Question

A vertical pole is 7√3 high and the length of its shadow is 21m. Find the angle of elevation of the source of light.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Angle\:of\:elevation=30\degree}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies Height\: of \: pole = 7\sqrt{3} \: m \\ \\ \tt: \implies Shadow \: of \: pole = 21 \: m \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Angle \: of \: elevation = ?$

• According to given question :

$\circ \: \tt{Let \: angle \: of \: elevation \: of \: \: pole \: be\: \alpha } \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies tan \: \alpha = \frac{Perpendicular}{Base} \\ \\ \tt: \implies tan \: \alpha = \frac{7\sqrt{3}}{21} \\ \\ \tt: \implies tan \: \alpha = \frac{\sqrt{3}}{3}$

$\tt: \implies tan \: \alpha = \frac{ \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } \\ \\ \tt: \implies tan \: \alpha = \frac{1}{ \sqrt{3} } \\ \\ \tt: \implies \alpha = {tan}^{ - 1} ( \frac{1}{ \sqrt{3} } ) \\ \\ \tt \circ \: tan \: 30 \degree = \frac{1}{ \sqrt{3} } \\ \\ \green{\tt: \implies \alpha = 30 \degree} \\ \\ \green{\tt \therefore Angle \: of \: elevation \: of \: pole \: to \: ground \: is \: 30 \degree}$

Answer 2

Step-by-step explanation:

HEY THERE!!! (Lovely Brother )

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\huge{\bold{SOLUTION:-}}SOLUTION:−

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

☔ Let ∠ACB =\thetaθ

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!

$</p><p>\begin{lgathered}\sf{ \frac{ \sqrt{3} }{3} \times \frac{ \sqrt{3} }{ \sqrt{3} } } \\ \\ \\ \implies \: \frac{3}{3 \sqrt{3} } \\ \\ \\ \implies \: \frac{ \cancel3}{ \cancel3 \sqrt{3} } \\ \\ \implies \: \frac{1}{ \sqrt{3} }\end{lgathered}.$

☔ Here, tan = 1/√3

➡ tan ∅ = 30°

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

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