A) 25m |
B) \( \Large 25\sqrt{3} \) m |
C) \( \Large \frac{25}{\sqrt{3}} \) |
D) 30m |
B) \( \Large 25\sqrt{3} \) m |
Let h be the height BC = x m.
In \( \Large \triangle BCA \),
\( \Large \tan 60 ^{\circ} = \frac{h}{x} => \sqrt{3} = \frac{h}{x} \)
\( \Large h = x\sqrt{3} \) ...(i)
Now in triangle ABD, \( \Large tan 30^{\circ} = \frac{h}{50 + x} \)
=> \( \Large \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{50+x} \)
\( \Large \left[ \because h = x\sqrt{3} \right] \)
\( \Large => 50 + x = 3x => x = 25 m \)
\( \Large \therefore\ h = 25\sqrt{3} m \) [from Eq. (i)]