OP Malhotra Class 10 Maths Solutions Chapter 17 Heights and Distances Ex 17

The availability of Class 10 ICSE Maths Solutions S Chand Chapter 17 Heights and Distances Ex 17 encourages students to tackle difficult exercises.

S Chand Class 10 ICSE Maths Solutions Chapter 17 Heights and Distances Ex 17

Question 1.
The angle of elevation of the top of a tower from a point at a distance of 100 metres from its foot on a horizontal plane is foundTo be 60°. Find the height of the tower.
Solution:
Let AB be the tower and O is the point from which the angle of elevation = 60° and distance AB = 100 m
Let AB = h m

∴ tan θ = \(\frac { AB }{ OB }\) ⇒ tan 60° = \(\frac { h }{ 100 }\)
⇒ h = 100 tan 60° = 100 × √3 = 100 (1.732)
= 173.2 m
Height of tower = 173.2 m

Question 2.
A vertical flagstaff stands on a horizontal plane. From a point distant 150 metres from its foot the angle of elevation of its top is found to be 30°; find the height of the flagstaff.
Solution:
Let AB be the flagstaff and a point O is given from where the top of the flagstaff makes an angle of elevation = 30° and distance from the foot B 150 m
LetAB be h

∴ tan θ = \(\frac { AB }{ OB }\)
⇒ tan 30° = \(\frac { h }{ 150 }\)
⇒ h = 150 × tan 30°
⇒ h = 150 × \(\frac{1}{\sqrt{3}}\) = \(\frac{150 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{150 \sqrt{3}}{3}\)m
= 50√3 = 50(1.732) = 86.600 = 86.6 m
∴ Height of flagstaff = 86.6 m

Question 3.
The string of a kite is 150 long and it makes an angle of 60° with the horizontal. Find the height of the kite from the ground.
Solution:
Let AB be the height of kite A from the ground and kite makes an angle of elevation with the horizontal ground of 60°
AC is the string 150 m long

Let AB = h m, Now
sin θ = \(\frac { AB }{ AC }\) ⇒ sin 60° = \(\frac { h }{ 150 }\)
⇒ h = 150 × sin 60° = 150 × \(\frac{\sqrt{3}}{2}\)
= 75√3 m = 75 × 1.732 m = 129.9 m
∴ Height of the kite = 129.9 m

Question 4.
If the shadow of a tower is 30 metres when the sun’s elevation is 30°, what is the length of the shadow when the sun’s elevation is 60° ?

Solution:
In (i), angle of sun’s elevation is 30° and length of shadow of the tower HB = AB, = 30 m
Let height of the sun be h
∴ tan θ = \(\frac { HB }{ AB }\) ⇒ tan 30° = \(\frac { h }{ 30 }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ 30 }\) ⇒ h = \(\frac{30}{\sqrt{3}}\) = \(\frac{30 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
⇒ h = \(\frac{30 \sqrt{3}}{3}\) = 10√3
Now when the elevation of sun be 60° Let the shadow of the tower =x
∴ 60° = \(\frac{\mathrm{BC}}{\mathrm{DB}}\) = \(\frac { h }{ x }\)
√3 = \(\frac{10 \sqrt{3}}{x}\) ⇒ x√3 = 10√3
x = \(\frac{10 \sqrt{3}}{\sqrt{3}}\) = 10
∴ Length of shadow = 10 m

Question 5.
The angle of elevation of the top of a tower (which is yet incomplete) at a point 120 m from its base is 45°. How much higher should it be raised so that the elevation at the same point may become 60° ?
Solution:
Let BC be the incomplete tower and DB be the complete
Let BC = x m and CD = h
At the top C,-the angle of elevation is 45° and at the top D, the angle of elevation is 60°

Now in right △ABC,
tan θ = \(\frac { CB }{ AB }\) ⇒ tan 45° = \(\frac { x }{ 120 }\)
⇒ 1 = \(\frac { x }{ 120 }\) ⇒ x = 120 m
In right △ABD,
tan 60° = \(\frac { DB }{ AB }\) ⇒ √3 = \(\frac{x+h}{120}\)
⇒ 120 × √3 = 120 + h ⇒ h = 120√3 – 120
= 120 (√3 – 1) = 120 (1.732 – 1)
= 120 × 0.732 = 87.84 m
∴ Tower should be raised more = 87.84 m

Question 6.
Find the angle of depression from the top of a tower 140 m high of an object on the ground at a distance of 200 m from the foot of the tower.
Solution:
Let AB be the tower and Pis a point on the ground such that PB = 200m and AB = 140m

Let θ be the angle of depression of P from A, the top of the tower
∴ ∠APB = θ (alternate angles)
∴ tan θ = \(\frac { AB }{ PB }\) = \(\frac { 140 }{ 200 }\) = \(\frac { 7 }{ 10 }\) = 0.7
From the table of tan θ,
tan 35° = 70021 wbich is nearest to 0.7
∴ Angle of depression = θ = 35°

Question 7.
A vertical tower Ia 20 m high. A man at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the loot of the tower?
Solution:
Let AB be the tower and M is a man standing at some distance from the foot of the tower B
Let MB = x m and cos θ = 0.53

∴ cos θ = \(\frac { MB }{ AM }\) = \(\frac{x}{\sqrt{20^2+x^2}}\)
⇒ 0.53 = \(\frac{x}{\sqrt{(20)^2+x^2}}\)
Squaring both sides,
(0.53)² = \(\frac{x^2}{400+x^2}\) ⇒ \(\frac { 2809 }{ 10000 }\) = \(\frac{x^2}{400+x^2}\)
⇒ 10000x² = 2809 × 400 + 2809x²
⇒ 10000x² – 2809x² = 1123600
⇒ 7191 x² = 11223600
⇒ x² = \(\frac { 1123600 }{ 7191 }\) = 156.25
∴ x = \(\sqrt{156.25}\) = 12.5
∴ The man is standing at a distance of 12.5 m from the foot of the tower

Question 8.
In the figure, it is given that AB is perp. to BD and is of length x metres. DC = 30m, ∠ADB = 30° and ∠ACB = 45° Without using tables, find x.

