NCERT Solutions for Class 10 Maths Chapter 9 Some Applications Of Trigonometry

Q1 ) A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30º (see figure).

Answer :

In \( ∆ \ ABC \) , \( \frac{AB}{AC} \ = \ sin30º \)

=> \( \frac{AB}{20} \ = \ \frac{1}{2} \) [ \( \because sin30º \ = \ \frac{1}{2} \) ]

=> \( AB \ = \ \frac{1}{2} \ × \ 20 \ = \ 10 \)

\(\therefore \) Height of the pole is 10 m.

Q1 ) A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30º with it. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree.

Answer :

In \( ∆ \ ABC \)



\( \frac{AB}{BC} \ = \ tan30º \ => \ \frac{AB}{8} \ = \ \frac{1}{ \sqrt{3}} \)

=> \( AB \ = \ \frac{8}{ \sqrt{3}} \)      (1)

And, \( \frac{AC}{BC} \ = \ sec30º \ => \ \frac{AC}{8} \ = \ \frac{2}{ \sqrt{3}} \)

=> \( AC \ = \ \frac{16}{ \sqrt{3}} \)      (2)

Now, Height of the tree = AB + AC

\( =\ \frac{8}{ \sqrt{3}} \ + \ \frac{16}{ \sqrt{3}} \ = \ \frac{24}{ \sqrt{3}} \)

\( = \ \frac{24}{ \sqrt{3}} \ × \ \frac{ \sqrt{3}}{ \sqrt{3}} \ = \ 8\sqrt{3} \) m

Q3 ) A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30º to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60º to the ground. What should be the length of the slide in each case?

Answer :

In \( ∆ \ BDE \)



\( \frac{DE}{BD} \ = \ cosec 30º \ => \ \frac{DE}{1.5} \ = \ 2 \)

=> \( DE \ = \ 2 × 1.5 \ = \ 3 \)

And in \( ∆ \ ABC \)

\( \frac{AC}{AB} \ = \ cosec 60º \ => \ \frac{AC}{3} \ = \ \frac{2}{ \sqrt{3}} \)

\( AC \ = \ \frac{2}{ \sqrt{3}} × 3 \ = \ 2\sqrt{3} \)

\(\therefore \) Length of slides are \( 3 \) m and \( 2\sqrt{3} \) m

Q4 ) 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 30º. Find the height of the tower.

Answer :

Let AB be the tower of height h metres and let C be a point at a distance of 30 m from the foot of the tower. The angle of elevation of the top of the tower from point C is given as 30º.



In \( ∆ \ CAB \), we have

\( \frac{AB}{CA} \ = \ tan30º \)

\( \frac{h}{30} \ = \ \frac{1}{ \sqrt{3}} \ => \ h \ = \ \frac{30}{ \sqrt{3}} \ = \ 10 \sqrt{3} \)

Hence, the height of the tower is \( 10 \sqrt{3} \) metres.

Q5 ) A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60º. Find the length of the string, assuming that there is no slack in the string.

Answer :

Let OA be the horizontal ground, and let K be the position of the kite at height 60 m above the ground. Let the length of the string OK be x metres. It is given \( ∠ \ KOA \ = \ 60º \)



In \( ∆ \ AOK \), we have

\( \frac{AK}{OK} \ = \ sin60º \ => \ \frac{60}{x} \ = \ \frac{ \sqrt{3}}{2} \)

\( x \ = \ \frac{120}{ \sqrt{3}} \ = \ 40 \sqrt{3} \)

\(\therefore \) the length of the string is \( 40 \sqrt{3} \) m

Q6 ) A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30º to 60º as he walks towards the building. Find the distance he walked towards the building.

Answer :

Let OA be the building and PL be the initial position of the man such that \( ∠ \ APR=30º \) and AO = 30 m. Let MQ be the position of the man at a distance PQ. Here \( ∠ \ AQR \ = \ 60º \).



Now from \( ∆ \ ARQ \) and \( ∆ \ ARP \) , we have

\( \frac{QR}{AR} \ = \ cot60º \ \)
\(=> \ \frac{QR}{AR} \ = \ \frac{1}{ \sqrt{3}} \ \)
\( => \ QR \ = \ \frac{AR}{ \sqrt{3}} \) (1)

and, \( \frac{PR}{AR} \ = \ cot30º \ \)
\( => \ \frac{PR}{AR} \ = \ \sqrt{3} \ \)
\( => \ PR \ = \ \sqrt{3}AR \) (2)

From (1) and (2), we get

\( PQ \ = \ PR \ - \ QR \ \)
\( = \ \sqrt{3}AR \ - \ \frac{AR}{ \sqrt{3}} \)
\( = \ \frac{(3-1)AR}{ \sqrt{3}} \ \)
\( = \ \frac{2 \sqrt{3}}{3} AR \)

\( = \ \frac{2 \sqrt{3}}{3} \ × \ 28.5 \ = \ 19 \sqrt{3} \)
[ \(\because AR \ = \ 30 \ - \ 1.5 \ = \ 28.5 \)m ]

\(\therefore \) the distance walked by the man towards the building is \( 19 \sqrt{3} \) metres.

