CGPET 2022 Solved Paper

Let AB be the tower of height hm and C and D are the points where the angle of elevation of the top of the tower be 30∘ and 60∘. Also, CD=10m and let AD be xm

In △ABD
tan‌60∘=

AB

AD

⇒√3=

h

x

⇒x=

h

√3

....(i)
Now, in
△ABC,
tan‌30∘=

AB

AC

⇒

1

√3

=

h

x+10

⇒x+10=√3h
⇒

h

√3

+10=√3h ...[From‌Eq.(i)]
⇒h+10√3=3h
⇒2h=10√3
⇒h=5√3
So, height of tower is 5√3m.