Question

ac is a straight shore line 5 km long. b is a boat off at sea. find the shortest distance from the boat to the shore.

Explanation:

Step1: Let the shortest distance be \( h \), and let \( AD = x \), then \( DC = 5 - x \).

In right triangle \( ABD \), \( \tan 60^\circ=\frac{h}{x}\), so \( x = \frac{h}{\tan 60^\circ}=\frac{h}{\sqrt{3}} \).
In right triangle \( CBD \), \( \tan 45^\circ=\frac{h}{5 - x}\), so \( 5 - x = h \), then \( x = 5 - h \).

Step2: Equate the two expressions for \( x \).

\( \frac{h}{\sqrt{3}}=5 - h \)
\( h = 5\sqrt{3}-h\sqrt{3} \)
\( h + h\sqrt{3}=5\sqrt{3} \)
\( h(1 + \sqrt{3})=5\sqrt{3} \)
\( h=\frac{5\sqrt{3}}{1 + \sqrt{3}} \)
Rationalize the denominator:
\( h=\frac{5\sqrt{3}( \sqrt{3}-1)}{(1 + \sqrt{3})(\sqrt{3}-1)}=\frac{15 - 5\sqrt{3}}{2}\approx\frac{15 - 8.66}{2}=\frac{6.34}{2}=3.17 \) (km)

Answer:

The shortest distance is \(\frac{15 - 5\sqrt{3}}{2}\text{ km}\) (or approximately \(3.17\text{ km}\))