the angle of elevation of the …

Let AB be the mountain of height h kilometres. Let C be a point at a distance of x km. from the base of the mountain such that the angle of elevation of the top at C is 30^o. Let D be a point at a distance of 10 km from C such that the angle of elevation at D is of 15^o.

In {tex}\triangle C A B,{/tex} we have
{tex}\tan 30 ^ { \circ } = \frac { A B } { A C }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { h } { x }{/tex}
{tex}\Rightarrow \quad x = \sqrt { 3 } h{/tex}  ...(i)
In {tex}\triangle D A B,{/tex} we have
{tex}\tan 15 ^ { \circ } = \frac { A B } { A D }{/tex}
{tex}\Rightarrow \quad 0.27 = \frac { h } { x + 10 }{/tex}
{tex}\Rightarrow{/tex} (0.27) (x +1) = h  ...(ii)
Substituting x = {tex}\sqrt { 3 }{/tex}h obtained from equation (i) in equation (ii), we get
{tex}0.27 ( \sqrt { 3 } h + 10 ) = h{/tex}
{tex}\Rightarrow \quad 0.27 \times 10 = h - 0.27 \times \sqrt { 3 } h{/tex}
{tex}\Rightarrow \quad h ( 1 - 0.27 \times \sqrt { 3 } ) = 2.7{/tex}
{tex}\Rightarrow{/tex} h (1 - 0.46) = 2.7
{tex}\Rightarrow \quad h = \frac { 2.7 } { 0.54 } = 5{/tex}
Hence, the height of the mountain is 5 km.