$A$ tower $B C D$ surmounted by a spire $D E$
stands on a horizontal plane. At the extremity $A$ of a horizontal
line $B A$ it is found that $B C$ and $D E$ subtend equal
angles. If $B C = 3 m, C D = 28 m$ and $D E = 5 m$ , then $B A$ is
equal to

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$\sqrt{18 \times 93}$

$\sqrt{36 \times 93}$

$\sqrt{34 \times 93}$

$\sqrt{34 \times 36}$

answer is A.

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Detailed Solution

Let $∠ B A C = ∠ D A E = θ$

$∠ D A B = α, A B = x . (Fig. 27.26)$ Now

Now $B C = 3 m, C D = 28 m, D E = 5 m .$

$\begin{array}{l} ∴ \tan (α + θ) = 36 / x \\ \tan (α) = 31 / x, \tan θ = 3 / x \end{array}$

So $\tan (α + θ) = \frac{\tan α + \tan θ}{1 - \tan α \tan θ}$

$\begin{array}{l} \Rightarrow \frac{36}{x} = \frac{\frac{31}{x} + \frac{3}{x}}{1 - \frac{31}{x} \times \frac{3}{x}} \\ \Rightarrow 36 (x^{2} - 93) = 34 x^{2} \Rightarrow 2 x^{2} = 36 \times 93 \\ \Rightarrow x^{2} = 18 \times 93 \\ \Rightarrow x = \sqrt{18 \times 93} . \end{array}$