Question

Construct ∆PQR such that, m$\angle$P = 80^°, m$\angle$Q = 70^°, l(QR) = 5.7 cm.

Answer 1

In Δ PQR,
∠P + ∠Q + ∠R = 180^∘ (Angle sum property)
⇒ 80^∘ + 70^∘ + ∠R = 180^∘
⇒ 150^∘ + ∠R = 180^∘
⇒ ∠R = 180^∘ − 150^∘
= 30^∘

​Steps of constructions:

(1) Draw seg QR of length 5.7 cm.
(2) Draw ray QA such that ∠RQA = 70°.
(3) Draw ray RB such that ∠QRB = 30°.
(4) Name the point of intersection of ray RB and QA as P.

Therefore, △PQR is the required triangle.