Solution:
In the figure, AB = x, DC = 30 m
∠ADB = 30°, ∠ACB = 45°
Let BC = y m
Now in right △ABC
tan θ = \(\frac { AB }{ BC }\) ⇒ tan 45° = \(\frac { x }{ y }\) ⇒ 1 = \(\frac { x }{ y }\)
⇒ x = y
Again in right △ABD, (∵ tan 45° = 1)
tan 30° = \(\frac { AB }{ BD }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{x}{y+30}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{x}{x+30}\) (∵ y = x)
⇒ √3 x = x + 30 ⇒ √3 x – x = 30
⇒ (1.732 – 1) x = 30 ⇒ 0.732x = 30
⇒ x = \(\frac{30}{0.732}\) = 40.98 m
∴ x = 40.98 m

Question 9.
In the figure, from a boat 200 in away from a erticaI cliff the angles of elevation of the top and the foot of a vertical concrete pillar at the edge of the cliff are 21° and 18° 30′. Calculate :
(i) the height of the cliff;
(ii) the height of the pillar.

Solution:
Let B be the boat which is 200 m away from a vertical cliff CA
PC is the pillar at the edge of cliff
Angles of elevation from B to the top of cliff C is 18° 3o’ and to the top of the pillar P is 21°
Let height of cliff CA = h m and pillar PC = x m
Now in right △CAB,
tan θ = \(\frac { CA }{ AB }\) ⇒ tan 18° 36′ = \(\frac { h }{ 200 }\)
⇒ 0.33460 = \(\frac { h }{ 200 }\) ⇒ h = 200 × 0.33460
⇒ h = 66.920 m = 66.92 m
Again in right △PAB.
tan 21° = \(\frac { PA }{ AB }\) ⇒ 0.38386 = 76.77200
⇒ x + 66.920 = 200 × 0.38386 = 76.77200
⇒ x = 76.772 – 66.920 = 9.852 = 9.85 m
(i) ∴ Height of cliff 66.92 m and
(ii) Height of pillar = 9.85 m

Question 10.
In the figure, the top of a tower is observed from two points on the same horizontal line through a point on the base of the tower. If the angles of elevation at the two points be 300 and 45°, and the distance between them is 20 m, find the height of the tower.

Solution:
Let TS be the tower and A and B are two points on the ground on the same line and distance between them is 20 in i.e.,
AB = 20 m
Angles of elevation from A is 30° and from B
is 45° to the top of the tower
Let height of the tower TS = h and BS = x
Now in right △TSB,
tan θ = \(\frac { TS }{ BS }\) ⇒ tan 45° = \(\frac { h }{ x }\)
⇒ 1 = \(\frac { h }{ x }\) ⇒ x = h
Similarly ¡n right △TSA,
tan 30° = \(\frac{h}{x+20}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+20}\) (∵ x = h)
⇒ h √3 = h + 20 ⇒ √3 h – h = 20
⇒ (1.732 – 1) h = 20 ⇒ 0.732h = 20
⇒ h = \(\frac { 20 }{ 0.732 }\) = \(\frac{20 \times 1000}{732}\) = 27.32
∴ Height of the tower = 27.32 m

Question 11.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank ¡s 60°. When he moves 40 m away
from the bank, he finds the angles of elevation to be 30°. Find (i) the height of the tree, correct to 2 decimal places ;
(ii) the width of river.
Solution:
Le TR be the tree on the bank of a river BR and A is another position of the rnar which is 40 m away from the first position B
∴ AB = 40m
and angles of elevation to B and A to the top of the tree are 60° and 30° respectively

Let height of tree TR = h m and width of
river BR = x m
Now in right △TRA
tan 30° = \(\frac { TR }{ RA }\) ⇒ tan 60° = \(\frac { h }{ x }\) ⇒ √3 = \(\frac { h }{ x }\)
∴ h = √3 x or x = \(\frac{h}{\sqrt{3}}\) = \(\frac{h \sqrt{3}}{3}\)
Similarly in right △TRA
tan 30° = \(\frac { TR }{ RA }\) ⇒ \(\frac{1}{\sqrt{3}}\) × 120 = \(\frac { h }{ x+40 }\)
⇒ √3 h = x + 40 ⇒ √3 h = \(\frac{h \sqrt{3}}{3}\) + 40
⇒ √3 h – \(\frac{\sqrt{3}}{3}\) h = 40 ⇒ \(\frac{(3 \sqrt{3}-\sqrt{3})}{3}\)h = 40
⇒ 2√3 h = 120 ⇒ h = \(\frac{1}{2 \sqrt{3}}\) = \(\frac{60}{\sqrt{3}}\)
⇒ h = \(\frac{60 \times \sqrt{3}}{\sqrt{3}+\sqrt{3}}\) = 20
(i) ∴ Height of tree 34.64 m
(ii) and width of river 20 m

Question 12.
In the figure. the a.gle of elevation of the top P of a vertical tower PQ from a point X Is 60°; at a point Y, 40m vertically above X, the angle of elevatIon Li 45°.
(i) Find the height of the tower PQ;
(ii) Find the distance XQ.
(Give yoar answers to the nearest metre)

