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Question
In ΔPQR, if 3∠P = 4∠Q = 6∠R, calculate the angles of the triangle.
Sum
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Solution
Given, 3∠P = 4∠Q = 6∠R
Then, ∠P = `6/3`∠R = 2∠R
∠Q = `6/4`∠R = `3/2`∠R

In ΔPQR,
∠P + ∠Q + ∠R = 180° ...[Angle sum property of a triangle]
⇒ 2∠R + `3/2`∠R + ∠R = 180°
⇒ 3∠R + `3/2`∠R = 180°
⇒ 6∠R + 3∠R = 180° × 2 ...[On taking LCM in LHS]
⇒ 9∠R = 360°
⇒ ∠R = `360^circ/9` = 40°
∴ ∠P = 2∠R = 2 × 40° = 80°
And ∠Q = `3/2`∠R = `3/2` × 40° = 60°
Hence, all the angles of the triangle are 80°, 60° and 40°.
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