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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)?

Verified Answer :

notice the picture below

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

thus, use the Law of Cosines

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

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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,

\(b^2=a^2+c^2-2ac\thinspace cosB\)

Substitute the values, we get

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

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

⇒ \(b=134.370157846\sim134miles\)

The distance from plane to airport is 134 miles.

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