The angle of elevation of the summit of a mountain from a point on the ground is 45^.

Question

The angle of elevation of the summit of a
mountain from a point on the ground is 45\^{\circ}.
After climbing up one km towards the summit
at an inclination of 30\^{\circ} from the ground, the
angle of elevation of the summit is found to be
60\^{\circ}. Then the height (in km) of the summit from
the ground is :

Accepted Answer

In $$\Delta $$CDF

sin 30o = $${z \over 1}$$

$$ \Rightarrow $$ z = $${1 \over 2}$$ km

cos 30o = $${Y \over 1}$$

$$ \Rightarrow $$ Y = $${{\sqrt 3 } \over 2}$$ km

Now in $$\Delta $$ABC,

tan 45o = $${h \over {X + Y}}$$

$$ \Rightarrow $$ h = X + Y

$$ \Rightarrow $$ X = h - $${{\sqrt 3 } \over 2}$$

Now in $$\Delta $$BDE

tan 60o = $${{h - z} \over X}$$

$$ \Rightarrow $$ $${\sqrt 3 }$$X = h - $${1 \over 2}$$

$$ \Rightarrow $$ $$\sqrt 3 \left( {h - {{\sqrt 3 } \over 2}} \right)$$ = h - $${1 \over 2}$$

$$ \Rightarrow $$ h = $${1 \over {\sqrt 3  - 1}}$$ km

Solutions · 1