CBSE Class 10 Some Applications of Trigonometry Sure Shot Questions Set 06

Question. If the height of a vertical pole is \( \sqrt{3} \) times the length of its shadow on the ground, then the angle of elevation of the Sun at that time is :*
(a) 30°
(b) 60°
(c) 45°
(d) 75°
Answer: (b) 60°
Explanation :
Let the length of the shadow be \( x \) ft.
Thus, the height of the vertical pole is \( \sqrt{3}x \) ft.
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \)
\( \tan \theta = \frac{\sqrt{3}x}{x} \)
\( = \sqrt{3} = \tan 60^\circ \)
or \( \theta = 60^\circ \)
Thus, the angle of elevation is 60°.

Question. From the top of a cliff 20 m 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) 20 m
(b) 40 m
(c) 60 m
(d) 80 m
Answer: (b) 40 m
Explanation :
Let the angle of elevation = angle of depression = \( x^\circ \)
Also, let the distance between the tower (BD) and the cliff (AC) be \( d \) m.
Thus, in \( \triangle ABC \),
\( \tan x = \frac{AC}{AB} = \frac{20}{d} \dots \text{(i)} \)
Also, in \( \triangle CDE \),
\( \tan x = \frac{ED}{CE} = \frac{h}{d} \dots \text{(ii)} \)
From equations (i) and (ii),
\( \frac{h}{d} = \frac{20}{d} \)
\( \Rightarrow h = 20 \text{ m} \)
\( \Rightarrow \text{Height of the tower is } 20 + 20 = 40 \text{ m} \).

Question. A ladder makes an angle of 60° with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder, in metres is :*
(a) \( \frac{2}{\sqrt{3}} \)
(b) \( 2\sqrt{3} \)
(c) \( 2\sqrt{2} \)
(d) 4
Answer: (d) 4
Explanation :
Given, the distance of the ladder AC from the base of the wall AB is 2 m.
So, \( \cos 60^\circ = \frac{2}{x} \)
\( \frac{1}{2} = \frac{2}{x} \)
\( \Rightarrow x = 2 \times 2 \)
The height of the ladder = 4 m

Question. The angle of depression of a car standing on the ground from the top of a 75 m high tower is 30°. The distance of the car from the base of the tower (in m) is :*
(a) \( 25\sqrt{3} \)
(b) \( 50\sqrt{3} \)
(c) \( 75\sqrt{3} \)
(d) 150
Answer: (c) \( 75\sqrt{3} \)
Explanation :
Let the distance of the car from the base of the tower be \( x \) m.
Height of tower = 75 m
\( \tan 30^\circ = \frac{\text{Perpendicular}}{\text{Base}} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{x} \)
\( \Rightarrow x = 75\sqrt{3} \)
Thus, distance of the parked car from the base of the tower is \( 75\sqrt{3} \).

Question. The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower (in metres) is :*
(a) \( 50\sqrt{3} \)
(b) \( 150\sqrt{3} \)
(c) \( 150\sqrt{2} \)
(d) 75
Answer: (b) \( 150\sqrt{3} \)
Explanation :
Let the distance of the parked car from the base of the tower be \( x \) m.
Height of tower = 150 m
\( \tan 30^\circ = \frac{\text{Perpendicular}}{\text{Base}} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{x} \)
\( \Rightarrow x = 150\sqrt{3} \text{ m} \)
Thus, distance of the parked car from the base of the tower is \( 150\sqrt{3} \).

Question. A kite is flying at a height of 30 m from the ground. The length of a string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is :*
(a) 45°
(b) 30°
(c) 60°
(d) 90°
Answer: (b) 30°
Explanation :
Given,
Perpendicular height = 30 m and hypotenuse = 60 m.
So, \( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \)
\( = \frac{30}{60} = \frac{1}{2} = \sin 30^\circ \)
Thus, \( \theta = 30^\circ \)

Question. The length of shadow of a tower on the plane ground is \( \sqrt{3} \) times the height of the tower. The angle of elevation of sun is :*
(a) 45°
(b) 30°
(c) 60°
(d) 90°
Answer: (b) 30°
Explanation :
Let the height of the tower be \( h \) m.
So, the length of the shadow will be \( \sqrt{3}h \) m.
Thus, \( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{h}{\sqrt{3}h} \)
\( = \frac{1}{\sqrt{3}} = \tan 30^\circ \)
Thus, \( \theta = 30^\circ \)

Question. At some point of time in 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°
(b) 60°
(c) 90°
(d) 45°
Answer: (d) 45°
Explanation :
Let the height of the tower be \( h \) m and the length of the shadow be \( L \) m.
Now, \( h = L \)
Thus, \( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \)
\( = \frac{h}{L} = 1 = \tan 45^\circ \)
Hence, \( \theta = 45^\circ \)

Question. A tower stands vertically on the ground. From a point on the ground which is 25 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 45°. Then the height (in metres) of the tower is :*
(a) \( 25\sqrt{2} \)
(b) \( 25\sqrt{3} \)
(c) 25
(d) 12.5
Answer: (c) 25
Explanation :
Let the height of the tower be \( H \) m.
Thus, \( \tan 45^\circ = \frac{\text{Perpendicular}}{\text{Base}} \)
or \( \frac{H}{25} = 1 \)
Hence, \( H = 25 \text{ m} \)

