**Topic**

A person walks along the beach from the position *A* to the location *REMOVE *and observe a ship *OLD* moored offshore. That person took measurements and got the results: \(AB = 30m,\widehat {CAB} = {60^0},\widehat {CBA} = {50^0}\) (Figure 23). Calculate distance from location *A* to the ship *OLD*(round the result to tenths of a meter)?

**Solution method – See details**

Step 1: Calculate the angle measure \(\widehat {ACB}\)

Step 2: Use the sine theorem to calculate the length *AC *by*ABC* then conclude* *

**Detailed explanation**

We have:** **\(\widehat {ACB} = {180^0} – (\widehat {CBA} + \widehat {CAB}) = {70^0}\)

Apply the sine theorem to*ABC* we have: \(\frac{{AC}}{{\sin \widehat {CBA}}} = \frac{{AB}}{{\sin \widehat {ACB}}} \Rightarrow AC = \frac{{ AB.\sin \widehat {CBA}}}{{\sin \widehat {ACB}}} = \frac{{30.\sin {{50}^0}}}{{\sin {{70}^0 }}} \approx 24.5\)

So the distance from the position *A* to the ship *OLD* is 24.5 m