Question

Construct:

$\Delta PQR$ if $PQ=5cm$, $\angle PQR={105}^{\circ }$ and $\angle QRP={40}^{\circ }$.
(Hint : Recall angle-sum property of a triangle),

Answer 1

Given:

$PQ=5cm$
$\angle PQR={105}^{\circ }$ and $\angle QRP={40}^{\circ }$
Since, we have $\angle PQR+\angle QRP+\angle RPQ={180}^{\circ }$
$\Rightarrow {105}^{\circ }+{40}^{\circ }+RPQ={180}^{\circ }$
$\Rightarrow {145}^{\circ }+\angle RPQ={180}^{\circ }$
$\Rightarrow \angle RPQ={180}^{\circ }-{145}^{\circ }$
$\therefore \angle RPQ={35}^{\circ }$
Steps of construction:

(i). Draw a line segment $PQ=5cm$
(ii) Using protractor make an angle of ${35}^{\circ }$ at vertex $P$.
(iii). Again using protractor make angle of ${105}^{\circ }$ at vertex $Q$.
(iv). the intersection of step(ii) and step(iii), denoted at $R$ with angle of ${40}^{\circ }$.
(v). Finally join $PR$ and $QR$

Which is the required triangle.