From the top of a lighthouse, the angles of depression of two stations on opposite sides and $a$ distance apart are $α$ and $β$ . The height of the lighthouse is:

\frac{a}{cot α cot β}

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\frac{a}{cot α + cot β}

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\frac{a cot α cot β}{cot α + cot β}

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\frac{a tan α tan β}{cot α + cot β}

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Solution

The correct option is B $\frac{a}{cot α + cot β}$
Let Height of Light House be $h = A D$ , as shown in the figure above.
and $∠ A B D = α (G i v e n)$
$∠ A C D = β (G i v e n)$

Now, in $△ A B D, B D = h cot α [cot θ = \frac{Adjacent Side}{Opposite Side}]$

Also, in $△ A D C, C D = h cot β [cot θ = \frac{Adjacent Side}{Opposite Side}]$

Given that, $B C = a$

$∴ a = B C = B D + C D$
$\Rightarrow a = h cot α + h cot β$
$\Rightarrow a = h (cot α + cot β)$
$∴ h = \frac{a}{cot α + cot β}$

Hence, option B is correct.

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From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be $α$ and $β$ . If the height of the lighthouse be h metres and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is $\frac{h (t a n α + t a n β)}{t a n α t a n β}$ metres. [3 MARKS]