These TS 10th Class Maths Chapter Wise Important Questions Chapter 12 Applications of Trigonometry given here will help you to solve different types of questions.

## TS 10th Class Maths Important Questions Chapter 12 Applications of Trigonometry

Previous Exams Questions

Question 1.

A person from the top of a building of height 25 m has observed another building top and bottom at an angle of elevation 45° and at an angle of depression 60° respectively. Draw a diagram for this data. (T.S. Mar. ’15)

Solution:

Question 2.

A ladder of 3.9 m length is laid against a wall. The distance between the foot of the wall and the ladder is 1.5 m. Find the height at which the ladder touches the wall. (T.S. Mar. ’15)

Solution:

h^{2} = (3.9)^{2} – (1.5)^{2}

= (3.9 + 1.5) (3.9 – 1.5)

= 5. 4 × 2.4

= (0.6 × 9) × (0.6 × 4)

= (0.6)^{2} × 6^{2}

∴ h = 6 × 0.6 = 3.6 m

Question 3.

An observer flying in an aeroplane at an altitude of 900 m observes two ships in front of him, which are in the same direction at an angles of depression of 60° and 30° respectively. Find the distance between the two ships. (T.S. Mar. ’15)

Solution:

In ∆ABC

Tan 60 = \(\frac{900}{\mathrm{x}}\)

\(\sqrt{3}\) = \(\frac{900}{\mathrm{x}}\)

⇒ x = \(\frac{900}{\sqrt{3}}\) = 300\(\sqrt{3}\)

In ∆ ABD

Tan 30 = \(\frac{900}{\mathrm{x+d}}\)

\(\frac{1}{\sqrt{3}}\) = \(\frac{900}{300 \sqrt{3}+d}\)

d = 600\(\sqrt{3}\) m

Question 4.

If the angle of elevation of sun increases from ‘O’ to 90 then the length of shadow of a tower decreases. Is this true ? Justify your answer. (T.S. Mar. ’16)

Solution:

Yes, this statement is true.

We observe this in day to day life.

AD – ground

BC – tower

AB – shadow

Question 5.

A boat has to cross a river. It crosses river by making an angle of 60° with bank, due to the stream of river it travels a distance of 450 m to reach another side of river. Draw a diagram to this data. (T.S. Mar. ’15)

Solution:

AB – width of river

AD, BC are river banks

AC – The distance travelled in river = 450 m

A – initial point, C – terminal point

Question 6.

Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road, the angles of elevation of top of the poles are 60° and 30°. Find the height of poles, (T.S. Mar. ’16)

Solution:

As shown in the figure

AD = width of road = 80 m.

AB, DE are two poles

AB = DE (∵ they have equal heights)

‘C’ is a point on road.

∠ACB = 30°, ∠DCE = 60°

Then in ∆ ACB

tan C = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)

⇒ tan 30 = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)

\(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)

⇒ AC = AB\({\sqrt{3}}\) ……………….. (1)

In ∆ CDE

tan C = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)

⇒ tan 60 = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)

\({\sqrt{3}}\) = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)

⇒ CD = \(\frac{\mathrm{DE}}{\sqrt{3}}\) ……………… (2)

but AC + CD = AD

⇒ AB\({\sqrt{3}}\) + \(\frac{\mathrm{DE}}{\sqrt{3}}\) = 80

But DE = AB

⇒ AB\({\sqrt{3}}\) + \(\frac{\mathrm{AB}}{\sqrt{3}}\) = 80

⇒ \(\frac{3 \mathrm{AB}+\mathrm{AB}}{\sqrt{3}}\) = 80

⇒ 4AB = 80\({\sqrt{3}}\)

⇒ AB = \(\frac{80 \sqrt{3}}{4}\) = 20 \({\sqrt{3}}\) m.

So height of the pole = 20 m.

Additional Questions

Question 1.

The angle of elevation of the top of a Tree from the foot of building is 60° and the angle of elevation of the top of the building from the foot of the tree is 30° what is the ratio of heights of tree and building.

