Question

two boats start their journey from the same point a and travel along directions ac and ad, as shown below:
what is the distance, cd, between the boats?
213.2 ft
115.5 ft
230.9 ft
346.4 ft

Explanation:

Step1: Find length of AC

In right - triangle ABC, $\sin60^{\circ}=\frac{AB}{AC}$. Given $AB = 300$ ft. Since $\sin60^{\circ}=\frac{\sqrt{3}}{2}$, we have $\frac{\sqrt{3}}{2}=\frac{300}{AC}$, so $AC=\frac{300\times2}{\sqrt{3}} = 200\sqrt{3}$ ft.

Step2: Find length of AD

In right - triangle ABD, $\sin30^{\circ}=\frac{AB}{AD}$. Given $AB = 300$ ft. Since $\sin30^{\circ}=\frac{1}{2}$, we have $\frac{1}{2}=\frac{300}{AD}$, so $AD = 600$ ft.

Step3: Use the cosine - law in triangle ACD

In $\triangle ACD$, $\angle CAD=60^{\circ}-30^{\circ}=30^{\circ}$. By the cosine - law, $CD^{2}=AC^{2}+AD^{2}-2\cdot AC\cdot AD\cdot\cos\angle CAD$. Substitute $AC = 200\sqrt{3}$ ft, $AD = 600$ ft and $\cos30^{\circ}=\frac{\sqrt{3}}{2}$ into the formula.
$CD^{2}=(200\sqrt{3})^{2}+600^{2}-2\times200\sqrt{3}\times600\times\frac{\sqrt{3}}{2}$
$CD^{2}=120000 + 360000-360000$
$CD^{2}=120000$
$CD = 200\sqrt{3}\approx346.4$ ft

Answer:

D. 346.4 ft