Question

In ΔPQR, if 3∠P = 4∠Q = 6∠R, calculate the angles of the triangle.

Answer 1

Given, 3∠P = 4∠Q = 6∠R

Then, ∠P = ${\displaystyle \frac{6}{3}}$∠R = 2∠R

∠Q = ${\displaystyle \frac{6}{4}}$∠R = ${\displaystyle \frac{3}{2}}$∠R

In ΔPQR, ∠P + ∠Q + ∠R = 180°  .....[Angle sum property of a triangle]

⇒ 2∠R + ${\displaystyle \frac{3}{2}}$∠R + ∠R = 180°

⇒ 3∠R + ${\displaystyle \frac{3}{2}}$∠R = 180°

⇒ 6∠R + 3∠R = 180° × 2  ......[On taking LCM in LHS]

⇒ 9∠R = 360°

⇒ ∠R = ${\displaystyle \frac{{360}^{\circ }}{9}}$ = 40°

∴ ∠P = 2∠R = 2 × 40° = 80°

And ∠Q = ${\displaystyle \frac{3}{2}}$∠R = ${\displaystyle \frac{3}{2}}$ × 40° = 60°

Hence, all the angles of the triangle are 80°, 60° and 40°.