In a prison wall there is a window of $1$ metre height, $24$ metres from the ground. An observer at a height of $10 m$ from ground, standing at a distance from the wall finds the angle of elevation of the top of the window and the top of the wall to be $45^{0}$ and $60^{0}$ respectively. The height of the wall above the window is

15 \sqrt{3}

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15 (1 - \frac{1}{\sqrt{3}})

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15 (\sqrt{3} - 1)

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14 (\sqrt{3} - 1)

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Solution

The correct option is D $14 (\sqrt{3} - 1)$
We can see that , $C E = D F = 10$

Therefore, $B C = B E - C E = 24 - 10 = 14$

in $△ B C D$
$tan 45^{o} = \frac{14}{b} \Rightarrow b = 14$

Now, In $△ A C D$
$tan 60^{o} = \frac{a + 14}{14} \Rightarrow \sqrt{3} = \frac{a + 14}{14} \Rightarrow a = 14 (\sqrt{3} - 1)$

Therefore, Answer is $14 (\sqrt{3} - 1)$

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The correct option is D 14(√3−1)We can see that , CE=DF=10Therefore, BC=BE−CE=24−10=14in △BCD tan45o=14b⇒b=14Now, In △ACD tan60o=a+1414⇒√3=a+1414⇒a=14(√3−1)Therefore, Answer is 14(√3−1)