From the top of a lighthouse 70 m high with its base at sea level, the angle of depression of a boat is `15^(@)`. The distance of the boat from the foot of the lighthouse is :

Options

A. `70(2-sqrt(3))m`
B. `70(2+sqrt(3))m`
C. `70(3-sqrt(3))m`
D. `70(3+sqrt(3))m`

Answer

B

Explanation

Let AB = lighthouse = 70 m
Let BC = distance of boat from lighthouse
i.e. BC=x m
`tan 15^(@)=(70)/(x) " " ...(i)`
`tan 15^(@)=tan [45^(@)-30^(@)]`
`=(tan 45^(@)-tan 30^(@))/(1+tan 45^(@) tan 306(@))=(1-(1)/(sqrt(3)))/(1+(1)/(sqrt(3)))`
`=(sqrt(3)-1)/(sqrt(3)+1)=((sqrt(3)-1)^(2))/(2)=(4-2sqrt(3))/(2)`
`tan 45^(@)= 2 - sqrt(3)`
From (i)
`x=(70)/(2-sqrt(3))xx(2+sqrt(3))/(2+sqrt(3))=70(2+sqrt(3))m`