The Heights of two tower are 180 metres and 60 metres respectively. If the angle of elevation of the top of the first tower from the foot of the second tower is `60^(@)`. Calculate the angle of elevation of the top of the second tower from the foot of the first.

Explanation

Let AB be the first tower and CD be the second tower.
As per question, AB = 180 m and CD = 60 m.
Also,`angleADB = 60^(@) and angleCBD = theta` (let).
`tan 60^(@) = (AB)/(BD)`
or, `" "sqrt(3) = (180)/(BD) " " or, " " BD = (180)/(sqrt(3))= 60sqrt(3)`
Again, from the right-angled triangle BCD we get,
`tan theta = (CD)/(BD)`
or,` " "tan theta = (60)/(60sqrt(3)) " " or, " " tan theta = (1)/(sqrt(3)) " "or, tan theta = tan 30^(@) " " or, " "theta = 30^(@)`
Hence the angle of elevation of the top of the second tower from the foot of the first tower is `30^(@)`.