## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Test - Page 354: 19

#### Answer

The plane's airspeed is 220 mph and the bearing is $357^{\circ}$

#### Work Step by Step

We can find the plane's resultant speed $a$ as it heads north: $a = \frac{630~mi}{3.0~hr} = 210~mph$ Let $b$ be the wind's speed. Then $b = 15~mph$ The angle between these two vectors is $C = 90^{\circ}+48^{\circ}$ which is $C = 138^{\circ}$ We can use the law of cosines to find $c$, the plane's airspeed: $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(210~mph)^2+(15~mph)^2-(2)(210~mph)(15~mph)~cos~138^{\circ}}$ $c = \sqrt{49006.8~mph^2}$ $c = 220~mph$ We can use the law of sines to find angle $B$, which is the angle the airspeed vector makes with the vertical: $\frac{c}{sin~C} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~C}{c}$ $B = arcsin(\frac{b~sin~C}{c})$ $B = arcsin(\frac{15~sin~138^{\circ}}{220})$ $B = arcsin(0.0456)$ $B = 3^{\circ}$ The plane's bearing is $360^{\circ}-3^{\circ}$ which is $357^{\circ}$ The plane's airspeed is 220 mph and the bearing is $357^{\circ}$

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