SOLUTION: point A(-a,b) is in quadrant ii and lies on the terminal arm of angle in standard position. Point B is the point of the terminal arm of angle and the unit circle centered at (O.O)

Algebra ->  Trigonometry-basics -> SOLUTION: point A(-a,b) is in quadrant ii and lies on the terminal arm of angle in standard position. Point B is the point of the terminal arm of angle and the unit circle centered at (O.O)       Log On


   

Question 759761: point A(-a,b) is in quadrant ii and lies on the terminal arm of angle in standard position. Point B is the point of the terminal arm of angle and the unit circle centered at (O.O) Determine the x-coordinate of B IN TERM OF a and b.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I constructed OAP and OBQ as right triangles.
AP=y and OP=x
The Pythagorean theorem says that OA=sqrt%28x%5E2%2By%5E2%29
Since OAP and OBQ have the same angle at O, and a 90%5Eo angle, they are similar right triangles.
Since they are similar, corresponding sides are proportional.
So OQ%2FOB=OP%2FOA and BQ%2FOB=AP%2FOA
We know the lengths of the sides of OAP (AP, OP, and OA).
We know OB=1 because it is the radius of the unit circle.
We can find the length of sides OQ and BQ:
OQ%2FOB=OP%2FOA means OQ%2F1=x%2Fsqrt%28x%5E2%2By%5E2%29 --> OQ=x%2Fsqrt%28x%5E2%2By%5E2%29
BQ%2FOB=AP%2FOA means BQ%2F1=y%2Fsqrt%28x%5E2%2By%5E2%29 --> BQ=y%2Fsqrt%28x%5E2%2By%5E2%29
The x-coordinate of B is -OQ=highlight%28-x%2Fsqrt%28x%5E2%2By%5E2%29%29.
The y-coordinate of B is BQ=highlight%28y%2Fsqrt%28x%5E2%2By%5E2%29%29.