Question

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

(Hint: Recall angle sum property of a triangle).

Answer 1

A rough sketch of the required $△PQR$ is as follows.

In order to construct $△PQR$, the measure of $\angle RPQ$ has to be calculated.

According to the angle sum property of triangles,

$\Rightarrow \angle PQR+\angle PRQ+\angle RPQ={180}^{\circ }$

$\Rightarrow {105}^{\circ }+{40}^{\circ }+\angle RPQ={180}^{\circ }$

$\Rightarrow {145}^{\circ }+\angle RPQ={180}^{\circ }$

$\Rightarrow \angle RPQ={180}^{\circ }-{145}^{\circ }={35}^{\circ }$

The steps of construction are as follows.

(i) Draw a line segment $PQ$ of length $5cm.$

(ii) At $P$, draw a ray $PX$ making an angle of ${35}^{\circ }$ with $PQ.$

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(iii) At point $Q$, draw a ray $QY$ making an angle of ${105}^{\circ }$ with $PQ.$

(iv) Point $R$ has to lie on both the rays, $PX$ and $QY$. Therefore, $R$ is the point of intersection of these two rays.

This is the required triangle $PQR.$