AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.2 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.2

### 10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.2 Textbook Questions and Answers

Question 1.

A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60°. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the road.

Answer:

Let the height of the tower = h mts say

Width of the road be = x m.

Distance between two points of observation = 10 cm.

Angles of elevation from the two points = 60° and 30°.

From the figure

tan 60° = \(\frac{h}{x}\)

√3 = \(\frac{h}{x}\)

⇒ h = √3x …….(1)

Also tan 30° = \(\frac{h}{10+x}\)

⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{10+x}\)

⇒ h = \(\frac{10+x}{\sqrt{3}}\) ………(2)

From equations (1) and (2) h

h = √3x = \(\frac{10+x}{\sqrt{3}}\)

∴ √3x = \(\frac{10+x}{\sqrt{3}}\)

√3 × √3x = 10 + x

⇒ 3x – x = 10

⇒ 2x = 10

⇒ x = \(\frac{10}{2}\) = 5m

∴ Width of the road = 5 m

Now Height of the tower = √3x = 5√3 m.

Question 2.

A 1.5 m tall boy is looking at the top of a temple which is 30 metre in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple.

Answer:

Height of the temple = 30 m

Height of the man = 1.5 m

Initial distance between the man and temple = d m. say

Let the distance walked = x m.

From the figure

tan 30° = \(\frac{30-1.5}{d}\)

⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{28.5}{d}\)

∴ d = 28.5 × √3m ………(1)

Also tan 60° = \(\frac{28.5}{d-x}\)

⇒ √3 = \(\frac{28.5}{d-x}\)

⇒ √3(d-x) = 28.5

⇒ √3(28.5 × √3-x) = 28.5

⇒ 28.5 × 3 – √3x = 28.5

⇒ √3x = 3 × 28.5-28.5

⇒ √3x = 2 × 28.5 = 57

∴ x = \(\frac{57}{\sqrt{3}}=\frac{19 \times 3}{\sqrt{3}}\) = 19√3

= 19 × 1.732

= 32.908 m.

∴ Distance walked = 32.908 m.

Question 3.

A statue stands on the top of a 2 m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue.

Answer:

Height of the pedestal = 2 m.

Let the height of the statue = h m. Angle of elevation of top of the statue = 60°.

Angle of elevation of top of the pedestal = 45°.

Let the distance between the point of observation and foot of the pedestal = x m.

From the figure

tan 45° = \(\frac{2}{x}\)

1 = \(\frac{2}{x}\)

∴ x = 2 m.

Also tan 60° = \(\frac{2+h}{x}\)

⇒ √3 = \(\frac{2+h}{x}\)

⇒ 2√3 = 2 + h

⇒ h = 2√3 – 2

= 2(√3-1)

= 2(1.732 – 1)

= 2 × 0.732

= 1.464 m.

Question 4.

From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7 m, then find the height of the tower.

Answer:

Angle of elevation of the top of the tower = 60°.

Angle of depression to the foot of the tower = 45°.

Distance between tower and building = 7 m.

Let the height of the building = x m and tower = y m.

From the figure

tan 45° = \(\frac{x}{7}\)

1 = \(\frac{x}{7}\)

∴ x = 7 m.

Also tan 60° = \(\frac{y-x}{7}\)

⇒ √3 = \(\frac{y-x}{7}\)

⇒ 7√3 = y – 7

∴ y = 7 + 7√3

= 7 (√3 + 1)

= 7(1.732 + 1)

= 2.732 × 7

= 19.124 m.

Question 5.

A wire of length 18 m had been tied with electric pole at an angle of eleva¬tion 30° with the ground. As it is covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut ?

Answer:

Length of the wire = 18 m

Let the length of the wire removed = x

Height of the pole be = h

From the figure

sin 30° = \(\frac{h}{18}\)

⇒ \(\frac{1}{2}\) = \(\frac{h}{18}\)

⇒ h = \(\frac{18}{2}\) = 9 m

Also sin 60° = \(\frac{h}{18-x}\)

\(\frac{\sqrt{3}}{2}\) = \(\frac{9}{18-x}\)

√3(18-x) = 9 × 2

18√3 – √3x = 18

√3x = 18√3 – 18

√3x = 18(√3-1)

x = \(\frac{18(\sqrt{3}-1)}{\sqrt{3}}\)

= \(\frac{6 \times 3(\sqrt{3}-1)}{\sqrt{3}}\)

= 6√3(√3-1)

= 6(3-√3)

= 18 – 6√3

= 18 – 10.392

= 7.608 m.