Question. The angle of elevation of the top of a tower from a point on the ground which is 30 m away from the foot of the tower is 45°. The height of the tower (in metres) is :*
(a) 15
(b) 30
(c) \( 30\sqrt{3} \)
(d) \( 10\sqrt{3} \)
Answer: (b) 30
Explanation :
Let the height of the tower be \( H \).
Thus, \( \tan 45^\circ = \frac{\text{Perpendicular}}{\text{Base}} \)
or \( \frac{H}{30} = 1 \)
Hence, \( H = 30 \text{ m} \)

Question. From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 200 m high, the distance of point P from the foot of the tower is :
(a) 340 m
(b) \( 200\sqrt{3} \) m
(c) 346 m
(d) 320 m
Answer: (b) \( 200\sqrt{3} \) m
Explanation :
In right angle \( \triangle ABP \)
\( \tan 30^\circ = \frac{AB}{PB} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{200}{PB} \)
\( \Rightarrow PB = 200\sqrt{3} \)
Distance of point P from the foot of the tower is \( 200\sqrt{3} \) m.

Question. The tops of two poles of height 24 m and 36 m are connected by a wire. If the wire makes an angle of 60° with the horizontal, then the length of the wire is :
(a) \( 8\sqrt{3} \) m
(b) 8 m
(c) \( 6\sqrt{3} \) m
(d) 6 m
Answer: (a) \( 8\sqrt{3} \) m
Explanation :
Let AB and DC be two poles and AD be the wire
In \( \triangle ADE \),
\( \sin 60^\circ = \frac{DE}{AD} \) [AD = Length of wire]
\( AD = \frac{DE}{\sin 60^\circ} = \frac{12}{\sqrt{3}/2} \)
\( = \frac{12 \times 2}{\sqrt{3}} = \frac{24 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
[On multiplying and dividing by \( \sqrt{3} \)]
\( = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ m} \)

Question. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops ?
(a) 13 m
(b) 17 m
(c) 18 m
(d) 23 m
Answer: (a) 13 m
Explanation :
Given that, AB = 6 m and EC = 11 m
\( \Rightarrow BC = 12 \text{ m} \)
\( \because BC = AD = 12 \text{ m} \)
and \( ED = EC - CD = EC - AB (\because AB = CD) \)
\( = 11 - 6 = 5 \text{ m} \)
In \( \triangle AED \), using Pythagoras theorem
\( (AE)^2 = (AD)^2 + (ED)^2 \)
\( = (12)^2 + (5)^2 \)
\( = 144 + 25 \)
\( = 169 = (13)^2 \)
\( \because AE = 13 \text{ m} \)
Then, the distance between their tops is 13 m.

Question. The length of shadow of a tree is 16 m when the angle of elevation of the Sun is 60°. What is the height of the tree ?
(a) 8 m
(b) 16 m
(c) \( 16\sqrt{3} \)
(d) \( \frac{16}{\sqrt{3}} \)
Answer: (c) \( 16\sqrt{3} \)
Explanation :
Let the height of the tree is \( h \) m.
\( \tan 60^\circ = \frac{h}{16} \)
\( \Rightarrow \sqrt{3} = \frac{h}{16} \)
\( h = 16\sqrt{3} \text{ m} \)

Question. What is the angle of elevation of the Sun, when the shadow of a pole of height \( x \) m is \( \frac{x}{\sqrt{3}} \text{ m} \) ?
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Answer: (c) 60°
Explanation :
Let angle of elevation be \( \theta \).
Height of the pole = X m
and Length of shadow = \( \frac{x}{\sqrt{3}} \text{ m} \)
In \( \triangle ABC \), \( \tan \theta = \frac{x}{x/\sqrt{3}} = \frac{\sqrt{3}x}{x} = \sqrt{3} \)
\( \tan \theta = \sqrt{3} = \tan 60^\circ \)
\( \theta = 60^\circ \)

Question. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 9 m from the surface of river, then find the width of the river.
(a) 24 m
(b) 25 m
(c) 27 m
(d) 29 m
Answer: (b) 25 m
Explanation :
Here, width of the river = DC
In \( \triangle ABC \),
\( \tan 30^\circ = \frac{AB}{BC} \)
\( BC = \frac{9}{\tan 30^\circ} = 9\sqrt{3} \text{ m} \)
Now, in \( \triangle ABD \)
\( \tan 45^\circ = \frac{AB}{BD} \)
AB = BD
BD = 9 m [... AB = 9 m]
\( DC = DB + BC = 9 + 9\sqrt{3} \)
\( = 9(\sqrt{3}+1) \)
= 24.5898 m [\( \sqrt{3} = 1.732 \)]
~ 25 m

Question. The angle of elevation of a cloud from a point 200 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. The height of the cloud is :
(a) 200 m
(b) 300 m
(c) 400 m
(d) 600 m
Answer: (c) 400 m
Explanation :
\( \tan 30^\circ = \frac{h}{PM} \)
\( \Rightarrow PM = \sqrt{3}h \dots \text{(i)} \)
\( \tan 60^\circ = \frac{h + 400}{PM} \)
\( \Rightarrow PM = \frac{h + 400}{\sqrt{3}} \dots \text{(ii)} \)
From equations (i) and (ii),
\( \sqrt{3}h = \frac{h + 400}{\sqrt{3}} \Rightarrow 3h - h = 400 \)
2h = 400
So, height of the cloud = 200 + 200 = 400 m.