Solution:
PQ is the tower, from a point X which is at some distance from Q the angle of elevation of the top P of the tower is 60 and from another point Y. which is 40 m above it, the angle of elevation of P is 45°
let height PQ = h and distance of QX = x
From Y, draw YZ || XQ, then
YZ = XQ = x, ZQ = YX = 40 m and PZ = (h-40) m
Now in right △PQX,

tan θ = \(\frac { PQ }{ QX }\)
⇒ tan 60° = \(\frac { h }{ x }\)
⇒ √3 = \(\frac { h }{ x }\)
⇒ h = √3 x or x = \(\frac{h}{\sqrt{3}}\) ….(i)
Similarly in right △PZY,
tan 45° = \(\frac { PZ }{ YZ }\) = \(\frac{h-40}{x}\)
⇒ 1 = \(\frac{h-40}{x}\) ⇒ x = h – 40 ….(ii)
\(\frac{h}{\sqrt{3}}\) = h – 40 {From (i)}
h = √3 h – 40√3
√3h – h = 40√3 ⇒ h (√3 – 1) = 40(1.732)
⇒ h(1.732 – 1) = 40 × 1.732
⇒ 0.732h = 40 × 1.732
h = \(\frac{40 \times 1.732}{0.732}\) = 94.64 m
= 95 m (in nearest metre)
and x = \(\frac{h}{\sqrt{3}}\) = \(\frac{94.64}{1.732}\) = 54.64 m
= 55 m (in nearest metre)
Hence height of tower PQ = 95 m
and distance XQ = 55 m

Question 13.
A guard observes an enemy boat, from an observation, tower at . height of 180 m above sea-level, to be at an angle 0f depression of 29°
(i) Calculate to the nearest metre. the distance of the boat from the tool of the observation tower.
(ii) After some time it is observed that the boat is 200 m from the foot of the observation tower. Calculate the new angle of depression.
Solution:
Let TR be the observation tower, B is the position of a boat which makes 290 of angle of depression, A is the second position of the boat which makes angle θ of angle of depression when the boat is 200 m from the tower

∴ TR = 180 m, AR = 200 m
Let BR = x m, then BA = (x – 200) m
∵ XT || BR
∴ ∠B = 29° and ∠A = θ (alternate angles)

(i) Now in right △TRB.
tan θ = \(\frac { TR }{ RB }\) ⇒ tan 29° = \(\frac{180^{\circ}}{x}\)
⇒ 0.55431 = \(\frac { 180 }{ x }\) ⇒ x = \(\frac { 180 }{ 0.55431 }\) = 324.728
⇒ x = 325 (nearest to metre)

(ii) Similarly in right△TRA
tan θ = \(\frac { TR }{ RA }\) = \(\frac { 180 }{ 200 }\) = \(\frac { 9 }{ 10 }\) = 0.9
From the tables, we see thai tan θ = 0.90040
which is nearest in 0.9
∴ θ = 42°

Question 14.
In the figure, the shadow of a vertical tower on level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. using the given figure, find the height or the tower, correct to one place of decimals.

Solution:
Let TR be the tower and RA ¡s its shadow at an angle of 45° of elevation and RB is the shadow at 30° of elevation such that AB = 10 m
Let h be the height of the tower TR and shadow RA = x m
Now in right △TRA,
tan θ = \(\frac { TR }{ RA }\) ⇒ tan 45° = \(\frac { h }{ x }\)
⇒ 1 = \(\frac { h }{ x }\) ⇒ h = x
and in right △TRB,
tan 30° = \(\frac { TR }{ RB }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+10}\)
⇒ √3 h = x + 10
⇒ √3 h = h + 10 ⇒ √3 h – h = 10 [from (i)]
⇒ h(√3 – 1) = 10 ⇒ h (1.732 – 1) = 10
⇒ 0.732h = 10 ⇒ h = \(\frac { 10 }{ 0.732 }\) = 13.66
∴ Height of tower = 13.66 m = 13.7 m

Question 15.
A player sifting on the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground as 60°. Find the distance between the foot of the tower and the ball. (Take √3 = 1.732)
Solution:
P is a player sitting on the top of a tower PQ and B is the ball lying on the ground and anlge of depression from P to ball B is 60° Height of the tower PQ = 20 m

Now, we have to find the distance BQ
Let BQ = x, then
In △PBQ, ∠Q = 90°
tan θ = \(\frac { P }{ B }\) = \(\frac { PQ }{ BQ }\) ⇒ tan 60° = \(\frac { 20 }{ x }\) ⇒ √3 = \(\frac { 20 }{ x }\)
⇒ x = \(\frac{20}{\sqrt{3}}\) = \(\frac{20 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{20(1.732)}{3}\)
= \(\frac{34.640}{3}\) = 11.546 = 11.55 m
∴ Distance between the foot of tower and ball = 11.55 m

Question 16.
The angle of elevation of the top of a tower from two points at distances a and b metres from the base and in the same straight line with it are complementary.
Prove that the height of the tower is √ab metres.

Solution:
In the figure. let the height of tower PQ = h
AQ = a and BQ = b
∠A = θ, then ∠B = 90° – θ (∵ Angles A and B are complementary)
Now in right △PQA.
tan θ = \(\frac { PQ }{ QA }\) = \(\frac { h }{ a }\) ⇒ h = a tan θ …(i)
and similarly in right △PQA,
tan (90° – θ) = \(\frac { PQ }{ QB }\)
⇒ cot θ = \(\frac { h }{ b }\) ⇒ h = b cot θ …(ii) {∵ tan (90° – θ) = cot θ}
Multiplying (i) and (ii)
h × h = a tan θ × b cot θ
⇒ h² = ab × 1 = ab (∵ tan θ cot θ = 1)
⇒ h = √ab
Hence height of the tower = √ab Hence proved.

