I need help with my math!!!!

A woman stood on the top of a 500-ft building and measured a 20 degree angle of elevation to an airplane flying at an altitude of 2500-ft. What as the straight-line distance from the woman to the airplane??

I really need help. The book explains it, but I don't understand what the book is trying to say.... Please help!

Think of this like a right triangle (or better yet draw it)

She is 500 ft high and the plane is at 2500 ft the distance between them would be the height of the triangle (2500-500=2000 ft)

now that we have the height and one angle we can use trigonometry to find our answer but before we can do that we need to know what side we are looking for. It's not the height and it's not the base,(the base would be the distance directly out from her) the plane is directly out and above her so it must be the hypotenuse.

now we must decide which equation we us, cos sin or tan.the height is opposite the angle. Now ask yourself what equation involves the opposite and hypotenuse? Sin. This equation is sinx=opposite/hypotenuse

now we plug our numbers into the equation and use a calculator to solve.
The equation becomes sin(20)=2000/hypotenuse
to solve for the hypotenuse it becomes hypotenuse=2000/cos(20)
I don't have a calculator with me so I can't do this right now

sin 20 = 2000/x
x sin 20 = 2000
x = 2000/sin 20
x = 5847.6 ft