Question

The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be 45° and 30°. Find the distance between the two objects. (Take `sqrt3 = 1.73`)

Options

• 200 m

• 150 m

• 107.5 m

• 73.2 m

Answer 1

73.2 m

Explanation:

Let C and D be the objects and CD be the distance between the objects.

In `Delta "ABC, tan"  45^circ = "AB"/"AC"`

AB = AC = 100 m

In `Delta "ABD, tan"  30^circ = "AB"/"AD"`

`"AD" xx 1/sqrt3 = 100`

`"AD" = 100 xx sqrt3 = 173.2  "m"`

`"CD" = "AD" - "AC"`

`= 173.2 - 100`

`= 73.2  "meters"`