Question 15 Marks
As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$ If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships (Use $\sqrt{3}=1.732)$
Answer
Let AB be the light house and C & D be positions of ships.
$\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{75}{x+y}$
$\Rightarrow x+y=75 \sqrt{3}$…………..(i)
$\tan 45^{\circ}=1=\frac{75}{y}$
$\Rightarrow y=75$………(ii)
Solving (i) & (ii) to get $x=75(\sqrt{3}-1)$
$\Rightarrow x=75 \times 0.732$
= 54.9m
Distance between the ships is 54.9m
View full question & answer→Let AB be the light house and C & D be positions of ships.
$\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{75}{x+y}$
$\Rightarrow x+y=75 \sqrt{3}$…………..(i)
$\tan 45^{\circ}=1=\frac{75}{y}$
$\Rightarrow y=75$………(ii)
Solving (i) & (ii) to get $x=75(\sqrt{3}-1)$
$\Rightarrow x=75 \times 0.732$
= 54.9m
Distance between the ships is 54.9m
