Question

Question 2
Construct $\Delta$ PQR if PQ=5cm, $\angle PQR={105}^{\circ }and\angle QRP={40}^{\circ }$.

Answer 1

$Given\angle PQR={105}^{\circ }and\angle QRP={40}^{\circ }\text{We know that sum of angles of a triangles is}{180}^{\circ }.\therefore \angle PQR+\angle QRP+\angle QPR={180}^{\circ }\Rightarrow {105}^{\circ }+{40}^{\circ }+\angle QPR={180}^{\circ }\Rightarrow {145}^{\circ }+\angle QPR={180}^{\circ }\Rightarrow \angle QPR={180}^{\circ }-{145}^{\circ }\Rightarrow \angle QPR={35}^{\circ }Toconstruct:\Delta PQRwhere\angle p={35}^{\circ },\angle Q={105}^{\circ }andPQ=5cm\text{Steps of construction:}$

(a) Draw a line segment PQ=5cm.
(b) At point P, draw $\angle XPQ={35}^{\circ }$ with the help of protractor.
(c) At point Q, draw $\angle YQP={105}^{\circ }$ with the help of protractor.
(d) XP and YQ intersect at point R.
It is the required triangle PQR.