Question

From the top of a light house 100 m high, the angles of depression of two ships are observed as ${48}^0$ and ${36}^0$ respectively. Find the distance between the two ships (in the nearest metre) if:
(i) the ships are on the same side of the light house.
(ii) the ships are on the opposite sides of the light house.

Given $tan{36}^0=1.1106$ and $tan{48}^0=0.7265$

Answer 1

Let us consider AB to be the lighthouse.
$h=100m$
And, let the two ships be $C$ and $D$ such that $\angle ADB={36}^0$ and $\angle ACB={48}^0$
In $△ABC$.
$\frac{AB}{BC}=tan{48}^0$

$BC=\frac{100}{1.1106}=90.04m$

In $△ABD$,
$\frac{AB}{BD}=tan{36}^0$
$BD=\frac{100}{0.7265}$

$=137.64m$
Now,
(i) If the ships are on the same side of the light house,
Then, the distance between the two ships $=BD-BC$

$=137.64-90.04$

$=47.6m$
(ii) If the ships are on the opposite sides of the light house,
Then, the distance between the two ships $=BD+BC$

$=137.64+90.04$

$=227.68m$