Answer:
(a) 463.13 km, approximately
(b) 15.09° north of west, approximately
(c) Is only approximately correct because its calculation involves trigonometric functions and square roots that, in this case, cannot been simplified, but rounded.
Explanation:
By depicting a triangle ABC,
(a) the length AC is given by [tex]\sqrt[2]{AB^2+BC^2-2*AB*BC*\cos{\theta_B}} =\sqrt[2]{200^2-275^2-2*200*275*\cos{154}}\approx463.13[/tex]
(b) [tex]\theta_A=\asin{\frac{BC}{AC}\sin{\theta_B}}=\asin{\frac{275}{463.13}\sin{154}}\approx 15.09\deg[/tex]
