The angles of depression from the top of a li | vedclass

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Question

(C) Let $MA$ be the lighthouse of height $h$ meters. Let $P$ and $Q$ be the positions of the two boats such that $PQ = 60 \, m$.In $	riangle AMP$,$

Vedclass · Accepted Answer

$30(\sqrt{3}+1) \, m$

Vedclass · Answer

(C) Let $MA$ be the lighthouse of height $h$ meters. Let $P$ and $Q$ be the positions of the two boats such that $PQ = 60 \, m$.In $	riangle AMP$,$	an 45^{\circ} = \frac{MA}{MP} \implies 1 = \frac{h}{MP} \implies MP = h$.In $	riangle AMQ$,$	an 30^{\circ} = \frac{MA}{MQ} \implies \frac{1}{\sqrt{3}} = \frac{h}{MP + PQ} \implies \frac{1}{\sqrt{3}} = \frac{h}{h + 60}$.Cross-multiplying,we get $h + 60 = h\sqrt{3}$.Rearranging,$h(\sqrt{3} - 1) = 60$.$h = \frac{60}{\sqrt{3} - 1}$.Rationalizing the denominator,$h = \frac{60(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{60(\sqrt{3} + 1)}{3 - 1} = \frac{60(\sqrt{3} + 1)}{2} = 30(\sqrt{3} + 1) \, m$.

The angles of depression from the top of a lighthouse to two boats,which are $60 \, m$ apart and in the same direction (due east),are $45^{\circ}$ and $30^{\circ}$. The height of the lighthouse is:

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