Answer to Question #136716 in Physics for Nielle

Question

A car transports its passengers between 3 buildings. It moves from the first building to the second building, 4.76km away, in a direction 37° north of east. It then moves from second building to the third building in a direction 69° west of north. Finally, it returns to the first building, sailing in a direction 28° east of south. Calculate the distance between (a) the second and third buildings and (b) the first and third buildings.

Answer 1

$$\alpha=28+(90-37)=81\degree\\\beta=28+(90-69)=49\degree\\\gamma=180-81-49=50\degree$$

a)

$$\frac{c}{\sin{\gamma}}=\frac{a}{\sin{\alpha}}\\\frac{4.76}{\sin{50}}=\frac{a}{\sin{81}}\\a=6.14\ km$$

b)

$$\frac{c}{\sin{\gamma}}=\frac{b}{\sin{\beta}}\\\frac{4.76}{\sin{50}}=\frac{b}{\sin{49}}\\b=4.69\ km$$