Question 17.
The length of a siring between a kite and a point on the ground is 90 metres. If the string makes an angle θ with the level ground such that tan θ = \(\frac { 15 }{ 8 }\), how high is the kite ? Assume, there is so slack in the string.
Solution:
Let K be the kite and AK be the string which makes an angle θ with the ground
i.e. ∠A = θ, AK = 90 m and tan θ = \(\frac { 15 }{ 8 }\)
Let h be the height of kite i.e. KT = h
Now in right △KTA

tan θ = \(\frac { KT }{ TA }\) = \(\frac { h }{ x }\)
∴ \(\frac { h }{ x }\) = \(\frac { 15 }{ 8 }\)
⇒ h = 15, x = 8
∴ AK = \(\sqrt{h^2+x^2}\) = \(\sqrt{15^2+8^2}\) = \(\sqrt{225+64}\) = \(\sqrt{289}\) = 17
∴ sin θ = \(\frac{h}{\mathrm{AK}}\) = \(\frac{15}{17}\)
If AK = 90, then h = \(\frac{15}{17}\) × 90 = \(\frac{1350}{17}\)
⇒ h = 79.4
Hence height of the kite = 79.4 m

Question 18.
An observer in figure, 1\(\frac { 1 }{ 2 }\) m tall, is 28.5 m away from a tower 30 m high. Determine the angle of elevation from his eye to the top of the tower.

Solution:
Let AB be the tower and CD is a man. The height of tower = 30 m and man CD
= 1\(\frac { 1 }{ 2 }\) m = 1.5 m
Who is 28.5 m away from the foot of the tower then CA = 28.5 m
Draw DE || CA, then
EA = 1.5 m,EB = 30 – 1.5 = 28.5 m and DE = CA = 28.5 m
and let angle of elevation from the eye of the man of the top B of the tower be θ
Now in △BED,
tan θ = \(\frac { BE }{ DE }\) = \(\frac { 28.5 }{ 28.5 }\) = 1
From the table, we see that
tan 45° = 1
∴ θ = 45°

Question 19.
Two men are on diametrically opposite side of a tower. They measure the angles of elevation of the top of the tower as 20° and 24° respectively. If the height of the tower is 40 m, find the distance between them.
Solution:
Let TR be the tower which is 40 m high and A and B are two men who are diametrically opposite sides of the tower making angles of elevation as 20° and 24° respectively with the top of the tower
Let AR = x and BR = y

Now in right △TRA
tan θ = \(\frac { TR }{ AR }\) ⇒ tan 20° = \(\frac { 40 }{ x }\)
⇒ 0.36397 = \(\frac { 40 }{ x }\) (From the tangents tables)
∴ x = \(\frac { 40 }{ 0.36397 }\) = 109.89
Similarly in right △TRB,
tan 24° = \(\frac { 40 }{ y }\) ⇒ 0.44523 = \(\frac { 40 }{ y }\)
⇒ y = \(\frac { 40 }{ 0.44523 }\) = 89.84
Distance between two men : x + y
= 109.89 + 89.84 = 199.73 m

Question 20.
The horizontal distance between the two towers is 60 m and the angular depression of the top of the first, as seen from the top of the second, which is 150 m high, is 30°; find the height of the first.
Solution:
Let AB and CD be the two towers which are
60 m apart i.e. BD = 60 m, CD = 150 m
Let AB = h, draw AE||BD
Angle of depression from C to A is 30°
∵ CX || AE || BD
∴ CAE = 30° (alternate angles)
and AE = 60 m,ED = h and CE = 150 – h

Now in right △CAE,
tan θ = \(\frac { CE }{ AE }\) ⇒ tan 30° = \(\frac{150-h}{60}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{150-h}{60}\)
150 – h = \(\frac{60}{\sqrt{3}}\) ⇒ h = 150 – \(\frac{60}{\sqrt{3}}\)
⇒ h = 150 – \(\frac{60 \times \sqrt{3}}{\sqrt{3}+\sqrt{3}}\) = 150 – \(\frac{60 \times \sqrt{3}}{3}\)
= 150 – 20√3
= 150 – 20 (1.732) = 150 – 34.640 = 115.36 m
∴ Height of first tower AB = 115.36 m

Question 21.
From the top of a cliff, 150 metres high, the angle of depression of the top and bottom of a tower are observed to be 30° and 60°, find the height of the tower.
Solution:
Let AB is cliff and CD is the tower
The angle of depression from A to the top C and bottom D of the tower are 30° and 60° respectively

∵ AX || CE || DB
∴ ∠ACE = 30°
and ∠ADB = 60° (altcrnate angles)
Now AB = 200 m
Let CD = h, then EB = h and CE = DB = x (suppose)
∴ AE = 200 – h
Now in right △ABD,
tan θ = \(\frac { AB }{ BD }\)
⇒ tan 60° = \(\frac { 200 }{ x }\)
⇒ √3 = \(\frac { 200 }{ x }\)
⇒ √3 x = 200
⇒ x = \(\frac{200}{\sqrt{3}}\)
Similarly in right △AEC,
tan 30° = \(\frac { AE }{ EC }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{200-h}{x}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{200-h}{\frac{200}{\sqrt{3}}}\)
= \(\frac{1}{\sqrt{3}}\) × \(\frac{200}{\sqrt{3}}\) = 200 – h
⇒ \(\frac{200}{3}\) = 200 – h
⇒ h = 200 – \(\frac{200}{3}\) = \(\frac{400}{3}\)
⇒ h = 133.33 m
∴ Height of tower = 133.33 m