Question. The shadow of a tower standing on a level plane is found to be 50 m longer when the Sun’s elevation is 30°, then when it was 60°. What is the height of the tower ?
(a) 25 m
(b) \( 25\sqrt{3} \) m
(c) \( \frac{25}{\sqrt{3}} \) m
(d) 30 m
Answer: (b) \( 25\sqrt{3} \) m
Explanation :
Let the height of tower be \( h \) and BC = \( x \) m.
In \( \triangle BCD \), \( \tan 60^\circ = \frac{h}{x} \)
\( \Rightarrow \sqrt{3} = \frac{h}{x} \)
\( \Rightarrow h = x\sqrt{3} \dots \text{(i)} \)
In \( \triangle ACD \), \( \tan 30^\circ = \frac{h}{50 + x} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{50 + x} \) [from (i)]
\( \Rightarrow 50 + x = 3x \)
\( \Rightarrow x = 25 \text{ m} \)
Now put the value of X in equation (i)
\( \therefore h = 25\sqrt{3} \text{ m} \)

Question. At the foot of a mountain, the elevation of its summit is 45°. After ascending 2 km towards the mountain upon an incline of 30°, the elevation changes to 60°. The height of the mountain is :
(a) \( (\sqrt{3}-1)\text{km} \)
(b) \( (\sqrt{3}+1)\text{km} \)
(c) \( (\sqrt{3}+1)\text{km} \)
(d) \( (\sqrt{3}+2)\text{km} \)
Answer: (b) \( (\sqrt{3}+1)\text{km} \)
Explanation :
Let AB be the mountain of the height \( h \) km.
In \( \triangle OAB \), \( \tan 45^\circ = \frac{AB}{OB} \)
OB = \( h \) km
In \( \triangle OLM \), \( \frac{OM}{OL} = \cos 30^\circ \)
OM = \( 2 \cos 30^\circ = \sqrt{3} \text{ km} \) [OL = 2 km]
LN = BM = OB – OM = \( (h - \sqrt{3}) \text{ km} \)
In \( \triangle OLM \), \( \sin 30^\circ = \frac{LM}{OL} \)
LM = \( 2 \sin 30^\circ = 1 \text{ km} \)
BN = LM = 1 km
In \( \triangle ALN \),
\( \tan 60^\circ = \frac{AN}{LN} \)
\( \Rightarrow \sqrt{3} = \frac{AB - BN}{LN} \)
\( \Rightarrow \sqrt{3} = \frac{h - 1}{h - \sqrt{3}} \)
\( \Rightarrow \sqrt{3}h - 3 = h - 1 \)
\( h = \frac{2}{\sqrt{3} - 1} = \frac{2}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \)
= \( (\sqrt{3}+1) \) km

Question. From the top of a cliff 200 m high, the angles of depression of the top and bottom of a tower are observed to be 30° and 45°, respectively. What is the height of the tower ?
(a) 400 m
(b) \( 400\sqrt{3} \)
(c) \( 400/\sqrt{3} \)
(d) None of the options
Answer: (d) None of the options
Explanation :
In \( \triangle ACB \), \( \tan 30^\circ = \frac{200 - h}{x} \)
\( = \frac{1}{\sqrt{3}} = \frac{200 - h}{x} \)
\( \Rightarrow X = (200 - h)\sqrt{3} \dots \text{(i)} \)
and in \( \triangle ADE \), \( \tan 45^\circ = \frac{200}{x} \)
\( \Rightarrow 1 = \frac{200}{x} \)
\( \Rightarrow x = 200 \text{ m} \dots \text{(ii)} \)
From equations (i) and (ii),
\( 200 = (200 - h)\sqrt{3} \)
\( \Rightarrow h = 200 - \frac{200}{\sqrt{3}} \text{ m} \)
\( = 200(\frac{\sqrt{3} - 1}{\sqrt{3}}) \times \frac{\sqrt{3}}{\sqrt{3}} \)
\( = 200 \frac{(3 - \sqrt{3})}{3} \)

Question. From the top of a building 60 m high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. The height of the tower is :
(a) 40 m
(b) 45 m
(c) 50 m
(d) 55 m
Answer: (a) 40 m
Explanation :
Let AB be a building and PQ be a tower.
Let height of tower PQ = \( h \) m
AM = (60 – h)
BQ = \( y \) = PM,
AB = 60m, BM = \( h \)m
Now, in \( \triangle AMP \),
\( \tan 30^\circ = \frac{AM}{PM} = \frac{60 - h}{y} \)
\( \frac{1}{\sqrt{3}} = \frac{60 - h}{y} \)
\( y = (60 - h)\sqrt{3} \dots \text{(i)} \)
Now, in \( \triangle AQB \),
\( \tan 60^\circ = \frac{AB}{BQ} = \frac{60}{y} \)
\( \sqrt{3} = \frac{60}{y} \)
\( y = \frac{60}{\sqrt{3}} \dots \text{(ii)} \)
From equations, (i) and (ii), we get
\( (60 - h)\sqrt{3} = \frac{60}{\sqrt{3}} \)
\( \Rightarrow 3(60 - h) = 60 \)
\( \Rightarrow 180 - 3h = 60 \)
\( \Rightarrow 3h = 180 - 60 \)
\( \Rightarrow 3h = 120 \)
\( h = 40 \text{ m} \)

