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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-> 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.
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?