Q7 ) From a point on the ground the angles of elevation of the bottom and top of a tower fixed at the top of a 20 m high building are 45º and 60º respectively. Find the height of the tower.

Answer :

Let BC be the building of height 20 m and CD be the tower of height x metres. Let A be a point on the ground at a distance of y metres away from the foot B of the building.



In \( ∆ \ ABC \), we have

\( \frac{BC}{AB} \ = \ tan45º \ \)
\( => \ \frac{20}{y} \ = \ 1 \)
\( => \ y \ = \ 20\)

In \( ∆ \ ABD \), we have

\( \frac{BD}{AB} \ = \ tan60º \ \)
\( => \ \frac{20+x}{20} \ = \ \sqrt{3} \)

\( => 20+x \ = \ 20 \sqrt{3} \ \)
\( => \ x \ = \ 20( \sqrt{3} -1) \) \(=> x \ = \ 20(1.732 - 1) \ \)
\( => \ x \ = \ 14.64 \)

\(\therefore \) the height of the tower is 14.64 metres.

Q8 ) A statue 1.6 m tall stands on the top a pedestal. From a point on the ground the angle of elevation of the top of the statue is 60º and from the same point the angle of elevation of the top of the pedestal is 45º. Find the height of the pedestal.

Answer :

Let BC be the pedestal of height h metres and CD be the statue of height 1.6 m. Let A be a point on the ground such that \( ∠ \ CAB \ = \ 45º \) and\( ∠ \ DAB \ = \ 60º \)



In \( ∆ \ ABC \) and \( ∆ \ ABD \), we have

\( \frac{AB}{BC} \ = \ cot45º \ \)
\( => \ \frac{AB}{h} \ = \ 1 \)
\( => \ AB \ = \ h \) (1)

and, \( \frac{BD}{AB} \ = \ tan60º \ \)
\( => \ \frac{BC + CD}{AB} \ = \ \sqrt{3} \)
\( => \ \frac{h+1.6}{h} \ = \ \sqrt{3} \ \)
\( => \ h+1.6 \ = \ h \sqrt{3} \)
\(=> h( \sqrt{3} \ - \ 1) \ = \ 1.6 \ \)
\(=> \ h \ = \ \frac{1.6}{ \sqrt{3} \ - \ 1} \ \)
\( = \ \frac{1.6}{ \sqrt{3} \ - \ 1} × \frac{ \sqrt{3} \ + \ 1}{ \sqrt{3} \ + \ 1} \)

\( = \ \frac{1.6( \sqrt{3} \ + \ 1)}{2} \ \)
\( = \ 0.8( \sqrt{3}+1) \) \( = \ 2.1856 \)

\(\therefore \) the height of the pedestal is \( 2.1856 \) m

Q9 ) The angle of elevation of the top of the building from the foot of the tower is 30º and the angle of elevation of the top of the tower from the foot of the building is 60º. If the tower is 50 m high, find the height of the building.

Answer :

Let AB be the building of height h and AC, the horizontal ground through C, the foot of the building. Since the building subtends an angle of 60º at C, hence \( ∠ \ ACB \ = \ 30º \). Let CD be the tower of height 50 m such that \( ∠ \ CAB \ = \ 60º \).



\( ∆ \ BAC \) and \( DCA \) , we have

\( \frac{AC}{AB} \ = \ cot30º \ \)
\( => \ \frac{AC}{h} \ = \ \sqrt{3} \)
\( => \ AC \ = \ \sqrt{3} \) (1)

and \( \frac{DC}{AC} \ = \ tan60º \ \)
\( => \ \frac{50}{AC} \ = \ \sqrt{3} \)
\(=> \ AC \ = \ \frac{50}{ \sqrt{3}} \) (2)

Equating the values of AC from (1) and (2), we get

\( \sqrt{3}h \ = \ 50 \sqrt{3} \ \)
\( => \ h \ = \ \frac{50}{ \sqrt{3}} × \frac{1}{ \sqrt{3}} \)

\( = \ \frac{50}{3} \ = \ 16.66 \)

\(\therefore \) the height of the building is 16.66 m.

Q10 ) Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angle of elevation of the top of the poles are 60º and 30º respectively. Find the height of the poles and the distances of the point from the poles

Answer :

Let AB and CD be two poles each of height h metres. Let P be a point on the road such that AP = x metres. Then CP = (80 – x) metres. Its is given that \( ∠ \ APB \ = \ 60º \) and \( ∠ \ CPD \ = \ 30º \).



