Let AB be a tower of height = 30 m. Two points on opposite sides are C and D. ∠ACB = 60°, ∠ADB = 45° Distance BC = x m. and BD = y m From ΔABD, tan(45°)=
AB
BD
=
30
y
1=
30
y
∴ BD = y = 30 m From ΔABC, tan(60°)=
AB
BC
=
30
x
√3=
30
x
⇒x=10√3m Distance between two points CD = BC + BD =x+y =(30+10√3)m =47.32m