(B) Let $h$ be the height of the object and $x$ be the distance from the base of the hill to the second observation point.In $\triangle BCD$, $\tan 60

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(B) Let $h$ be the height of the object and $x$ be the distance from the base of the hill to the second observation point.In $\triangle BCD$, $\tan 60^{\circ} = \frac{h}{x}$ $\Rightarrow \sqrt{3} = \frac{h}{x}$ $\Rightarrow x = \frac{h}{\sqrt{3}}$.In $\triangle ACD$, $\tan 30^{\circ} = \frac{h}{120 + x}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{120 + x}$ $\Rightarrow 120 + x = h\sqrt{3}$.Substituting $x = \frac{h}{\sqrt{3}}$ into the equation:$120 + \frac{h}{\sqrt{3}} = h\sqrt{3}$$120 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left( \frac{3 - 1}{\sqrt{3}} \right) = \frac{2h}{\sqrt{3}}$$h = \frac{120 \times \sqrt{3}}{2} = 60\sqrt{3} \ m$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

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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 $60^{\circ}$. Then the height of the object (in metres) is:

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A
$120$
B
$60 \sqrt{3}$
C
$120 \sqrt{3}$
D
$60$

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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 $60^{\circ}$. Then the height of the object (in metres) is:

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$A$ vertical pole subtends an angle $\tan ^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the angles subtended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$, then $(\tan \alpha, \tan \beta)$ is equal to

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