The angle of elevation of the top of a tower a point A due south of it is `30^@` and from a point B due west of it is `45^@` .If the height of the tower is 100 meters ,then find the distance AB.

Options

A. 400
B. 50
C. 100
D. 200

Answer

D

Explanation

In the following figure OP is the tower .

Op= 100 m
In right angled triangle :POA ,
`tan 30 ^@ = 100 /(OA)`
`therefore OA= 100 cot 30 ^@ = 100 sqrt(3)`
In right angled triangle POB,
`tan 45 ^@=(100)/(OB)`
`therefore OB = 100 cot 45 ^@= 100`
In right angled triangle AOB, using Pythagoras theorem ,we get
`AB^2=OA^2+ OB^2`
`= 3xx 100 ^2+100^2`
`=4xx 100^2`
`therefore AB = 2 xx 100 = 200 m `