Question 22.
From a point P on level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, how far is P from the foot of the tower?
Solution:
Let AB be tower which is 100 m long P is a point on the ground such that angle of elevation from A is 30°

Let PB = x m
Now in right △ABP,
tan θ = \(\frac { AB }{ PB }\) ⇒ tan 30° = \(\frac { 100 }{ x }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 100 }{ x }\) ⇒ x = 100 × √3 = 100 × 1.732
x = 173.2
∴ Distance between P and the foot of tower = 173.2 m

Question 23.
From the top of a building AB, 60 metres high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find
(i) the horizontal distance between AB and CD;
(ii) the height of the lamp post CD.

Solution:
AB and CD are the lamp posts
AB = 60 m
and let CD = h
Angles of depression from A to the top and bottom of CD are 30° and 60°

∴ Draw CE || DB
∴ ∠ACE = 30° and
∠ADB = 60° (alternate anSIes)
∴ EB = CD = h, then AE = (60 – h) m
Let CE = BD = x
Now in right △ABD
tan θ = \(\frac { AB }{ BD }\) ⇒ tan 60° = \(\frac { 60 }{ x }\)
⇒ √3 = \(\frac { 60 }{ x }\) ⇒ x = \(\frac{60}{\sqrt{3}}\) = \(\frac{60 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{60 \sqrt{3}}{3}\)
x = 20√3 m
Similarly in right △AEC,
tan 30° = \(\frac { AE }{ EC }\) = \(\frac{60-h}{x}\) = \(\frac{60-h}{20 \sqrt{3}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{60-h}{20 \sqrt{3}}\) ⇒ \(\frac{20 \sqrt{3}}{\sqrt{3}}\) = 60 – h
⇒ 20 = 60 – h ⇒ h = 60 – 20 = 40

(i) ∴ Distance between AB and
CD = BD = x = 20√3 m
= 20 (1.732) = 34.640 = 34.64 m

(ii) Height of CD = 40 m

Question 24.
An aeroplane when 34000 in high passes vertically above another aeroplane at an instance when their angles of elevation at the same observation pout are 60° and 45°. How many metres higher is the one than the other?

Solution:
An aeroplane A is flying ax the height of 3000 m above the ground and passes vertically above another aeroplane C
Let distance between AC = h
Then CD = 3000 – h
The angle of elevation from B to C and A are 45° and 60°. respectively
Let BD = x

Now in right △ABD,
tan θ = \(\frac { AD }{ BD }\)
⇒ tan 60° = \(\frac { 3000 }{ x }\)
⇒ √3 = \(\frac { 3000 }{ x }\)
⇒ x = \(\frac{3000}{\sqrt{3}}\) m
Similarly in right △CBD,
tan 45° = \(\frac { CD }{ BD }\) = \(\frac{3000-h}{x}\)
⇒ 1 = \(\frac{3000-h}{x}\)
⇒ x = 3000 – h
⇒ h = 3000 – \(\frac{3000 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
= 3000 – \(\frac{3000 \times \sqrt{3}}{3}\)
= 3000 – 1000(√3)
= 3000 – 1000(1.732)
= 3000 – 1732
= 1268 m
∴ First aeroplane is 1268 m higher than the second

Question 25.
The angle of elevation of a cloud from a point 60 m above the lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud.

Solution:
Let C be the cloud and RS is the surface of the lake and P is the point of observation such that RP = SQ = 60 m
T is the reflection of cloud C in the lake, then CS = ST
Let PQ ⊥ CT, then ∠CPQ = 30° and ∠TPQ = 60°
Let CQ = h, then CS = h + 60
∴ ST = h + 60
Now in △CPQ,
tan θ = \(\frac { CQ }{ PQ }\) ⇒ tan 30° = \(\frac { h }{ PQ }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ PQ }\) ⇒ PQ = h√3 …(i)
Similarly in △CPQ,
tan 60° = \(\frac { QT }{ PQ }\) ⇒ √3 = \(\frac{60+60+h}{\mathrm{PQ}}\)
PQ = \(\frac{120+h}{\sqrt{3}}\) ….(ii)
From (i) and (ii)
√3 h = \(\frac{120+h}{\sqrt{3}}\)
3h = 120 + h ⇒ 3h – h = 120
⇒ 2h = 120 ⇒ h = \(\frac { 120 }{ 2 }\) = 60
∴ Height of the cloud = CS = h + 60 = 60 + 60 = 120 m

Question 26.
The angle of elevation of the top of a h at the foot of a tower is 60° and the angle of elevation of the top of the tower fro the foot of the hill is 30°. If the tower 50 m high, what is the height of the hill?

Solution:
Let AB be the hill and CD be the tower
Height of tower CD = 50
Let height of the hill AB = h
Angle of elevation of A at D = 60°
and of C at B = 30°
Now in right △BCD,
tan θ = \(\frac { 50 }{ BD }\)

⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 50 }{ BD }\) ⇒ BD = 50√3 …(i)
Again in right △ABD,
tan 60° = \(\frac { AB }{ BD }\) ⇒ √3 = \(\frac { h }{ BD }\)
⇒ BD = \(\frac{h}{\sqrt{3}}\) …(ii)
From (i) and (ii)
\(\frac{h}{\sqrt{3}}\) = 50√3 ⇒ h = 50√3 × √3 = 50 × 3 = 150
⇒ Height of hill = 150 m

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
From a window A, 10 m above ground, the angle of elevation of the top C of a tower is x°, where tan x =\(\frac { 5 }{ 2 }\), and the angle of depression of the foot D of the tower is y°, where tan y= \(\frac { 1 }{ 4 }\).
See figure given here.
Calculate the height CD of the tower in metres.

