The angle of elevation of an object on a hill | vedclass

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(B) Let $h$ be the height of the object and $x$ be the distance from the second point to the base of the hill.In $	riangle ACD$,$	an 30^{\circ} = \f

Vedclass · Accepted Answer

$60 \sqrt{3}$

Vedclass · Answer

(B) Let $h$ be the height of the object and $x$ be the distance from the second point to the base of the hill.In $	riangle ACD$,$	an 30^{\circ} = \frac{h}{120 + x}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{120 + x}$ $\Rightarrow 120 + x = h\sqrt{3} \quad \dots(i)$In $	riangle BCD$,$	an 60^{\circ} = \frac{h}{x}$ $\Rightarrow \sqrt{3} = \frac{h}{x}$ $\Rightarrow x = \frac{h}{\sqrt{3}} \quad \dots(ii)$Substituting $(ii)$ into $(i)$:$120 + \frac{h}{\sqrt{3}} = h\sqrt{3}$$120 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left( \frac{3 - 1}{\sqrt{3}} ight) = \frac{2h}{\sqrt{3}}$$h = \frac{120 	imes \sqrt{3}}{2} = 60\sqrt{3} \ m$.

The angle of elevation of an object on a hill is observed from a certain point in the horizontal plane through its base to be $30^{\circ}$. After walking $120 \ m$ towards it on level ground,the angle of elevation is found to be $60^{\circ}$. The height of the object (in metres) is:

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