Mathematics Question

Question

Find the missing coordinate of P, using the fact that P lies on the unit circle in the given quadrant.

a) P( ,-15/17) Quadrant IV

b) P(2/5, ) Quadrant I

Answer 1

equation for the unit circle is: x^2 + y^2 = 1

plug in y = 15/17 and solve for x

x^2 = 1 - 225/289 = 64/289

x = sqr(64/289) = 8/17

P( 8/17, -15/17) ignore the negative square root since x >0 in quadrant IV

(8/17)^2 + (-15/17)^2 = 64/17^2 + 225/17^2 = 289/289 = 1

P(2/5, y)

(2/5)^2 + y^2 = 1

y^2 = 1 -4/25 =21/25 or 1-.16 = .84

y = sqr(21/25) = (sqr21)/5 =sqr.84 = about .917

ignore the negative square root as y>0 in Quadrant I

P(.4, .2sqr21) = about (.4, .92)

(.4)^2 + (.2sqr21)^2 = .16 +.04(21) = .16 + .84 = 1