Solution:
Let CD is the tower and A is window 10 m above the ground BD
The angle of elevation of the top of the lower C is x° and angle of depression from A to the foot of the tower is y°
tan x = \(\frac { 5 }{ 2 }\) and tan y = \(\frac { 1 }{ 4 }\)
From A, draw AE || BD such that
ED = AB = 10 m, BD = AE = x (suppose)
Let CD = h, then CE = h – 10
Now in right △ACE,
tan x = \(\frac { CE }{ AE }\) ⇒ \(\frac { 5 }{ 2 }\) = \(\frac { CE }{ BD }\)
BD = \(\frac { 2 }{ 5 }\) CE
Similarly in right △ABD,
tan y = \(\frac { AB }{ BD }\) ⇒ \(\frac { 1 }{ 4 }\) = \(\frac { 10 }{ BD }\)
⇒ BD = 40 m
∴ CE = \(\frac { 5 }{ 2 }\) BD = \(\frac { 5 }{ 2 }\) × 40 = 100 m
∴ CD = CE + ED = 100 + 10 = 110 m

Question 2.
In the figure, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm, RQ = 9 cm. Calculate the length of PR.

Solution:
In the figure, ∠PSR 90°, PQ = 1o cm
SQ = 6cm, RQ = 9 cm
∴ RS = RQ + QS = 9 + 6 = 15 cm
Let PR = x
In right △PQS,
PQ² = PS² + QS² (Pythagoras Theorem)
(10)² = PS² + (6)² ⇒ 100 = PS² + 36
⇒ PS² = 100 – 36 = 64 = (8)²
∴ PS = 8 cm
Now in right △PRS,
PR² = PS² + RS²
= (8)² + (15)² = 64 + 225 = 289 = (17)²
∴ PR = 17 cm

Question 3.
A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far ¡s he standing from the foot of the tower ?

Solution:
Let BC be the vertical tower whose height is 20 m
A is a man at some distance from the tower and angle of elevation is θ with the top of the tower and cos θ = 0.53
In the right △ABC,
cos θ = 0.53
∴ θ = 58° (using the cos tables)
∴ tan θ = \(\frac { AB }{ BC }\)
⇒ tan 58° = \(\frac { 20 }{ BC }\) ⇒ 1.6000 = \(\frac { 20 }{ BC }\)
⇒ BC = \(\frac { 20 }{ 1.6 }\) = \(\frac { 200 }{ 16 }\) = \(\frac { 25 }{ 2 }\) = 12.5 m
∴ The man is away from the foot of the tower
= 12.5 m

Question 4.
The shadow of a vertical tower AB on level ground is increased by 10 m, when the altitude of the sun changes from 45° to 30°, as shown in the adjoining figure. Find the height of the tower and give your answer correct to \(\frac { 1 }{ 10 }\) of a metre.

Solution:
AB is the vertical tower and CB is its shadow when the angle of elevation is 45°. When the elevation changes from 45° to 30°, the length of its shadow is increased by 10
m, i.e., DB = CB + 10 m
Let height of tower AB = h m
Now in right △ACB,
tan θ = \(\frac { AB }{ CB }\) ⇒ tan 45° = \(\frac { h }{ CB }\) ⇒ 1 = \(\frac { h }{ CB }\)
∴ CB = h
Similarly in right △ADB,
tan 30° = \(\frac { AB }{ DB }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{\mathrm{CB}+10}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{h+10}\)
⇒ h + 10 = √3 h ⇒ √3 h – h = 10
⇒ h (√3 – 1) = 10 ⇒ h(1.732 – 1) = 10
⇒ 0.732h = 10 ⇒ h = \(\frac { 10 }{ 0.732 }\)
h = \(\frac{10 \times 1000}{732}\) = \(\frac { 10000 }{ 732 }\) = 13.66 m
∴ Height of tower = 13.66 m = 13.7 m (upto one decimal place)

Question 5.
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°.
Calculate :
(i) the width of the river and
(ii) the height of the tree.
Solution:
Let TR be the tree on the opposite bank of a river whose width is AR and the angle of elevation of the top T of the tree is 60° After moving 50 m away, the angle of elevation of T at B will be 30° Let TR = h and AR = x

Now in right △TAR,
tan θ = \(\frac{\mathrm{TR}}{\mathrm{AR}}\) ⇒ tan 60° = \(\frac { h }{ r }\) ⇒ √3 = \(\frac { h }{ x }\)
⇒ h = √3 x ⇒ x = \(\frac{h}{\sqrt{3}}\)
Similarly in right △TBR,
tan 30° = \(\frac{\mathrm{TR}}{\mathrm{BR}}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+50}\)
⇒ x + 50 = √3 h ⇒ \(\frac{h}{\sqrt{3}}\) + 50 = √3 h
⇒ h + 50√3 = 3h ⇒ 3h – h = 50√3
⇒ 2h = 50√3 ⇒ h = \(\frac{50 \sqrt{3}}{2}\) = 25√3 m