Solution:

Let the height of the tree = x m = AB

Let the height of the building = y m = CD

distance between the tree and building = d = BD

Angle of elevation of the top of the tree = 60°

From the figure tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\) = \(\frac{\mathrm{x}}{\mathrm{d}}\)

\(\sqrt{3}\) = \(\frac{\mathrm{x}}{\mathrm{d}}\)

d = \(\frac{\mathrm{x}}{\sqrt{3}}\) ……………. (1)

∴ tan 30° = \(\frac{\mathrm{CD}}{\mathrm{BD}}\) = \(\frac{\mathrm{y}}{\mathrm{d}}\)

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

d = y\(\sqrt{3}\) ………………. (2)

\(\frac{\mathrm{x}}{\sqrt{3}}\) = y\(\sqrt{3}\)

\(\frac{\mathrm{x}}{\mathrm{y}}\) = \(\sqrt{3}\) × \(\sqrt{3}\)

\(\frac{\mathrm{x}}{\mathrm{y}}\) = \(\frac{3}{1}\)

x : y = 3 : 1

Hence, the ratio of the heights of the tree and building = 3 : 1.

Question 2.

The angle of the top of a pillar from two points are at a distance of 7m and 12m Find the height of the pillar from the base of the pillar and in the same straight line with it are complementary.

Solution:

From the figure,

Let AB be a height of the pillar = h m

Let the two points on the ground be ’c and d’ such that they make a distance 7 m and 12 m

From foot of the pillar

AC = 7m ; AD = 12m

Angles of elevation are ∠ACB = 0 ;

∠ADB = (90 – θ)

In the right angled ∆ABC, we have

tan θ = \(\frac{\mathrm{AB}}{\mathrm{AD}}\)

tan θ = \(\frac{\mathrm{h}}{7}\) …………… (1)

From.the right angled ∆ABC we have

tan (90 – θ) = \(\frac{\mathrm{AB}}{\mathrm{AD}}\)

cot θ = \(\frac{\mathrm{h}}{12}\)

\(\frac{1}{\tan \theta}\) = \(\frac{\mathrm{h}}{12}\)

tan θ = \(\frac{12}{\mathrm{~h}}\) …………….. (2)

From (1) and (2)

\(\frac{\mathrm{h}}{12}\) = \(\frac{12}{\mathrm{~h}}\)

h^{2} = 84

h = \(\sqrt{4 \times 21}\)

h = 2 \(\sqrt{21}\) m

∴ The height of the pillar = h = 2 \(\sqrt{21}\)

Question 3.

A wire of length 25m had been tied with electric pole at an angle of elevation 30° with the ground. Because it was 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.

Solution:

In the figure

Let PQ be a height of the electric pole = h m

given QS be the actual length of the wire = 25 m

let Q is length of the wire was cut S, R are the first and second points of observations.

Let PS = a + b ; PR = b

Angles of elevations are ∠PSQ = 30°

∠PRQ = 60°

From ∆PSQ

sin 30° = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)

\(\frac{1}{2}\) = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)

2 PQ = 25

PQ = 12.5

From ∆PRQ

tan 60° = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)

\(\sqrt{3}\) = \(\frac{12.5}{\mathrm{b}}\)

PR = b = \(\frac{12.5}{\sqrt{3}}\)

∆PQR, From

cos 60° = \(\frac{\mathrm{PR}}{\mathrm{QR}}\)

\(\frac{1}{2}\) = \(\frac{\frac{12.5}{\sqrt{3}}}{\mathrm{QR}}\)

\(\frac{1}{2}\) = \(\frac{12.5}{\sqrt{3 \mathrm{QR}}}\)

QR = \(\frac{25}{\sqrt{3}}\)

∴ Length of the wire was cut = \(\frac{25}{\sqrt{3}}\)

Question 4.

Two boys are on opposite sides of a tower 200m height. They measure the angle of elevation of the top of the tower as 45°, 60° respectively. Find the distance through which the boys are separated.

Solution:

Given height of Tower = 200 m

We say ‘x’ is the distance between 1^{st} person and base of the tower and ‘y’ is the distance between 2^{nd} person and base of the tower.

From ∆ ABD

Tan 45° = \(\frac{\mathrm{BD}}{\mathrm{AD}}\)

1 = \(\frac{200}{\mathrm{x}}\)

x = 200m

From ∆ BDC

Tan 60° = \(\frac{\mathrm{BD}}{\mathrm{DC}}\)

Question 5.

The tower height is l5mts and length of shadow is 15 \(\sqrt{3}\) m what is the angle of elevation of the sun.

Solution:

Let AC be the height of the tower = 15 m

Length of shadow = 15\(\sqrt{3}\) m

Let angle of elevation be θ.

From ∆ ABC Tan θ = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac{15}{15 \sqrt{3}}\)

Tan θ = \(\frac{1}{\sqrt{3}}\)

Tan θ = Tan 30°

θ = 30°

∴ the angle of elevation of the sun θ = 30°