Question

The height of the vertical pole is √3 times the length of its shadow on the ground, then angle of elevation of the sun at that time is
(a)30º
(b)60º
(b)45º
(b)75º

Answer 1

$\huge\bigstar\mathfrak\blue{\underline{\underline{SOLUTION:}}}$

Let the height of tower be h

then the height of shadow will be √3h

In ∆ABC,

$tan \angle ACB = \frac{h}{ \sqrt{3h} } \\ \\ = > tan \angle ACB = \frac{1}{ \sqrt{3} } \\ \\ = > \angle ACB = 30 \degree$

Therefore,

Angle of elevation is 30°

Option (a)✓ 30°

hope it helps ☺️

Answer 2

Step-by-step explanation:

Let the height of tower be h

then the height of shadow will be √3h

In ∆ABC,

\begin{lgathered}tan \angle ACB = \frac{h}{ \sqrt{3h} } \\ \\ = > tan \angle ACB = \frac{1}{ \sqrt{3} } \\ \\ = > \angle ACB = 30 \degree\end{lgathered}

tan∠ACB=

3h

h

=>tan∠ACB=

3

1

=>∠ACB=30°

Therefore,

Angle of elevation is 30°

Option (a)✓ 30°