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.

Let CD be the lighthouse whose height is 75 m. Let the two ships be at A and B such that their angles of depression from D are 30° and 45° respectively.

Let AB = x m and BC = y m
In right triangle BCD, we have

$tan space 45 degree space equals space CD over BC rightwards double arrow space space space 1 space equals space 75 over straight y rightwards double arrow space space space straight y space equals space 75 space straight m space space space space.... left parenthesis straight i right parenthesis$
In right triangle ACD, we have
$tan space 30 degree space equals space CD over AC rightwards double arrow space space tan space 30 degree space equals space fraction numerator CD over denominator AB plus BC end fraction rightwards double arrow space space space fraction numerator 1 over denominator square root of 3 end fraction equals fraction numerator 75 over denominator straight x plus straight y end fraction rightwards double arrow space space space straight x plus straight y space equals space 75 square root of 3 rightwards double arrow space space space straight y space equals space left parenthesis 75 square root of 3 minus straight x right parenthesis space straight m space space space space space.... left parenthesis ii right parenthesis$
Comparing (i) and (ii) we get

$75 space equals space 75 square root of 3 minus straight x rightwards double arrow space space straight x space equals space 75 square root of 3 minus 75 rightwards double arrow space space space space space equals space 75 space left parenthesis square root of 3 minus 1 right parenthesis space straight m$
Hence, the distane between two ships be $equals space 75 space left parenthesis square root of 3 minus 1 right parenthesis space space straight m.$

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