Question. The angles of depression of two ships from the top of a light – house are 60° and 45° towards east. If the ships are 300 m apart, the height of the light – house is :
(a) \( 200(3+\sqrt{3})\text{m} \)
(b) \( 250(3+\sqrt{3})\text{m} \)
(c) \( 150(3+\sqrt{3})\text{m} \)
(d) \( 160(3+\sqrt{3})\text{m} \)
Answer: (c) \( 150(3+\sqrt{3})\text{m} \)
Explanation :
Let \( x \) be the distance BC.
In \( \triangle ABC \),
\( \frac{h}{x} = \tan 60^\circ \)
\( h = \sqrt{3}x \dots \text{(i)} \)
In \( \triangle ABD \),
\( \frac{h}{x + 300} = \tan 45^\circ \)
\( h = x + 300 \)
\( h = \frac{h}{\sqrt{3}} + 300 \) [From eqn. (i)]
\( h\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) = 300 \)
\( \Rightarrow h = \frac{300\sqrt{3}}{\sqrt{3}-1} \)
\( \Rightarrow h = \frac{300\sqrt{3} \times (\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \)
\( \Rightarrow h = \frac{300 \times (3+\sqrt{3})}{(3-1)} \)
\( \Rightarrow h = \frac{300 \times (3+\sqrt{3})}{2} \)
\( h = 150 (3 + \sqrt{3}) \)
Height of the light house = \( 150 (3 + \sqrt{3}) \text{ m} \)

Question. From the top of a tower, the angles of depression of two objects P and Q (situated on the ground on the same side of the tower) separated at a distance of \( 100(3 - \sqrt{3})\text{m} \) are 45° and 60° respectively. The height of the tower is :
(a) 200 m
(b) 250 m
(c) 300 m
(d) None of the options
Answer: (c) 300 m
Explanation :
Let AB is tower whose height is \( h \text{ m} \).
Distance between objects P and Q are \( 100(3 - \sqrt{3}) \text{ m} \) and BP is \( X \text{ m} \).
In \( \triangle ABP \),
\( \tan 60^\circ = \frac{AB}{X} = \frac{h}{X} \)
\( X = \frac{h}{\sqrt{3}} \dots \text{(i)} \)
In \( \triangle ABQ \),
\( \tan 45^\circ = \frac{AB}{BQ} = \frac{h}{100(3 - \sqrt{3}) + X} \)
\( \Rightarrow h = 100(3 - \sqrt{3}) + X \)
Now, from equation (i), we put value of X
\( h = 100(3 - \sqrt{3}) + \frac{h}{\sqrt{3}} \)
\( \Rightarrow h - \frac{h}{\sqrt{3}} = 100(3 - \sqrt{3}) \)
\( \Rightarrow \frac{h(\sqrt{3}-1)}{\sqrt{3}} = 100(3 - \sqrt{3}) \)
\( \Rightarrow h = \frac{100\sqrt{3}(3 - \sqrt{3})}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \)
\( \Rightarrow h = \frac{100\sqrt{3}(3\sqrt{3} + 3 - 3 - \sqrt{3})}{3-1} \)
\( \Rightarrow h = \frac{100 \times 2\sqrt{3} \times \sqrt{3}}{2} \)
\( \Rightarrow h = 100 \times 3 = 300 \text{ m} \).

Question. An observer, 1.5 m tall, is 28.5 m away from a 30 m high tower, then the angle of elevation of the top of the tower from the eye of the observer is :
(a) 60°
(b) 30°
(c) 15°
(d) 45°
Answer: (d) 45°
Explanation :
Here AB = Tower
CD = Observer
Here \( BD = EC = 28.5 \)
\( AE = AC - EC \)
\( = 30 - 1.5 \)
\( = 28.5 \)
In \( \triangle ADE \)
\( \tan \theta = \frac{AE}{DE} = \frac{28.5}{28.5} \)
\( = 1 \)
\( D = 45^\circ \).

Question. The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°, then the distance of the car from the tower (in meter) is :
(a) \( 150\sqrt{3} \)
(b) \( 100\sqrt{3} \)
(c) 50
(d) \( \sqrt{3} \)
Answer: (a) \( 150\sqrt{3} \)
Explanation :
Let distance of car = \( x \)
In \( \triangle ABC \)
\( \tan 30^\circ = \frac{AB}{BC} = \frac{150}{x} \)
\( \frac{1}{\sqrt{3}} = \frac{150}{x} \)
\( x = 150\sqrt{3} \)

Question. A ladder 10 m long reaches a window 8 m above the ground. The distance of the foot of the ladder from the base of the wall is ................. m.
(a) 3
(b) 4
(c) 6
(d) 2
Answer: (c) 6
Explanation :
Let AC be a ladder and AB be a window.
In \( \triangle ABC \),
\( AC^2 = AB^2 + BC^2 \)
\( 100 = 64 + BC^2 \)
\( BC^2 = 36 \)
\( BC = 6 \text{ cm} \).

Question. The height of the tower is 10 m. Then the length of its shadow when sun’s altitude is 45º is ...........
(a) 15 m
(b) 10 m
(c) 5 m
(d) 2 m
Answer: (b) 10
Explanation :
Let, Shadow = \( x = BC \)
From triangle ABC,
\( \tan 45^\circ = \frac{AB}{BC} \)
\( AB = BC = 10 \text{ m} \).

