Given∠PQR=105∘and∠QRP=40∘We know that sum of angles of a triangles is 180∘.∴∠PQR+∠QRP+∠QPR=180∘⇒105∘+40∘+∠QPR=180∘⇒145∘+∠QPR=180∘⇒∠QPR=180∘−145∘⇒∠QPR=35∘Toconstruct:ΔPQRwhere∠p=35∘,∠Q=105∘andPQ=5cmSteps of construction:
(a) Draw a line segment PQ=5cm.
(b) At point P, draw ∠XPQ=35∘ with the help of protractor.
(c) At point Q, draw ∠YQP=105∘ with the help of protractor.
(d) XP and YQ intersect at point R.
It is the required triangle PQR.