In \( ∆ \ APB \), we have

\( \frac{AB}{AP} \ = \ tan60º \ \)
\( => \ \frac{h}{x} \ = \ \sqrt{3} \)
\( => \ h \ = \ \sqrt{3}x \) (1)

In \( ∆ \ CPD \), we have

\( \frac{CD}{CP} \ = \ tan30º \ \)
\( => \ \frac{h}{80-x} \ = \ \frac{1}{ \sqrt{3}} \)
\( => \ h \ = \ \frac{80-x}{ \sqrt{3}} \) (2)

Equating the values of h from (1) and (2), we get

\( \sqrt{3}x \ = \ \frac{80-x}{ \sqrt{3}} \ \)
\( =>\ 3x \ = \ 80-x \)
\( => \ x \ = \ 20 \)

Putting x = 20 in (1), we get

\( h \ = \ \sqrt{3} × 20 \ \)
\( = \ (1.732) × 20 \ = \ 34.64 \)

\(\therefore \) The point is at a distance of 20 metres from the first pole and 60 metres from the second pole. And the height of the pole is 34.64 metres.

Q11 ) A T.V. tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60º. From a point 20 m away from this point on the same bank, the angle of elevation of the top of the tower is 30º (see figure). Find the height of the tower and the width of the river.

Answer :

Let AB be the T.V. tower of height h standing on the bank of a river. Let C be the point on the opposite bank of the river such that BC = x metres. Let D be another point away from C such that CD = 20, and the angles of levation of the top of the T.V. tower at C and D are 60º and 30º respectively. i.e., \( ∠ \ ACB \ = \ 60º \) and \( ∠ \ AOB \ = \ 30º \)



In \( ∆ \ ABC \) , we have

\( \frac{AB}{BC} \ = \ tan60º \)

\( => \ \frac{h}{x} \ = \ \sqrt{3} \ \)
\( => \ h \ = \ \sqrt{3}x \) (1)

In \( ∆ \ ABD \), we have

\( \frac{AB}{BD} \ = \ tan30º \ \)
\( => \ \frac{h}{x+20} \ = \ \frac{1}{ \sqrt{3}} \)
\( => \ h \ = \ \frac{x+20}{ \sqrt{3}} \) (2)

Equating values of h from (1) and (2), we get

\( \sqrt{3}x \ = \ \frac{x+20}{ \sqrt{3}} \ \)
\( => \ 3x \ = \ x + 20 \)
\( => \ x = 10 \)

Putting x = 10 in (1), we get

\( h \ = \ \sqrt{3} × 10 \ = \ 17.32 \)

\(\therefore \) the height of the T.V. tower is 17.32 metres and the widith of the river is 10 metres.

Q12 ) From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60º and the angle of depression of its foot is 45º. Determine the height of the tower.

Answer :

Let AB be the building of height 7 metres and let CD be the cable tower. It is given that the angle of elevation of the top D of the tower observed from A is 60º and the angle of depression of the base C of the tower observed from A is 45º. Then \( ∠ \ EAD \ = \ 60º \) and \( ∠ \ BCA \ = \ 45º \) . Also AB = 7 m



In \( ∆ \ EAD \), we have

\( \frac{DE}{EA} \ = \ tan60º \ \)
\( => \ \frac{h}{x} \ = \ \sqrt{3} \)
\( => \ h \ = \ \sqrt{3} x \) (1)

In \( ∆ \ ABC \), we have

\( \frac{AB}{BC} \ = \ tan45º \ \)
\( => \ \frac{7}{x} \ = \ 1 \)
\( => \ x \ = \ 7 \)
(2)

Putting x = 7 in (1), we get

\( h \ = \ 7 \sqrt{3} \ \)
\( => \ DE \ = \ 7 \sqrt{3} \)m

∴ \( CD \ = \ CE \ + \ ED \ \)
\( = \ 7 \ + \ 7\sqrt{3} \ \)
\( = \ 19.124 \)m

\(\therefore \) The height of the cable tower is 19.124 m.

Q13 ) As observed from the top of a 75 m tall lighthouse, the angles of depression of two ships are 30º and 45º. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer :

Let AB be the lighthouse of height 75 m and let two ships be at C and D such that the angles of depression from B are 45º and 30º respectively.



Let AC = x and CD = y.