Question. If the angles of elevation of a tower from two points distant a and b (a > b) from its foot and in the same straight line from it are 30° and 60°, then the height of the tower is :
(a) \( \sqrt{a+b} \)
(b) \( \sqrt{ab} \)
(c) \( \sqrt{a-b} \)
(d) \( \sqrt{ab} \)
Answer: (b) \( \sqrt{ab} \)
Explanation :
Let AB be the tower and P and Q are such points that PB = \( a \), QB = \( b \) and angles of elevation at P and Q are 30° and 60° respectively.
Let \( AB = h \)
Now in right DAPB,
\( \tan q = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AP}{PB} \)
\( \Rightarrow \tan 30^\circ = \frac{h}{a} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{a} \dots \text{(i)} \)
Similarly, in right DAQB,
\( \tan 60^\circ = \frac{AB}{QB} \)
\( \Rightarrow \sqrt{3} = \frac{h}{b} \dots \text{(ii)} \)
Multiplying (i) and (ii)
\( \frac{1}{\sqrt{3}} \times \sqrt{3} = \frac{h}{a} \times \frac{h}{b} \)
\( \Rightarrow 1 = \frac{h^2}{ab} \)
\( \Rightarrow h^2 = ab \)
\( \Rightarrow h = \sqrt{ab} \)
\( \therefore \) Height of the tower = \( \sqrt{ab} \)

Question. The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart. the height of the light house is:
(a) \( 50\sqrt{3+1} \text{ m} \)
(b) \( 50\sqrt{3-1} \text{ m} \)
(c) \( 50(\sqrt{3}-1) \text{ m} \)
(d) \( 50(\sqrt{3}+1) \text{ m} \)
Answer: (c) \( 50(\sqrt{3}-1) \text{ m} \)
Explanation :
Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively.
... XAY || CD
\( \therefore \angle ACB = \angle XAC = 30^\circ \)
and \( \angle ADB = \angle YAD = 45^\circ \),
CD = 100m
Let AB = \( h \) and CB = \( x \), then
BD = \( (100 - x) \)
Now in right DACB,
\( \tan q = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AP}{CB} \)
\( \tan 30^\circ = hx \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x} \)
\( \Rightarrow x = \sqrt{3}h \dots \text{(i)} \)
Similarly, in right DADB,
\( \tan 45^\circ = \frac{AB}{BD} \)
\( \Rightarrow 1 = \frac{h}{100 - x} \)
\( \Rightarrow 100 - x = h \)
\( \Rightarrow x = 100 - h \dots \text{(ii)} \)
From (i) and (ii)
\( \sqrt{3}h = 100 - h \)
\( \Rightarrow \sqrt{3}h + h = 100 \)
\( \Rightarrow (\sqrt{3}+1)h = 100 \)
\( h = \frac{100}{\sqrt{3}+1} = \frac{100(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \)
\( = \frac{100(\sqrt{3}-1)}{3-1} = \frac{100(\sqrt{3}-1)}{2} \)
\( = 50(\sqrt{3}-1) \)
\( \therefore \) Height of light house = \( 50(\sqrt{3}-1) \text{ m} \)

Question. If the angle of elevation of a cloud from a point 200 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°, then the height of the cloud above the lake is:
(a) 200 m
(b) 500 m
(c) 30 m
(d) 400 m
Answer: (d) 400 m
Explanation :
Let AB be the surface of the lake and P be the point of observation.
So AP = 60 m.
Here, C is the position of the cloud and C' is the reflection in the lake. Then CB = C'B.
Let PM be the perpendicular from P on CB.
Then \( \angle CPM = 30^\circ \) and \( \angle C'PM = 60^\circ \)
Let \( CM = h \), \( PM = x \), then \( CB = h + 200 \) and \( C'B = h + 200 \)
Here, we have to find the height of cloud.
So we use trigonometric ratios.
In DCMP,
\( \Rightarrow \tan 30^\circ = \frac{CM}{PM} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = hx \)
\( \Rightarrow x = \sqrt{3}h \)
Again in DPMC',
\( \Rightarrow \tan 60^\circ = \frac{C'M}{PM} \)
\( \Rightarrow \sqrt{3} = \text{C'B} + \text{BMPM} \)
\( \Rightarrow \sqrt{3} = h + 200 + 200x \)
\( \Rightarrow \sqrt{3}x = h + 400 \)
Put, \( x = \sqrt{3}h \)
\( \Rightarrow 3h = h + 400 \)
\( \Rightarrow 2h = 400 \)
\( \Rightarrow h = 200 \)
Now,
\( \Rightarrow CB = h + 200 \)
\( \Rightarrow CB = 200 + 200 \)
\( \Rightarrow CB = 400\text{m} \)

Question. If the height of a vertical pole is \( \sqrt{3} \) times the length of its shadow on ground then the angle of elevation of the sun at that time is:
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Answer: (c) 60°
Explanation :
Here, AO be the pole; BO be its shadow and \( \theta \) be the angle of elevation of the sun.
Let BO = \( x \)
Then, AO = \( x\sqrt{3} \)
In DAOB,
\( \tan q = \frac{AO}{BO} \)
\( \Rightarrow \tan q = \frac{x\sqrt{3}}{x} \)
\( \Rightarrow \tan q = \sqrt{3} \)
\( \Rightarrow \tan q = \tan 60^\circ \)
\( \Rightarrow q = 60^\circ \)

