Question

From the top of a cliff ,150 m high, the angles of depression of two boats are 60 degree and 30 degree. find the distance between the two boats if the boats are on the same side of the cliffs

Answer 1

Given:

• A cliff, 150 m high.
• The angles of depression of two boats being 60° and 30°.

To find: The distance between the boats.

Answer:

(Diagram for reference attached below.)

$\sf (\ tan\ \theta\ =\ \dfrac{Opposite}{Base}\ )$

In triangle ABC,

$\sf tan\ {60}^{\circ}\ =\ \dfrac{AB}{BC}\\ \\\\(tan\ {60}^{\circ}\ =\ \sqrt{3})\\ \\\\\sqrt{3}\ =\ \dfrac{150}{BC}\\\\\\BC\ =\ 150\sqrt{3}.$

In triangle ABD,

$\sf tan\ {30}^{\circ}\ =\ \dfrac{AB}{BD}\\ \\\\\\(tan\ {30}^{\circ}\ =\ \dfrac{1}{\sqrt{3}})\\\\\\\\\dfrac{1}{\sqrt{3}}\ =\ \dfrac{150}{BC\ +\ CD}\\\\\\\\\dfrac{1}{\sqrt{3}}\ =\ \dfrac{150}{50\sqrt{3} \ +\ CD}\\\\\\\\50\sqrt{3}\ +\ CD\ =\ 150\sqrt{3}\\ \\\sf CD\ \ =\ \sf 150\sqrt{3}\ -\ \sf 50\sqrt{3}\\\\\\CD\ =\ \sqrt{3} \bigg(150\ -\ 50\bigg)\\\\\\CD\ =\ \sqrt{3}(100)$

Therefore, the distance between the two ships is 100√3 m.