Question

as observed from the top of a 100 M high Lighthouse from the sea level the angle of depression of two ships are 30 degree and 40 degree if once it is exactly behind the other on the same side of the Lighthouse find the distance between the two ships use under root 3 is equal to 1.7 32

Answer 1

The distance between the two ships is 54 meter.

Explanation

According to below diagram, AB is the height of the lighthouse which is 100 meter. C and D are the positions of two ships, for which the angles of depression from the top of the lighthouse are 40° and 30° respectively.

The distance between two ships is CD.

In right angle triangle ABC....

$tan(40)=\frac{AB}{AC}\\ \\ 0.839=\frac{100}{AC}\\ \\ AC= \frac{100}{0.839}=119.2$

Now, in right angle triangle ABD.....

$tan(30)=\frac{AB}{AD}\\ \\ \frac{1}{\sqrt3} =\frac{100}{AD} \\ \\AD= 100\sqrt3 = 100*1.732=173.2$

So,  $CD = AD - AC = (173.2 -119.2)meter = 54 meter$

Thus, the distance between the two ships is 54 meter.

Answer 2

Find the distance BC:

tan θ = opp/adj

tan (50) = BC/100

BC = 100tan(50)

Find the distance BD:

tan θ = opp/adj

tan(60) = BD/100

BD = 100 tan (60)

Find the distance between the two ships:

Distance = 100 tan(60) - 100 tan (50)

Distance = 54.03 m

Answer: The distance between the two ships is 54.03 m