Question 6.

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.

Answer:

Height of the tower = 30 m

Angle of elevation of the top of the tower = 60°.

Angle of elevation of the top of the building = 30°.

Let the distance between the foot of the tower and foot of the building be d m and height of the building be x m.

From the figure

tan 60° = \(\frac{30}{d}\)

√3 = \(\frac{30}{d}\)

⇒ d = \(\frac{30}{\sqrt{3}}=\frac{10 \times 3}{\sqrt{3}}\) = 10√3m

Also tan 30° = \(\frac{x}{d}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{x}{10 \sqrt{3}}\)

⇒ x = \(\frac{10 \sqrt{3}}{\sqrt{3}}\) = 10 m

∴ Height of the building = 10 m

Question 7.

Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.

Answer:

Width of the road = 120 f.

Angle of elevation of the top of the 1st tower = 60°.

Angle of elevation of the top of the 2 tower = 30°.

Let the distance of the point from the 1st pole = x.

Then the distance of the point from

the 2nd pole = 120 – x.

and height of each pole = h say.

From the figure

tan 60° = \(\frac{h}{x}\)

⇒ √3 = \(\frac{h}{x}\)

⇒ h = √3x ……..(1)

Also tan 30° = \(\frac{\mathrm{h}}{120-\mathrm{x}}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{120-\mathrm{x}}\)

⇒ h = \(\frac{120-x}{\sqrt{3}}\)

From (1) and (2)

√3x = \(\frac{120-x}{\sqrt{3}}\)

⇒ √3.√3x = 120-x

⇒ 3x = 120 – x

⇒ 3x + x = 120

⇒ 4x = 120

⇒ x = \(\frac{120}{4}\) = 30 ft

Now h = √3x = √3 × 30 = 1.732 x 30 = 51.960 feet

∴ Distances of the poles = 30 ft. and 120 – 30 fts = 90 ft.

Height of each pole = 51.96 ft.

Question 8.

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.

Answer:

Let the height of the tower = h m.

Angles of elevation of the top of the tower from two points = x° and (90° – x)

From the figure

tan x = \(\frac{h}{4}\) ……. (1)

Also tan (90° – x) = \(\frac{h}{9}\)

⇒ cot x = \(\frac{h}{9}\)

⇒ \(\frac{1}{\tan x}\) = \(\frac{h}{9}\)

∴ tan x = \(\frac{9}{h}\) …….. (2)

From (1) and (2)

tan x = \(\frac{h}{4}\) = \(\frac{9}{h}\)

∴ \(\frac{h}{4}\) = \(\frac{9}{h}\)

h × h = 9 × 4

⇒ h^{2} = 36

⇒ h = 6 m

Question 9.

The angle of elevation of a jet plane from a point A on the ground is 60°. After 4 flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 meter, find the speed of the jet plane. (√3 = 1.732)

Answer:

Height of the plane from the ground PM = RN = 1500√3 m.

Angle of elevation are 30° and 60°.

From the figure

tan 60° = \(\frac{PM}{QM}\)

√3 = \(\frac{1500 \sqrt{3}}{\mathrm{QM}}\)

QM = \(\frac{1500 \sqrt{3}}{\sqrt{3}}\) = 1500 m

Also tan 30° = \(\frac{RN}{QN}\)

\(\frac{1}{\sqrt{3}}=\frac{1500 \sqrt{3}}{\mathrm{QM}+\mathrm{MN}}\)

QM + MN = 1500√3 × √3

1500 + MN = 1500 × 3

MN = 4500 – 1500

MN = 3000 mts.

∴ Distance travelled in 15 seconds = 3000 mts.

∴ Speed of the jet plane = \(\frac{\text { distance }}{\text { time }}=\frac{3000}{15}\) = 200 m/s

= 200 × \(\frac{18}{5}\) kmph

= 720 kmph

Speed = 200 m/sec. or 720 kmph.