Question

plz sol it friends..​

Answer 1

PROCESS

LET the height is h m

now AB=h m

now

$\tan(30) = \frac{h}{bd} \\ bc = \sqrt{3} \: h \: m \\ now \\ bc = ( \sqrt{3} h - 40)m$

now

$\tan(60) = \frac{h}{ \sqrt{3}h - 40 } \\ \sqrt{3} \times ( \sqrt{3} h - 40) = h \\ 3h - 40 \sqrt{3} = h \\ 2h = 40 \sqrt{3} \\ h = 20 \sqrt{3} \: m \\ therefore \: the \: height \: of \: the \: tower \: is \: = 20 \sqrt{3} \: m$