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:
173 m
200 m
273 m
300 m
Explanation:
Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100 m, $\angle$ACB = 30° and $\angle$ADB = 45°.
$\dfrac{AB}{AC}$=tan 30°=$\dfrac{1}{\sqrt{3}}\Rightarrow$ AC = AB x $\sqrt{3}$ = 100$\sqrt{3}$ m.
$\dfrac{AB}{AD}$=tan 45°=1 $\Rightarrow$AD = AB = 100 m.
$\therefore$ CD = (AC + AD) = (100$\sqrt{3}$ + 100) m
= 100($\sqrt{3}$ + 1)
= (100 x 2.73) m
= 273 m.
Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100 m, $\angle$ACB = 30° and $\angle$ADB = 45°.
$\dfrac{AB}{AC}$=tan 30°=$\dfrac{1}{\sqrt{3}}\Rightarrow$ AC = AB x $\sqrt{3}$ = 100$\sqrt{3}$ m.
$\dfrac{AB}{AD}$=tan 45°=1 $\Rightarrow$AD = AB = 100 m.
$\therefore$ CD = (AC + AD) = (100$\sqrt{3}$ + 100) m
= 100($\sqrt{3}$ + 1)
= (100 x 2.73) m
= 273 m.