Question

Construct $△PQR$ such that $PQ=5$ cm, $QR=3$ cm and $\angle Q={95}^{\circ }$.

Answer 1

$1$. Draw a line segment QR of length $3$cm.

$2$. Now, we draw ${95}^o$ degrees from point Q.

$3$. Taking Q as centre, $5$cm radius, we draw an arc. Let the point where arc intersects the ray be point p.

$4$. Joint PR and label the sides.

Thus, PQR is the required triangle.