Question

As observed from the top of a $75\mathrm{m}$ high lighthouse from the sea level, the angles of the depression of the 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

$\Rightarrow$Let AB be the lighthouse of height '$75\mathrm{m}$'. Let C and D be the positions of the ships.

'$30°$' and '$45°$' are the angles of depression of the two ships.

So, from the figure, $\angle \mathrm{EAD}=30°\mathrm{and}\angle \mathrm{EAC}=45°$

$\Rightarrow \angle \mathrm{EAD}=\angle \mathrm{ADC}=30°$ ($\because$ Alternate Angles)

$\Rightarrow \angle \mathrm{EAC}=\angle \mathrm{ACB}=45°$ ($\because$ Alternate Angles)

$\Rightarrow$Height of lighthouse AB$=75\mathrm{m}$

Step 1:Find the distance BD:

$\Rightarrow$In right$∆\mathrm{ABD}$, $\tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}}$

$\frac{1}{\sqrt{3}}=\frac{75}{\mathrm{BD}}$ ($\because \tan 30°=\frac{1}{\sqrt{3}}$)

$\mathrm{BD}=75\sqrt{3}$

$=75\times 1.732$ ($\because \sqrt{3}=1.732$)

$=129.9\mathrm{m}$

Step 2: Find the distance BC:

$\Rightarrow$In right$∆\mathrm{ABC}$, $\tan 45°=\frac{\mathrm{AB}}{\mathrm{BC}}$

$1=\frac{75}{\mathrm{BC}}$

$\mathrm{BC}=75\mathrm{m}$

Step 3: Subtract BC from BD to get CD:

$\Rightarrow$$\mathrm{CD}=\mathrm{BD}-\mathrm{BC}$

$=129.9-75$

$=54.9\mathrm{m}$

Therefore, the distance between the two ships is $54.9\mathrm{m}$