Question

$\huge\pink{Question }$


Starfish swimming around the the first corner of the pool at deft of 90 m and looking at other corner. For the starfish the angle of elevation to the surface of water is 60 and angle of depression to the bottom of the pool is 45°.
How deep the pool is ? ​

Answer 1

$\huge\sf\pink{Answer}$

☞ Your Answer is 60√3 m

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$\huge\sf\blue{Given}$

✭ Dept of fish is 90 m

✭ From the fish the angle elevation to the surface is 60°

✭ Angle of depression to the bottom of the pool is 45°

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$\huge\sf\gray{To \:Find}$

◈ Depth of the pool?

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$\huge\sf\purple{Steps}$

Refer the attachment for the diagram!!

In ∆ABD

➝ $\sf tan\theta = \dfrac{Opposite}{Adjacent}$

➝ $\sf tan \ 60^{\circ} = \dfrac{AD}{BD}$

➝ $\sf tan \ 60^{\circ} = \dfrac{90}{BD}$

$\sf \bigg\lgroup tan \ 60^{\circ} = \sqrt{3} \bigg\rgroup$

➝ $\sf \sqrt{3} = \dfrac{90}{BD}$

➝ $\sf BD \sqrt{3} = 90$

➝ $\sf BD = \dfrac{90}{\sqrt{3}}$

Rationalise the denominator,

➝ $\sf BD = \dfrac{90}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$

➝ $\sf BD = \dfrac{90\sqrt{3}}{3}$

➝ $\sf \green{BD = 30\sqrt{3} \ m}$

In ∆ BDC

➳ $\sf tan\theta = \dfrac{Opposite}{Adjacent}$

➳ $\sf tan \ 45^{\circ} = \dfrac{DC}{BD}$

➳ $\sf tan \ 45^{\circ} = \dfrac{DC}{30\sqrt{3}}$

$\sf \bigg\lgroup tan \ 45^{\circ} = 1 \bigg\rgroup$

➳ $\sf 1 = \dfrac{DC}{30\sqrt{3}}$

➳ $\sf \red{DC = 30\sqrt{3}}$

So then the dept will be,

»» $\sf Depth = BD+DC$

»» $\sf Depth = 30\sqrt{3} + 30\sqrt{3}$

»» $\sf \orange{Depth = 60\sqrt{3} \ m}$

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