3) In APQR,< R=<Q +9°,<Q = (3x - 2)<P = xº,find < P,<Q,< R of triang

1.	3) In APQR,< R=<Q +9°,<Q = (3x - 2)<P = xº,find < P,<Q,< R of triangle PQR
Answer» Given, <R = <Q + 9, <Q = (3x - 2), <P = x <R = (3x - 2) + 9 = (3x + 7) Using angle sum property of triangle <P + <Q + <R = 180x + (3x - 2) + (3x + 7) = 1807x + 5 = 1807x = 175x = 175/7 = 25 Therefore, angles are<P = x = 25°<Q = 3x - 2 = 325 - 2 = 73°<R = 3x + 7 = 325 + 7 = 82°

3) In APQR,< R=<Q +9°,<Q = (3x - 2)<P = xº,find < P,<Q,< R of triangle PQR

Answer»

Given, <R = <Q + 9, <Q = (3x - 2), <P = x

<R = (3x - 2) + 9 = (3x + 7)

Using angle sum property of triangle

<P + <Q + <R = 180x + (3x - 2) + (3x + 7) = 1807x + 5 = 1807x = 175x = 175/7 = 25

Therefore, angles are<P = x = 25°<Q = 3x - 2 = 3*25 - 2 = 73°<R = 3x + 7 = 3*25 + 7 = 82°