SOLUTION: please help me with this problem 1.) a plane left C and flew in the direction 270degrees for 3hrs. at 279mph and landed at B. a second plane flew from C in the direction 222degree

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Question 202258: please help me with this problem
1.) a plane left C and flew in the direction 270degrees for 3hrs. at 279mph and landed at B. a second plane flew from C in the direction 222degrees30minutes for 4hrs. and landed at A. If A is due south of B, find the distance between the two points and the average speed of the second plane.

kindly please give me
some tips on how to
answer this easily..
thanks a lot..
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
1.) a plane left C and flew in the direction 270degrees for 3hrs. at 279mph
and landed at B.
a second plane flew from C in the direction 222degrees30minutes for 4hrs. and
landed at A.
If A is due south of B, find the distance between the two points and the average speed of the second plane
:
It will help to draw this out, a right triangle ABC, Angle B is a right angle
:
Change 222 deg, 30 min to 222.5 degrees
:
270 - 222.5 = 47.5 degrees is angle C:
:
279 * 3 = 837 mi is the distance C to B flown by the 1st plane
:
Let s = speed of the 2nd plane
then
4s = distance from C to A (also the hypotenuse of the triangle)
:
Use the cosine of 47.5 to find S:
Cos(47.5) = 837/4s
:
.67559 = 837/4s
:
.67559*4s = 837
:
2.70236s = 837
s =
s = 309.73 mph speed of the 2nd plane
:
Find the distance from C to A; (Traveled for 4 hrs at the above speed):
4 * 309.73 = 1238.9 miles from C to A
:
:
Find the distance from A to B (call it d) by using the Sine of 47.5
:
Sin(47.5) =
.73728 =
d = .73728 * 1238.9
d = 913.4 mi from A to B
:
:
Did all this make sense to you?