Question. The shadow of a 5 m long stick is 2 m long. At the same time the length of the shadow of a 12.5 m high tree is:
(a) 3 m
(b) 3.5 m
(c) 4.5 m
(d) 5 m
Answer: (d) 5 m
Explanation :
Let AB be the stick and BC be its shadow and PQ be the tree and QR be its shadow.
We have,
AB = 5 m, BC = 2 m, PQ = 12.5 m
In DABC,
\( \tan q = \frac{AB}{BC} \)
\( \Rightarrow \tan q = \frac{5}{2} \)
Now, In DPQR,
\( \tan q = \frac{PQ}{QR} \)
\( \Rightarrow \frac{5}{2} = 12.5 \text{ QR}[Using (i)] \)
\( \Rightarrow QR = 12.5 \times 25 = 255 \)
\( \Rightarrow QR = 5 \text{ m} \)

Question. If a 1.5 m tall girl stands at a distance of 3 m from a lamp post and casts a shadow of length 4.5 m on the ground, then the height of the lamp post is:
(a) 1.5 m
(b) 2 m
(c) 2.5 m
(d) 2.8 m
Answer: (c) 2.5 m
Explanation :
Let AB be the lamp-post; CD be the girl and DE be her shadow.
We have,
CD = 1.5 m, AD = 3 m, DE = 4.5 m
Let \( \angle E = q \)
In DCDE,
\( \tan q = \frac{CD}{DE} \)
\( \Rightarrow \tan q = 1.54.5 \)
\( \Rightarrow \tan \theta = 13\dots \text{(i)} \)
Now, in DABE,
\( \tan q = \frac{AB}{AE} \)
\( \Rightarrow 13 = ABAD + DE [Using (i)] \)
\( \Rightarrow 13 = AB3 + 4.5 \)
\( \Rightarrow AB = 7.53 \)
\( \Rightarrow AB = 2.5 \text{ m} \)

Question. The height of the vertical pole is \( \sqrt{3} \) times the length of its shadow on the ground, then angle of elevation of the sun at that time is:
(a) 30°
(b) 60°
(c) 45°
(d) 75°
Answer: (b) 60°
Explanation :
Let the angle of elevation of the sun be q.
Suppose AB is the height of the pole and BC is the length of its shadow.
It is given that,
AB = \( \sqrt{3}BC \)
\( \tan q = \frac{AB}{BC} \)
\( \Rightarrow \tan q = \frac{\sqrt{3}BC}{BC} = \sqrt{3} \)
\( \Rightarrow \tan q = \tan 60^\circ \)
\( \Rightarrow q = 60^\circ \)
Thus, the angle of elevation of the sun is 60°.

Question. From a light house the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the light house is h metres, the distance between the ships is :
(a) \( (\sqrt{3}+1)h \) metres
(b) \( (\sqrt{3}-1)h \) metres
(c) \( \sqrt{3}h \) metres
(d) \( 1+(1+1\sqrt{3})h \) metres
Answer: (a) \( (\sqrt{3}+1)h \) metres
Explanation :
Let AB be light house and P and Q are two ships on its opposite sides which form angle of elevation of A as 45° and 30° respectively
AB = \( h \).
Let PB = \( x \) and QB = \( y \)
Now in right DAPB,
\( \tan q = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{PB} \)
\( \Rightarrow \tan 45^\circ = \frac{h}{x} \)
\( \Rightarrow 1 = \frac{h}{x} \)
\( \Rightarrow x = h \dots \text{(i)} \)
Similarly in right DAQB,
\( \tan 30^\circ = \frac{AB}{QB} = y \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{y} \)
\( \Rightarrow y = \sqrt{3}h \dots \text{(ii)} \)
Adding (i) and (ii)
PQ = \( x + y = h + \sqrt{3}h \)
= \( (\sqrt{3}+1)h \) meters

Question. A tower subtends an angle of 30° at a point on the same level as its foot. At a second point ‘h’ metres above the first, the depression of the foot of the tower is 60°. The height of the tower is:
(a) \( 2h^2 \text{ m} \)
(b) \( \sqrt{3}h \text{ m} \)
(c) \( h \text{ m} \)
(d) \( h\sqrt{3}\text{m} \)
Answer: (c) \( h \text{ m} \)
Explanation :
Let AB be the tower and C is a point on the same level as its foot such that \( \angle ACB = 30^\circ \)
The given situation can be represented as the above figure.
Here D is a point \( h \text{ m} \) above the point C.
In DBCD,
\( \Rightarrow \tan B = \frac{CB}{BC} \)
\( \Rightarrow \tan 60^\circ = \frac{h}{BC} \)
\( \Rightarrow \sqrt{3} = \frac{h}{BC} \)
\( \Rightarrow CB = \frac{h}{\sqrt{3}} \dots \text{(i)} \)
Again in DABC,
\( \tan C = \frac{AB}{CB} \)
\( \Rightarrow \tan 30^\circ = \frac{AB}{(h/\sqrt{3})} \) [Using (i)]
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB\sqrt{3}}{h} \)
\( \Rightarrow AB = h \)

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 there is no slack in the string, the angle of elevation of the kite at the ground is:
(a) 45°
(b) 30°
(c) 60°
(d) 90°
Answer: (b) 30°
Explanation :
Let AB be the height of the kite above the ground.
Now, AB = 30m
Let the length of the string = AC = 60 m
Now, \( \sin \theta = \frac{AB}{AC} \)
\( \Rightarrow \sin q = \frac{30}{60} \)
\( \Rightarrow \sin q = \frac{1}{2} \Rightarrow \sin q = \sin 30^\circ \)
\( \Rightarrow q = 30^\circ \)

