Question

In a triangle PQR, ∠P + ∠Q = 130° and ∠P + ∠R = 120°. Calculate each angle of the triangle.

Answer 1

In ΔPQR,
∠P + ∠Q = 130°              ....(given)
Now, ∠P + ∠Q = ∠PRY ....(Exterior angle property)
⇒ ∠PRY = 130°
∠PRY + ∠R = 180°         ....(Linear pair)
⇒ 130° + ∠R = 180°
⇒ ∠R = 180° - 130° = 50°

Also, ∠P + ∠R = 120°     ....(given)
Now, ∠P + ∠R = ∠PQX ....(Exeterior angle property)
⇒ ∠PQx = 120°
∠PQX +∠Q = 180°       ....(Linear pair)
⇒ 120°+ ∠Q = 180°
⇒ ∠Q = 180° - 120° = 60°

In ΔPQR,
∠P + ∠Q + ∠R = 180°     ....(Angle sum property of a triangle)
⇒ ∠P + 60° + 50° = 180°
⇒ ∠P = 180° - 110° = 70°

Thus, the angles of ΔPQR are as follows:
∠P = 70°, ∠Q = 60° and ∠R = 50°.