Question

A vertical pole is $\sf 7\sqrt{3} m$ high and the length of its shadow is $\sf 21 \: m$ . Find the angle of elevation of the source of light.

Provide the figure also.

Answer 1

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$\huge{\bold{SOLUTION:-}}$

☔ Let AB be the pole and AC be it's Shadow.

☔ Let ∠ACB =$\theta$

Then, AB measure = 7√3 metres

☔ And , AC = 21 metres

☔Using Trigonometry Ratio!!

➡Here, Perpendicular and Base Given, So we must be user Tangent!!

➡ tan = Perpendicular/Base

tan = 7√3/21

➡√3/3

Now, Rationalizing it!

$\sf{ \frac{ \sqrt{3} }{3} \times \frac{ \sqrt{3} }{ \sqrt{3} } } \\ \\ \\ \implies \: \frac{3}{3 \sqrt{3} } \\ \\ \\ \implies \: \frac{ \cancel3}{ \cancel3 \sqrt{3} } \\ \\ \implies \: \frac{1}{ \sqrt{3} }$

☔ Here, tan = 1/√3

➡ tan ∅ = 30°

☔ Hence, 30°, it is angle of elevation of the source of light.

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Answer 2

Answer:

Step-by-step explanation:

Let the pole be AB.Let the shadow of pole be BC due to the source of light.

A diagram is framed for the given condition.

According to the question,AB=7✓3mBC=21m.

We know the formula of tangent of an angle of a right angled triangle is ratio of height to the base.

Applying that,we get:

tan theta=1/✓3

Applying tan inverse both sides,we get angle of elevation=30°.

Hope you understood