Question

The angle of elevation of the top of a
mountain from the top of a 100 m high
hill is 30° and from the base of the hill
is 45°. Find the height of the mountain.​

Answer 1

Solution:-

:- Take ∆ABC

$\rm \to \tan45 \degree = \dfrac{BC}{AB}$

$\rm \to \: 1 = \dfrac{h}{x}$

$\rm \to \: h = x$

:- Take ∆EDC

$\rm \to \: \tan30 \degree = \dfrac{CD}{ED}$

$\rm \to \dfrac{1}{ \sqrt{3} } = \dfrac{h - 100m}{x}$

Now put the value h = x

$\rm \to \dfrac{1}{ \sqrt{3} } = \dfrac{h - 100m}{h}$

Now using cross multiplication

$\rm \to \: h = \sqrt{3} (h - 100m)$

$\rm \to \:h = \sqrt{3} h - 100 \sqrt{3}$

$\rm \to \: 100 \sqrt{3} = \sqrt{3} h - h$

$\rm \to \: 100 \sqrt{3} = ( \sqrt{3} - 1)h$

$\rm \to \: \dfrac{100 \sqrt{3} }{ \sqrt{3} - 1} = h$

Now take √3 = 1.732

$\rm \to \: h = \dfrac{100 \times 1.732}{1.732 - 1}$

$\rm \to \: h = \dfrac{173}{0.732} cm$

$\rm \to \: h \: = 236.9 \approx237m$

High of mountain is 237 m