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The angles of depression from the top of a light house of two boats are \( \Large 45 ^{\circ} \) and \( \Large 30 ^{\circ} \) towards the west. If the two boats are 5 m apart, then the height of the light house is

A) \( \Large \left(2.5\sqrt{3}-1\right) m \)
B) \( \Large 2.5 \left(\sqrt{3}-1\right) m \)
C) \( \Large \left(2.5\sqrt{3}+1\right) m \)
D) \( \Large 2.5 \left(\sqrt{3}+1\right) m \)

Correct Answer:




D) \( \Large 2.5 \left(\sqrt{3}+1\right) m \)

Description for Correct answer:

Let height of the light house

AB = h

Then, in \( \Large \triangle ABC, \)

\( \Large \tan 45 ^{\circ} = \frac{AB}{BC}=>AB=BC=hz \)

Now, in \( \Large \triangle ABD, \)

\( \Large \frac{1}{\sqrt{3}} = \frac{h}{DC+CB} \)

\( \Large \tan 30 ^{\circ} = \frac{AB}{DB} \)

=> \( \Large \frac{1}{\sqrt{3}} = \frac{h}{DC+CB} \)

=> \( \Large \frac{1}{\sqrt{3}}=\frac{h}{h+5}=>h+5=\sqrt{3}h \)

=> \( \Large h \left(\sqrt{3}-1\right) = 5 \)

=> \( \Large h = \frac{5}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \)

= \( \Large \frac{5}{2} \left(\sqrt{3}+1\right) m = 2.5 \left(\sqrt{3}+1\right) m \)

Part of solved Height and Distance questions and answers : >> Elementary Mathematics >> Height and Distance

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