Question

Construct a $\Delta ABC$ in which $BC=3.8$ cm, $(AB+AC)=5$ cm and $\angle ABC={60}^{\circ }$.

Answer 1

Given : $ABC$ is a triangle, $BC=3.8$ cm,
$(AB+AC)=5$ cm and $\angle ACB={60}^{\circ }$.
To construct : $\Delta ABC$
Steps of construction :
1. Draw a straight line $BC=3.8$ cm.
2. At $C$, make $\angle BCX={60}^{\circ }$.
3. At $CX$, cut off $CD=5$ cm.
4. Join $BD$.
5. At $B$, make $\angle DBA=\angle BDA$ where $A$ lies in $DC$.
Hence, $\Delta ABC$ is the required triangle.