An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W. How far is the plane from the airport (round to the nearest mile)?

notice the picture below

what we have, is really an angle, encroached by two sides

thus, use the Law of Cosines

[tex]\bf \textit{Law of Cosines}\\ \quad \\ c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\\\ -----------------------------\\\\ \textit{so the distance "c" from the airplane to the airport}\\ \textit{using angle C, and the sides "a" and "b"} \\\\\\ c=\sqrt{150^2+170^2-(2\cdot 150\cdot 170)cos(49.17^o)}\qquad \begin{cases} a=150\\ b=170\\ C=49.17^o \end{cases}[/tex]

SerenaBochenek SerenaBochenek · Answer 2 · 2019-02-21T08:53:38+01:00

Answer:

The distance from plane to airport is 134 miles.

Step-by-step explanation:

Given an airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W. we have to find the distance of plane from the airport.

Given the sides c and a that are

c=150 miles and a=170 miles also ∠B=49.17°

we have to calculate b

By law of cosines,

[tex]b^2=a^2+c^2-2ac\thinspace cosB[/tex]

Substitute the values, we get

[tex]b^2=(170)^2+(150)^2-2\times170\times150cos(49.17^{\circ})[/tex]

⇒ [tex]b^2=51400-51000cos(49.17^{\circ})[/tex]

⇒ [tex]b=134.370157846\sim134miles[/tex]

The distance from plane to airport is 134 miles.