(ii) ∴ Height of tree = h = 25(1.732) = 43.3 m
and width of river AR = x = \(\frac{h}{\sqrt{3}}\) = \(\frac{25 \sqrt{3}}{\sqrt{3}}\) = 25 m
Question 6.
Two people standing on the same side of a tower in straight line with it, measure the angles of elevation of the top of the tower as 25° and 50° respectively. If the height of the tower is 70°, find the distance between the two people.
Solution:
Let TR be the tower and A and B are two people which makes angles of elevation with the top T of the tower as 25° and 50° respectively
Height of the tower TR = 70 m

Now in right △TBR,
tan θ = \(\frac { TR }{ AR }\) ⇒ tan 50° = \(\frac { 70 }{ BR }\)
1.19175 = \(\frac { 70 }{ BR }\) = BR = \(\frac { 70 }{ 1.19175 }\) = 58.737 m
Similarly in right △TAR,
tan 25° = \(\frac { TR }{ AR }\) ⇒ 0.46631 = \(\frac { 70 }{ AR }\)
AR = \(\frac { 70 }{ 0.46631 }\) = 150.115 m
∴ Distance between two people = AB
= AR – BR = 150.115 – 58.737 = 91.378 m

Question 7.
From the top of a cliff 92 m high, the angle of depression of a boy is 20°. Calculate to the nearest metre, the distance of the boy from the foot of the cliff.
Solution:
Let AB be cliff which is 92 m in height and makes an angle of 20° of depression with a boy C
Let CB = x

∵AX || BC
∴ ∠C = 20° (alternate angles)
Now in right △ABC,
tan θ = \(\frac { AB }{ BC }\) ⇒ tan 20° = \(\frac { 92 }{ x }\)
⇒ 0.36397 = \(\frac { 92 }{ x }\) ⇒ x = \(\frac { 92 }{ 0.36397 }\)
⇒ x = 252.768
∴ Distance of boy from the foot of the cliff = 252.768 m

Question 8.
The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places.
Solution:
Let AB be vertical tower
the altitudes of sun changes frcm 45° to 30° when the back of ground increased by 10 m
Let height of AB tower = h

Now in right △ACB,
tan θ = \(\frac { AB }{ CB }\)
⇒ tan 45° = \(\frac { h }{ CB }\) ⇒ 1 = \(\frac { h }{ CB }\)
⇒ CB = h …(i)
Similarly in right △ADB,
tan 30° = \(\frac { AB }{ DB }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ DB }\)
⇒ DB = h√3
∴ DC = DB – CB
⇒ 10 = h√3 – h ⇒ h (√3 – 1) = 10
⇒ h = (1.732 – 1) = 10 ⇒ h × 0.732 = 10
⇒ h = \(\frac { 10 }{ 0.732 }\) = 13.66
∴ Height of tower = 13.66 m

Question 9.
From the top of a hill the angles of depression of two consecutive killometre stones, due east are found to be 30° and 45° respectively. Find the distance of the two stones from the foot of the hill.
Solution:
PQ is a hill whose top is P and A and B are the consecutive kilometre stones which make angle of depression of 45° and 30°
Let height of hill PQ = h
and QA = x km

∴ QB = (x + 1) km
Now in right △PQA,
tan θ = \(\frac { PQ }{ QA }\) ⇒ tan 45° = \(\frac { h }{ x }\)
⇒ 1 = \(\frac { h }{ x }\) ⇒ h = x …(i)
similarly in right △PQB
tan 30° = \(\frac { PQ }{ QB }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+1}\)
⇒ x + 1 = √3 h ⇒ x = √3 h – 1 …(ii)
From (i) and (ii)
√3 h – 1 = h ⇒ √3 h – h = 1
(1.732 – 1) h = 1 ⇒ 0.732h = 1
∴ h = \(\frac { 1 }{ 0.732 }\) = \(\frac { 1000 }{ 732 }\) = 1.366
∴ QA = h = 1.366 km and QB = (1 + 1.366)
= 2.366 km

Question 10.
A vertical pole and a vertical tower are on the same level ground. From the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of the tower is 30°. Find the height of the tower if the length of pole is 20 m.
Solution:
Let AE is the tower and BC is the pole standing on the same level ground
BC = 20 m
and angle of elevation of the top of tower = 60°
and angle of depression of the foot of the tower = 30°
From B, draw BD || CE
Such that DE = BC = 20 cm
and ∠BEC = ∠EBD = 30° (alternate angles)
BD = CE
Let height of tower AE = h

and CE = DB = x
∴ AD = h – 20
Now in right △BCE,
tan θ = \(\frac { BC }{ CE }\) ⇒ tan 30° = \(\frac { 20 }{ x }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 20 }{ x }\) ⇒ x = 20√3 m …(i)
∴ BD = x = 20√3 m
Now in right △ABD,
tan 60° = \(\frac { AD }{ BD }\) ⇒ √3 = \(\frac{h-20}{x}\)
⇒ √3 = \(\frac{h-20}{20 \sqrt{3}}\) ⇒ 20√3 × √3 = h – 20 m
⇒ 60 = h – 20 ⇒ h = 60 + 20 = 80 m
∴ Height of tower = 80 m

Question 11.
From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B correct to two decimal places.
Solution:

In △DBC, tan 60° = \(\frac{10}{BC}\)
⇒ √3 = \(\frac{10}{BC}\)
⇒ BC = \(\frac{10}{\sqrt{3}}\)
In △DAC, tan 30° = \(\frac{10}{B C+A B}\)

Question 12.
The top of a light house 100 m high the angle of depression of two ships on opposite sides of it are 48° and 36° respectively. Find the distance between the two ships to the nearest meter.

