Question

The shadow of a tower standing on level ground is found to be 40 m longer when Sun’s altitude is 30° than when it was 60°. Find the height of the tower.

Answer 1

When the sun’s altitude is the angle of elevation of the top of the tower from the tip of the shadow.

Let AB be h m and BC be x m. From the question, DB is 40 m longer than BC.

So, BD = (40 + x) m

And two right triangles ABC and ABD are formed.

In ΔABC,

tan 60^o = AB/ BC

√3 = h/x

x = h/√3 … (i)

In ΔABD,

tan 30^o = AB/ BD

1/ √3 = h/ (x + 40)

x + 40 = √3h

h/√3 + 40 = √3h [using (i)]

h + 40√3 = 3h

2h = 40√3

h = 20√3

Therefore, the height of the tower is 20√3 m.