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Height and Distance
Height and Distance
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Question

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

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Solution

In Δ\Delta CDF sin 30 o = z1{z \over 1} ⇒\Rightarrow z = 12{1 \over 2} km cos 30 o = Y1{Y \over 1} ⇒\Rightarrow Y = 32{{\sqrt 3 } \over 2} km Now in Δ\Delta ABC, tan 45 o = hX+Y{h \over {X + Y}} ⇒\Rightarrow h = X + Y ⇒\Rightarrow X = h - 32{{\sqrt 3 } \over 2} Now in Δ\Delta BDE tan 60 o = h−zX{{h - z} \over X} ⇒\Rightarrow 3{\sqrt 3 }X = h - 12{1 \over 2} ⇒\Rightarrow 3(h−32)\sqrt 3 \left( {h - {{\sqrt 3 } \over 2}} \right) = h - 12{1 \over 2} ⇒\Rightarrow h = 13−1{1 \over {\sqrt 3 - 1}} km

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