Solving Lesson 18 Page 80 SBT Math 10 – Kite>

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 byABC then conclude

Detailed explanation

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

Apply the sine theorem toABC 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