Answer:
Quadratic
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Translate the words to English please<span />
Since sin(2x)=2sinxcosx, we can plug that in to get sin(4x)=2sin(2x)cos(2x)=2*2sinxcosxcos(2x)=4sinxcosxcos(2x). Since cos(2x) = cos^2x-sin^2x, we plug that in. In addition, cos4x=cos^2(2x)-sin^2(2x). Next, since cos^2x=(1+cos(2x))/2 and sin^2x= (1-cos(2x))/2, we plug those in to end up with 4sinxcosxcos(2x)-((1+cos(2x))/2-(1-cos(2x))/2)
=4sinxcosxcos(2x)-(2cos(2x)/2)=4sinxcosxcos(2x)-cos(2x)
=cos(2x)*(4sinxcosx-1). Since sinxcosx=sin(2x), we plug that back in to end up with cos(2x)*(4sin(2x)-1)
Answer: The plane is 6,547 feet from the closer edge of the runway.
This is a trigonometry problem that will involve setting up 2 different triangles. One going to the closer edge of the runway and the other going to the far end of the runway.
The constant between the 2 triangles is the height of the plane. You can solve for this is both triangles and set them equal to each other.
Do that will give you the equation:
x/tan(70) = (x+3000)/tan(76) where x is the distance to the closer end of the runway.
If you solve that equation, you will get x = 6547 feet.
The answer is 5.7 because you can only round down not up
