document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #462216 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!

I constructed OAP and OBQ as right triangles.
\n" ); document.write( " $\"AP=y\"$ and $\"OP=x\"$
\n" ); document.write( "The Pythagorean theorem says that $\"OA=sqrt%28x%5E2%2By%5E2%29\"$
\n" ); document.write( "Since OAP and OBQ have the same angle at O, and a $\"90%5Eo\"$ angle, they are similar right triangles.
\n" ); document.write( "Since they are similar, corresponding sides are proportional.
\n" ); document.write( "So $\"OQ%2FOB=OP%2FOA\"$ and $\"BQ%2FOB=AP%2FOA\"$
\n" ); document.write( "We know the lengths of the sides of OAP (AP, OP, and OA).
\n" ); document.write( "We know $\"OB=1\"$ because it is the radius of the unit circle.
\n" ); document.write( "We can find the length of sides OQ and BQ:
\n" ); document.write( " $\"OQ%2FOB=OP%2FOA\"$ means $\"OQ%2F1=x%2Fsqrt%28x%5E2%2By%5E2%29\"$ --> $\"OQ=x%2Fsqrt%28x%5E2%2By%5E2%29\"$
\n" ); document.write( " $\"BQ%2FOB=AP%2FOA\"$ means $\"BQ%2F1=y%2Fsqrt%28x%5E2%2By%5E2%29\"$ --> $\"BQ=y%2Fsqrt%28x%5E2%2By%5E2%29\"$
\n" ); document.write( "The x-coordinate of B is $\"-OQ=highlight%28-x%2Fsqrt%28x%5E2%2By%5E2%29%29\"$ .
\n" ); document.write( "The y-coordinate of B is $\"BQ=highlight%28y%2Fsqrt%28x%5E2%2By%5E2%29%29\"$ . \n" ); document.write( "

\n" );