Solution:

As AD = 100 m
In right angled △ABD
tan 48° = \(\frac{AD}{BD}\) ⇒ BD = \(\frac{\mathrm{AD}}{\tan 48^{\circ}}\)
BD = \(\frac{100}{\tan 48^{\circ}}\) …(i)
In right angled △ADC
tan 36° = \(\frac{AD}{DC}\) ⇒ DC = \(\frac{\mathrm{AD}}{\tan 36^{\circ}}\)
DC = \(\frac{100}{\tan 36^{\circ}}\) …(ii)
From (i) and (ii)
BD + DC = \(\frac{100}{\tan 48^{\circ}}\) + \(\frac{100}{\tan 36^{\circ}}\)
BC = \(\frac{100}{\tan \left(90^{\circ}-42^{\circ}\right)}\) + \(\frac{100}{\tan \left(90^{\circ}-54^{\circ}\right)}\)
= \(\frac{100}{\cot 42^{\circ}}\) + \(\frac{100}{\cot 54^{\circ}}\)
= 100 × tan 42° + 100 × tan 54°
= 100 × 0.9004 + 100 × 1.3764
= 90.04 + 137.64 = 227.68 m = 228 m

Question 13.
A man observes the angles of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.
Solution:
AB is a building CD = 60 m (given)
In △ABC:
tan 60° = \(\frac{AB}{BC}\) ⇒ √3 = \(\frac{AB}{BC}\)
∴ BC = \(\frac{\mathrm{AB}}{\sqrt{3}}\) …(i)

In right angled △ABD:
tan 30° = \(\frac{AB}{BD}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}+60}\) ⇒ BC + 60 = √3 AB
∴ BC = √3 AB – 60
From equation (i) and (ii) we have
\(\frac{\mathrm{AB}}{\sqrt{3}}\) = √3 AB – 60 …(ii)
⇒ AB = 3AB – 60√3 ⇒ 3AB – AB = 60 × 1.732
⇒ AB = \(\frac{60 \times 1.732}{2}\) = 51.96 m = 52 m

Question 14.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest metre.
Solution:
Let AB be the lighthouse and C and D be the two ships.

Then, in △ADB
tan 30° = \(\frac { AB }{ BD }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{80}{\mathrm{BD}}\) ⇒ BD ⇒ 80√3 m …(i)
In △ABC, tan 40° = \(\frac{80}{\mathrm{BC}}\)
0.84 = \(\frac{80}{\mathrm{BD}}\) ⇒ BC = \(\frac { 80 }{ 0.84 }\) = 95.25 m
From (i), BD = 80√3 m
⇒ BC + DC = 80 × 1.73 m
⇒ DC = 138.4 – 95.25 m
= 43.15 m
Hence, distance between the two ships is = 43 m

Question 15.
In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :
(i) the horizontal distance between AB and CD.
(ii) the height of the lamp post.

Solution:
In △ABC
∠ACB = 60°
\(\frac { AB }{ BC }\) = tan 60°

BC = \(\frac{\mathrm{AB}}{\tan 60^{\circ}}\)
BC = \(\frac{60}{1.73}\) = 34.68 m
In △AXD
\(\frac { AX }{ XD }\) = tan 30°
⇒ XD = BC
∴ \(\frac { AX }{ BC }\) = tan 30°
\(\frac{\mathrm{AX}}{34.68}\) = tan 30°
⇒ AX = 34.68 × \(\frac{1}{\sqrt{3}}\)
AX = 34.68 × \(\frac { 1 }{ 1.732 }\) = 20.02 m
∴ CD = AB – AX
= 60 – 20.02 = 39.98 m
= 40 m (approx)

Question 16.
An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number.
Solution:
Let aeroplane is at position O and ship 1 is at point A and ship 2 is at point B.
⇒ OM is the altitude drawn from point O towards the river.
In △OMA
45° = \(\frac{\mathrm{OM}}{\mathrm{AM}}\)

l = \(\frac{250}{x}\) {∵ tan 45° = 1}
⇒ x = 250 m
In △OMB, tan 60° = \(\frac{250}{y}\)
⇒ √3 = \(\frac{250}{y}\) ⇒ y = \(\frac{250}{\sqrt{3}}\) = \(\frac{250}{1.73}\)
y = 144.34
∴ Width of river = x + y = 250 + 144.34 = 394.34 m

Question 17.
The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is 30° and 24° respectively.
Find the height of the two towers. Give your answer correct to 3 significant figures.

Solution:
Consider the following figure:

Tan 30° = \(\frac { AE }{ EC }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { AE }{ 120 }\)
⇒ \(\frac{120}{\sqrt{3}}\) = AE
∴ AE = 69.28 m
tan 24° = \(\frac { EB }{ EC }\)
⇒ 0.445 = \(\frac { EB }{ 120 }\)
∴ EB = 53.43 m
Thus, height of first tower, AB = AE + EB = 69.28 + 53.43 = 122.71 m
And, height of second tower, CD = EB = 53.427 m

Question 18.
An aeroplane at an altitude of 1500 metres finds that two ships are sailing towards it in the same direction. The angles of depression as observed from the aeroplane are 45° and 30° respectively. Find the distance between the two ships.
Solution:
A is the aeroplane, D and C are the ships sailing towards A. Ships are sailing towards the aeroplane in the same direction.
In the figure, height AB = 1500 m
C and D are the positions of two ships

To find: Distance between the ships, that is CD
In the right-angled △ABC,
tan 45° = \(\frac { 1500 }{ BD }\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 1500 }{ BD }\)
⇒ BD = 1500√3 m
⇒ BD = 1500(1.732) m
⇒ BD = 2598 m
∴ Distance between the ships = CD
= BC – BC
= 2598 – 1500 = 1098 m
So, the distance between the two ships is 1098 m.