CBSE Class 10 - Applications of Trigonometry - Short Answer Based Questions #class10Maths #cbsenotes #eduvictors

CBSE Class 10 - Applications of Trigonometry - Short Answer Based Questions

Q1: In the figure below which option a, b and c refers to line of sight?

Answer: option a

Q2: If a pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground, then find the Sun’s elevation.

Answer: As shown in figure below:

Height of the pole BC = 6m

Shadow length = $2\sqrt{3}$ m

Let angle of elevation = θ

Now in ∆BAC, tan θ = $\frac{6}{2\sqrt{3}} = \frac{2 \times 3}{2\sqrt{3}} = \sqrt{3}$

tan θ = tan 60°

θ = 60° (Answer)

Q3: A vertical straight tree, 15 m high, is broken by the wind, in such a way that its top just touches the ground and makes an angle of 60° with the ground. At what height from the ground did the tree break? (Use $\sqrt{3}$ = 1.73)

Answer: 

Height of the tree AB = 15 m

It broke at C. Its top A touches the ground at D.

Now, AC = CD,∠BDC = 60°

Let BC = x

AC = AB – BC

∴ AC = 15 – x

⇒ CD = 15 – x [∵ AC = CD]

In rt. ∆CBD,

sin 60° = $\frac{BC}{CD}$

⇒$\frac{\sqrt{3}}{2}$

⇒$\frac{\sqrt{3}}{2} = \frac{x}{15 - x}$

⇒$2x = \sqrt{3}(15-x)$

⇒$2x = 15\sqrt{3} - \sqrt{3}x$

⇒$2x + \sqrt{3}x = 15\sqrt{3} $

⇒$x = \frac{15\sqrt{3}}{2 + \sqrt{3}}$

On rationalising,

⇒$x = \frac{15\sqrt{3}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$

⇒$x = \frac{30\sqrt{3} - 45}{4 - 3} = $

⇒ x = 51.9 - 45 = 6.9m

Hence, the tree broke at the height of 6.9 m.

Q4: If the length of the shadow of a tower is increasing, does then the angle of elevation of the Sun increase?

Answer: No the angle of elevation of the Sun decreases.

Q5: The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

Answer: 

Let h = Height of tower AB

Angle of elevation at C = 30°

Angle of elevation at D = 30 + 15 = 45°

In ∆ADB, tan 45° = $\frac{h}{x}$

⇒ h = x         ...(i)

In ∆ACB, tan 30° = $\frac{h}{20+x}$

From equation (i),

⇒ $\frac{1}{\sqrt{3}} = \frac{h}{20+h}$  

⇒ $20 + h = \sqrt{3}h$ 

⇒ $20 = (\sqrt{3} - 1)h$

⇒ $h = \frac{20}{\sqrt{3} - 1}$

On rationalising,

⇒ $h = \frac{20(\sqrt{3} + 1)}{3 - 1}$

⇒ $h = 10(\sqrt{3} + 1)m$

👉See Also:

CH 9: Applications of Trigonometry (MCQs) or Online Quiz 

Ch 10: Quiz On Circles

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