In \( ∆ \ ABC \), we have

\( \frac{AB}{AC} \ = \ tan45º \ \)
\( => \ \frac{75}{x} \ = \ 1 \)
\( => \ x \ = \ 75 \) (1)

In \( ∆ \ ABD \), we have

\( \frac{AB}{AD} \ = \ tan30º \ \)
\( => \ \frac{75}{x+y} \ = \ \frac{1}{ \sqrt{3}} \)
\( => \ x+y \ = \ 75 \sqrt{3} \) (2)

From (1) and (2), we have

\( 75+y \ = \ 75 \sqrt{3} \ \)
\( => \ y \ = \ 75( \sqrt{3} - 1) \ \) T
\( => \ y \ = \ 75(1.732 - 1) \ = \ 54.9 \)

\(\therefore \) the distance between the two ships is 54.9 metres.

Q14 ) A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60º. After some time, the angle of elevation reduces to 30º. Find the distance travelled by the balloon during the interval.

Answer :

Let P and Q be the two positions of the balloon and let A be the point of observation. let ABC be the horizontal through A. It is given that angles of elevation of the balloon in two position P and Q from \( ∠ \ PAB \ = \ 60º \) , \( ∠ \ QAB \ = \ 30º \). It is also given that MQ = 88.2

\( => \ CQ \ = \ MQ \ - \ MC \ \)
\( = \ 88.2 \ - \ 1.2 \ = \ 87 \)m .



In \( ∆ \ ABP \), we have

\( \frac{BP}{AB} \ = \ tan60º \ \)
\( => \ \frac{87}{AB} \ = \ \sqrt{3} \)
\( => \ AB \ = \ \frac{87}{ \sqrt{3}} \ = \ \frac{87 \sqrt{3}}{3} \ = \ 29 \sqrt{3} \)      (1)

In \( ∆ \ ACQ \), we have

\( \frac{CQ}{AC} \ = \ tan30º \ \)
\( => \ \frac{87}{AC} \ = \ \frac{1}{ \sqrt{3}} \)
\( => \ AC \ = \ 87 \sqrt{3} \) (2)

\(\therefore \) \( PQ \ = \ BC \ = \ AC \ - \ AB \ \)
\( = \ 87 \sqrt{3} \ - \ 29 \sqrt{3} \ \)
\( = \ 58 \sqrt{3} \)
\( = \ 100.456 \)m

\(\therefore \) the balloon travels \( 100.456 \)m

Q15 ) A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30º, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60º. Find the time taken by the car to reach the foot of the tower.

Answer :

Let AB be the tower of height h. Let C be the initial position of the car and let after 6 seconds, the car be at D. It is given that the angles of depression at C and D are 30º and 60º respectively. Let the speed of the car be v m/s. Then ,

CD = Distance travelled by the car in 6 seconds

\(=> CD \ = \ 6v \) metres
[\(\because \) Distance = Speed × Time]
Let the car takes t seconds to reach the tower AB from C. Then,

\( DA \ = \ vt \) metres



In \( ∆ \ ABC \), we have

\( \frac{AB}{AC} \ = \ tan60º \)
\( => \frac{h}{vt} \ = \ \sqrt{3} \)
\( => \ h \ = \sqrt{3} vt \) (1)

In \( ∆ \ ABD \) , we have

\( \frac{AB}{AD} \ = \ tan30º \)
\( => \frac{h}{vt + 6v} \ = \frac{1}{\sqrt{3}} \)
\( => \ h\sqrt{3} \ = vt + 6v \) .... (2)

Substituting the value of h from (1) in (2), we get

\( \sqrt{3} \ × \ \sqrt{3}vt \ = \ vt + 6t \)
\( => 3vt \ = \ vt \ + \ 6v \)
\( => \ 3t \ - \ t \ = \ 6 \)
\( => t \ = \ 3 \)

\(\therefore \) the car will reach the tower from C in 3 seconds.

Q16 ) The angle of elevation of the top of a tower from two points at a distance of 4 m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer :

Let AB be the tower. Let C and D be the two points at distances 9 m and 4 m respectively from the base of the tower. Then AC = 9 m, AD = 4m.



Let \( ∠ \ ACB \ = \ \theta \) and \( ∠ \ ADB \ = \ 90º \ - \ \theta \)

Let h be the height of the tower AB.



In \( ∆ \ ACB \), we have

\( \frac{AB}{AC} \ = \ tan \theta \ \)
\( => \ \frac{h}{9} \ = \ tan \theta \) ...... (1)

In \( ∆ \ ADB \), we have

\( \frac{AB}{AD} \ = \ tan(90º \ - \ \theta) \ \)
\( => \ \frac{h}{4} \ = \ cot \theta \) .. (2)

From (1) and (2), we have

\( \frac{h}{9} \ × \ \frac{h}{4} \ = \ tan \theta \ × \ cot \theta \ \)
\( => \ \frac{h^2}{36} \ = \ 1 \)
\( => \ h^2 \ = \ 36 \ \)
\( => \ h \ = \ 6 \)

∴ the height of the tower is 6 metres.