Question. The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30°. The distance of the car from the base of the tower (in m.) is :
(a) \( 25\sqrt{3} \)
(b) \( 50\sqrt{3} \)
(c) \( 75\sqrt{3} \)
(d) 150
Answer: (a) \( 25\sqrt{3} \)
Explanation :
Given that the angle of depression is 30°, then angle of elevation is (90° – 30° = 60°).
We have,
\( \tan 60^\circ = \frac{\text{height of tower}}{\text{distance of the car from the base of the tower}} \)
\( \Rightarrow \sqrt{3} = \frac{75}{\text{Distance of the car from the base of the tower}} \)
\( \Rightarrow \text{distance of the car from the base of the tower} \)
\( = \frac{75}{\sqrt{3}} = 25\sqrt{3} \)

Question. The length of the shadow of a tower standing on level ground is found to be \( 2x \) metres longer when the sun's elevation is 30° than when it was 45°. The height of the tower is :
(a) \( (2\sqrt{3})x \) m
(b) \( (3\sqrt{2}x) \) m
(c) \( (\sqrt{3}-1)x \) m
(d) \( (\sqrt{3}+1)x \) m
Answer: (d) \( (\sqrt{3}+1)x \) m
Explanation :
Let CD = \( h \) be the height of the tower.
We have, AB = \( 2x \), \( \angle DAC = 30^\circ \) and \( \angle DBC = 45^\circ \)
In \( \triangle BCD \),
\( \tan 45^\circ = \frac{CD}{BC} \)
\( \Rightarrow 1 = \frac{h}{BC} \)
\( \Rightarrow BC = h \)
Now, in \( \triangle ACD \),
\( \tan 30^\circ = \frac{CD}{AC} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BC} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{2x + h} \)
\( \Rightarrow 2x + h = h\sqrt{3} \)
\( \Rightarrow h\sqrt{3} - h = 2x \)
\( \Rightarrow h(\sqrt{3}-1) = 2x \)
\( \Rightarrow h = \frac{2x}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \)
\( \Rightarrow h = \frac{2x(\sqrt{3}+1)}{2} \)
\( \Rightarrow h = x(\sqrt{3}+1) \text{ m} \)

Question. A ladder makes an angle of 60° with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder is:
(a) \( 4/\sqrt{3} \) m
(b) \( 4\sqrt{3} \) m
(c) \( 2\sqrt{2} \) m
(d) 4 m
Answer: (d) 4 m
Explanation :
Let AB be the wall and AC be the ladder
We have, BC = 2m and \( \angle ACB = 60^\circ \)
In \( \triangle ABC \),
\( \cos 60^\circ = \frac{BC}{AC} \)
\( \Rightarrow \frac{1}{2} = \frac{2}{AC} \)
\( \Rightarrow AC = 4\text{m} \)

Question. From the top of a cliff 20 m 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) 20 m
(b) 40 m
(c) 60 m
(d) 80 m
Answer: (b) 40 m
Explanation :
Let AB be the cliff and CD be the tower.
We have, AB = 20 m
Also, CE = AB = 20 m
Let \( \angle ACB = \angle CAE = \angle DAE = q \)
In \( \triangle ABC \),
\( \tan q = \frac{AB}{BC} \)
\( \Rightarrow \tan q = \frac{20}{BC} \)
\( \Rightarrow \tan q = \frac{20}{AE} \) (As, BC = AE)
\( \Rightarrow AE = \frac{20}{\tan q} \dots \text{(i)} \)
Also, in \( \triangle ADE \),
\( \tan q = \frac{DE}{AE} \)
\( \Rightarrow \tan q = \frac{DE}{(20/\tan q)} \) [Using (i)]
\( \Rightarrow DE = 20 \times \frac{\tan q}{\tan q} \)
\( \Rightarrow DE = 20\text{m} \)
Now, CD = DE + CE = 20 + 20 = 40 m

Question. The tops of two towers of heights \( x \) and \( y \), standing on a level ground subtend angles of 30° and 60°, respectively at the centre of the line joining their feet. Then, \( x : y \) is:
(a) 1 : 2
(b) 2 : 1
(c) 1 : 3
(d) 3 : 1
Answer: (c) 1 : 3
Explanation :
Let AB and CD be the two towers such that AB = \( x \) and CD = \( y \).
We have, \( \angle AEB = 30^\circ \), \( \angle CED = 60^\circ \) and BE = DE
In \( \triangle ABE \),
\( \tan 30^\circ = \frac{AB}{BE} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{BE} \)
\( \Rightarrow BE = x\sqrt{3} \)
Also, in \( \triangle CDE \),
\( \tan 60^\circ = \frac{CD}{DE} \)
\( \Rightarrow \sqrt{3} = \frac{y}{DE} \)
\( \Rightarrow DE = \frac{y}{\sqrt{3}} \)
As, BE = DE
\( \Rightarrow x\sqrt{3} = \frac{y}{\sqrt{3}} \)
\( \Rightarrow x : y = 1 : 3 \)

