Question

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships

Answer 1

Given Data

• Height of lighthouse (hh) = 75 m

• Angle of depression to ship 1 (θ1θ1) = 30°

• Angle of depression to ship 2 (θ2θ2) = 45°

• Both ships are on the same straight line, one behind the other, on the same side of the lighthouse.

Approach

Let the base of the lighthouse be point DD, the top be AA, the nearer ship (angle 45°) be at CC, and the farther ship (angle 30°) be at BB. Both ships are at sea level.

Let the horizontal distance from the base DD to CC be xx, and from DD to BB be yy. We need to find the distance BC=y−xBC=y−x.

Using the tangent of the angle of depression (which equals the angle of elevation from the ship to the top of the lighthouse):

tan⁡(θ)=oppositeadjacent=hdistance from basetan(θ)=adjacentopposite=distance from baseh

So,

• For ship at CC (45°):

  tan⁡(45∘)=75x  ⟹  x=75tan(45∘)=x75⟹x=75

• For ship at BB (30°):

  tan⁡(30∘)=75y  ⟹  y=75tan⁡(30∘)tan(30∘)=y75⟹y=tan(30∘)75tan⁡(30∘)=13  ⟹  y=75×3tan(30∘)=31⟹y=75×3

Distance Between the Two Ships

BC=y−x=753−75=75(3−1) mBC=y−x=753−75=75(3−1) m

Final Answer

The distance between the two ships is:

75(3−1) meters75(3−1) meters