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.