Question

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30 and 45 respectively. If the lighthouse is 100 m high, the distance between the two ships is:

Answer 1

Answer:

The distance between the two ships is 273.2 m

Step-by-step explanation:

Let P and Q be the position of two ships and AB be the light house.

Given:

AB = 100 m

In right Δ ABP,

$cot\:45=\frac{BP}{AB}$

$1=\frac{BP}{100}$

$BP=100\:m$

In right Δ ABQ,

$cot\:30=\frac{BP}{AB}$

$\sqrt{3}=\frac{BQ}{100}$

$\sqrt{3}*100=BQ$

$1.732*100=BQ$

$BQ=173.2\:m$

Therefore,

the distance between two ships is

BP+BQ

100+173.2

273.2 m

Answer 2

Answer:

The distance between the two ships is 273.2 m

Step-by-step explanation:

Let P and Q be the position of two ships and AB be the light house.

Given:

AB = 100 m

In right Δ ABP,

In right Δ ABQ,

Therefore,

the distance between two ships is

BP+BQ

100+173.2

273.2 m