Question

Find the approximate length of AB from the given image. (Take $\sqrt{3}=1.73$)

• A. 25 m
• B. 26 m
• C. 27 m
• D. 30 m

Answer 1

The correct option is B 26 m

Let the height of tower is h m.
In $\Delta$ ABC.
$tan{60}^{\circ }=\frac{BC}{AB}=\frac{(x+30)}{h}$
$\sqrt{3}=\frac{x+30}{h}\Rightarrow x=\sqrt{3}h-30$ -(1)
In $\Delta$ ABO,
$tan{30}^{\circ }=\frac{BD}{AB}=\frac{x}{h}$
$\frac{1}{\sqrt{3}}=\frac{x}{h}\Rightarrow h=\sqrt{3}x$
Replace h in (1)
$x=\sqrt{3}\left(\sqrt{3}x\right)-30$
$x=3x-30\Rightarrow x=15m$
$h=\sqrt{3}\times 15=15\times 1.73m$
$h=26m$ (approximately)