Question

If the angle of elevation of an object from a point 200 meters above the lake is found to be ${30}^0$ and the angle of depression of its image in the lake is ${45}^0,$ then the height of the object above the lake is

• A. $\frac{200(\sqrt{3}-1)}{(\sqrt{3}+1)}$ meters
• B. $\frac{200(\sqrt{3}-1)}{\sqrt{3}}$ meters
• C. $\frac{200(\sqrt{3}+1)}{\sqrt{3}}$meters
• D. $\frac{200(\sqrt{3}+1)}{(\sqrt{3}-1)}$ meters

Answer 1

The correct option is D $\frac{200(\sqrt{3}+1)}{(\sqrt{3}-1)}$ meters

Let the height of the object above the lake is ${}^{\prime }{h}^{\prime }m.$

Hence, the reflection of object in the lake will be at the same distance ${}^{\prime }{h}^{\prime }m$ from the surface of lake.

Let $AD=BC=xm$

Also, $AB+DC+200m$

$DE=(h-200)$

In, $△ADE,$

$\Rightarrow \tan 30°=\frac{ED}{DA}=\frac{(h-200)}{x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{(h-200)}{x}$

$\Rightarrow x=\sqrt{3}(h-200)⟶\left(1\right)$

In, $△ADF,$

$\Rightarrow \tan 45°=\frac{FD}{DA}=\frac{(h+60)}{x}$

$\Rightarrow 1=\frac{(h+60)}{x}$

$\Rightarrow x=h+200⟶\left(2\right)$

From $\left(1\right)\xi \left(2\right),$ we get

$\Rightarrow h+200=\sqrt{3}h-200\sqrt{3}$

$\Rightarrow \sqrt{3}h-h=200\sqrt{3}+200$

$\Rightarrow (\sqrt{3}-1)h=200(\sqrt{3}+1)$

$\Rightarrow h=\frac{200(\sqrt{3}+1)}{(\sqrt{3}-1)}$

Hence, height of object from the lake is $\frac{200(\sqrt{3}+1)}{(\sqrt{3}-1)}m.$

Hence, the answer is $\frac{200(\sqrt{3}+1)}{(\sqrt{3}-1)}m.$