Question

The angle of elevation of the top of a 100 m high tree from two points A and B on the opposite side of the tree are 52° and 45° respectively. Find the distance AB, to the nearest metre.

Answer 1

In ΔADC,

tan 52° = `"DC"/"AC" = 100/"AC"`

`\implies` 1.2799 = `100/"AC"`   ...(From table)

`\implies` AC = `100/1.2799`

`\implies` AC = 78.13 m

In ΔBCD,

tan 45° = `"CD"/"BC"`

`\implies` 1 = `100/"BC"`

BC = 100 m

∴ AB = AC + BC

= 78.13 + 100

= 178.13 m

Hence, the distance AB is 178 m ...(approx)