Question

( begin{aligned} tan 60^{circ} &=frac{25 sqrt{3}}{a} sqrt{3} &=frac{25 sqrt{3}}{a} therefore & a=25 end{aligned} )
NOW in ( triangle A B D )
( tan x^{circ}=frac{25 sqrt{3}}{(50+25)}=frac{25 sqrt{3}}{75} )
( tan x=frac{sqrt{3}}{3} )
( tan x=frac{1}{sqrt{3}} )
( therefore x^{prime}=30^{circ} operatorname{ans}(2) 30^{circ} )

# 41. The shadow of a standing tower of height 25/3 m is found to be 50 m longer when the sun's elevation changes from 60° to xº. Find the measure of X RRB NTPC (Phase I) 2016 (1) 45° (2) 30 (3) 75º (4) 90°

Solution