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