Question. In a rectangle, the angle between a diagonal and a side is 30° and the length of this diagonal is 8 cm. The area of the rectangle is:
(a) 16 cm²
(b) \( 16\sqrt{2} \text{ cm}^2 \)
(c) \( 16\sqrt{3} \text{ cm}^2 \)
(d) \( 8\sqrt{3} \text{ cm}^2 \)
Answer: (c) \( 16\sqrt{3} \text{ cm}^2 \)
Explanation :
Let ABCD be the rectangle in which \( \angle BAC = 30^\circ \) and AC = 8 cm.
In \( \triangle ABC \),
We have, \( \frac{AB}{AC} = \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \Rightarrow \frac{AB}{8} = \frac{\sqrt{3}}{2} \)
\( \Rightarrow AB = 4\sqrt{3} \) cm
and \( BC = \frac{8}{2} = 4 \) cm
\( \therefore \) Area of the rectangle = \( (AB \times BC) = (4\sqrt{3} \times 4) \)
= \( 16\sqrt{3} \text{ cm}^2 \)

Question. If a 1.5m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is :
(a) 1.5 m
(b) 2 m
(c) 2.5 m
(d) 2.8 m
Answer: (c) 2.5 m
Explanation :
Let AB is girls and CD is lamp-post. AB = 1.5 which casts her shadow EB.
\( \Rightarrow EB = 4.5 \text{ m}, BD = 3 \text{ m} \)
Now in \( \triangle AEB \),
\( \tan q = \frac{AB}{BE} = \frac{1.5}{4.5} = \frac{1}{3} \)
and in \( \triangle CED \),
\( \tan q = \frac{CD}{ED} \)
\( \Rightarrow \frac{1}{3} = \frac{h}{4.5 + 3} \)
\( \Rightarrow \frac{1}{3} = \frac{h}{7.5} \)
\( \Rightarrow h = \frac{7.5}{3} = 2.5 \text{ m} \)
\( \therefore \) Height of lamp-post = 2.5 m

Question. The tops of two poles of height 20m and 14m are connected by a wire. If the wire makes an angle of 30° with horizontal, then the length of the wire is :
(a) 12 m
(b) 10 m
(c) 8 m
(d) 6 m
Answer: (a) 12 m
Explanation :
Let AB and CD be two poles AB = 20 m, CD = 14 m. A and C are joined by a wire CE || DB and angle of elevation of A is 30°.
Let CE = DB = \( x \) and AC = \( l \)
Now, AE = AB – EB = AB – CD = 20 – 14 = 6m
Now in right \( \triangle ACE \),
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AE}{AC} \)
\( \Rightarrow \sin 30^\circ = \frac{6}{AC} \)
\( \Rightarrow \frac{1}{2} = \frac{6}{AC} \)
\( \Rightarrow AC = 2 \times 6 = 12 \text{ m} \)
\( \Rightarrow \text{Length of AC} = 12 \text{ m} \)

Question. From a light house the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the light house is \( h \) metres, the distance between the ships is:
(a) \( (\sqrt{3}+1)h \) metres
(b) \( (\sqrt{3}-1)h \) metres
(c) \( \sqrt{3}h \) metres
(d) \( 1+(1+1\sqrt{3})h \) metres
Answer: (a) \( (\sqrt{3}+1)h \) metres
Explanation :
Let AB be light house and P and Q are two ships on its opposite sides which form angle of elevation of A as 45° and 30° respectively. AB = \( h \).
Let PB = \( x \) and QB = \( y \)
Now in right \( \triangle APB \),
\( \tan q = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{PB} \)
\( \Rightarrow \tan 45^\circ = \frac{h}{x} \)
\( \Rightarrow 1 = \frac{h}{x} \)
\( \Rightarrow x = h \dots \text{(i)} \)
Similarly in right \( \triangle AQB \),
\( \tan 30^\circ = \frac{AB}{QB} = \frac{h}{y} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{y} \)
\( \Rightarrow y = \sqrt{3}h \dots \text{(ii)} \)
Adding (i) and (ii)
PQ = \( x + y = h + \sqrt{3}h \)
= \( (\sqrt{3}+1)h \text{ metres} \)

Question. If \( x = r \sin q \cos f \), \( y = r \sin q \sin f \) and \( z = r \cos q \), then:
(a) \( x^2 + y^2 + z^2 = r^2 \)
(b) \( x^2 + y^2 - z^2 = r^2 \)
(c) \( x^2 - y^2 + z^2 = r^2 \)
(d) \( z^2 + y^2 - x^2 = r^2 \)
Answer: (a) \( x^2 + y^2 + z^2 = r^2 \)
Explanation :
\( x = r \sin q \cos f \)
\( y = r \sin q \sin f \)
\( z = r \cos q \)
Squaring and adding these equations, we get
\( x^2 + y^2 + z^2 = (r \sin q \cos f)^2 + (r \sin q \sin f)^2 + (r \cos q)^2 \)
\( \Rightarrow x^2 + y^2 + z^2 = r^2 \sin^2 q \cos^2 f + r^2 \sin^2 q \sin^2 f + r^2 \cos^2 q \)
\( \Rightarrow x^2 + y^2 + z^2 = r^2 \sin^2 q (\cos^2 f + \sin^2 f) + r^2 \cos^2 q \)
\( \Rightarrow x^2 + y^2 + z^2 = r^2 \sin^2 q (1) + r^2 \cos^2 q \)
\( \Rightarrow x^2 + y^2 + z^2 = r^2 (\sin^2 q + \cos^2 q) \)
\( \Rightarrow x^2 + y^2 + z^2 = r^2(1) \)
\( \Rightarrow x^2 + y^2 + z^2 = r^2 \)