.
Nearer gun distance \( = h = 3000(\sqrt{3} - 1) \) m.

Question. Assertion (A): If the angle of elevation of Sun, above a perpendicular line (tower) decreases, then shadow of the tower increases.
Reason (R): It is due to the decrease in the slope of line of sight.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation of A.