SOLUTION: Airport B is 320 miles from airport A on a bearing of S40°E. A pilot wishes to fly from A to B, but to avoid a storm must first fly due East at a speed of 210 mph for an hour, and

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Question 1173747: Airport B is 320 miles from airport A on a bearing of S40°E. A pilot wishes to fly from A to B, but to avoid a storm must first fly due East at a speed of 210 mph for an hour, and then from this point (call it C) turns to fly to B. Find the distance, to the nearest mile, and the bearing, to the nearest degree, that the pilot must fly to airport B.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Nobody seems to want this one, here is may effort
:
Airport B is 320 miles from airport A on a bearing of S40°E.
A pilot wishes to fly from A to B, but to avoid a storm must first fly due East at a speed of 210 mph for an hour, and then from this point (call it C) turns to fly to B.
Find the distance, to the nearest mile, and the bearing, to the nearest degree, that the pilot must fly to airport B?
:
drawing this out will help it make sense
In the triangle ABC find the interior angle A: 90 - 40 = 50 degrees.
Find the side opposite which is the dist from C to B, using the law of Cosines
a%5E2+=+b%5E2+%2B+c%5E2+-+2%28b%2Ac%29%2Acos%28A%29
a%5E2+=+210%5E2+%2B+320%5E2+-+2%28210%2A320%29%2Acos%2850%29
Some tedious math
a^2 = 44100+102400 - 2(67200)*.6428
a^2 = 146500 -(134400*.6428)
a^2 = 146500 - 86390.65
a = sqrt%2860109.35%29
a = 245.17 mi from C to B
:
To find the bearing we have to find angle C
Use the law of sines to find C
320%2Fsin%28C%29 = 245%2Fsin%2850%29
cross multiply
245*sin(C) = 320*sin(50)
245*sin(C) = 245
sin(C) = 245%2F245
sin(C) = 1
C = 90 degrees
This is parallel to the north south line, so the direction to B is South (180 degrees)