A pilot flies 720 miles from Dallas, Texas, to his first destination. After dropping off his cargo, he flies southeast 290 miles to his second destination. If the angle formed by his trip is 125°, what is the distance he will fly from the second destination back to Dallas?

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stokholm stokholm · Answer 1 · 2018-08-19T20:27:22+02:00

Alright, lets get started.

Please refer the diagram I have attached.

Plane flies from point A to point B then point C.

We have a triangle in which two sides AB and BC known and the angle between these two lines.

We could use cosine law to find the third side AC

[tex] b^2 = a^2 + c^2 - 2 a c*cos B [/tex]

[tex] b^2 = 290^2 + 720^2 - 2 * 720 * 290 * cos 125 [/tex]

[tex] b^2 = 84100 + 518400 - 417600 cos 125 [/tex]

[tex] b^2 = 602500 - 417600 * (-0.5735) [/tex]

[tex] b^2 = 602500 + 239493.6 [/tex]

[tex] b^2 = 841993.6 [/tex]

Taking square root both sides

b = 917.6 miles

It means plane has to fly 917.6 miles from the second destination back to Dallas. : Answer

Hope it will help :)

thetdman thetdman · Answer 2 · 2019-10-11T02:02:00+02:00

Answer:

918 miles is the answer

Step-by-step explanation:

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