If the angle of elevation of a cloud from a point ‘h’ metres above a lake is $θ_{1}$ and the angle of depression of its reflection in the lake is $θ_{2}$ . Find the height at which the cloud is located from the ground.

\frac{h (t a n θ_{1} + t a n θ_{2})}{t a n θ_{2} - t a n θ_{1}}

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\frac{h (t a n θ_{1} - t a n θ_{2})}{t a n θ_{2} + t a n θ_{1}}

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\frac{h (t a n θ_{1} - t a n θ_{2})}{t a n θ_{2} - t a n θ_{1}}

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\frac{h (t a n θ_{1} + t a n θ_{2})}{t a n θ_{1} - t a n θ_{2}}

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Solution

The correct option is A $\frac{h (t a n θ_{1} + t a n θ_{2})}{t a n θ_{2} - t a n θ_{1}}$
Let P be the cloud and Q be its reflection.
Let A be the point of observation such that AB = h
Let the height of the cloud be x. (PS = x)
PR = x – h and QR = x + h

Let AR = y
In the right ∆ARP,
$t a n θ_{1} = \frac{P R}{A R} = \frac{x - h}{y} \dots . (1)$
In the ∆ AQR, $t a n θ_{2} = \frac{Q R}{A R} = \frac{x + h}{y} \dots . (2)$

Add (1) and (2),
$t a n θ_{1} + t a n θ_{2} = \frac{x - h}{y} + \frac{x + h}{y} = \frac{2 x}{y}$

Subtracting (1) from (2),
$t a n θ_{2} - t a n θ_{1} = \frac{x + h}{y} - \frac{x - h}{y} = \frac{2 h}{y}$
$\frac{t a n θ 1 + t a n θ 2}{t a n θ 2 - t a n θ 1} = \frac{2 x}{y} \div \frac{2 h}{y} = \frac{2 x}{y} \times \frac{y}{2 h} = \frac{x}{h}$
$∴ x = \frac{h (t a n θ_{1} + t a n θ_{2})}{t a n θ_{2} - t a n θ_{1}}$

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If the angle of elevation of a cloud from a point ‘h’ metres above a lake is θ1 and the angle of depression of its reflection in the lake is θ2. Find the height at which the cloud is located from the ground.

If the angle of elevation of a cloud from a point ‘h’ metres above a lake is $θ_{1}$ and the angle of depression of its reflection in the lake is $θ_{2}$ . Find the height at which the cloud is located from the ground.