Question

Construct a $△ABC$ in which $BC=3.6cm,AB+AC=4.8cm$ and $\angle B={60}^{\circ }$.

Answer 1

Steps of construction :

(i) Draw a line segment $BC=3.6cm$.

(ii) At $B$, draw a ray $BX$ making an angle of ${60}^{\circ }$ and cut off $BE=4.8cm$.

(iii) Join $EC$.

(iv) Draw perpendicular bisector of $CE$ which intersects $BE$ at $A$.

(v) Join $AC$.

$△ABC$ is the required triangle.