Question

Construct a $△ABC$ in which:

$AB+BC+AC=11.8cm$, $\angle B={60}^{\circ }$ and $\angle C={45}^{\circ }$.

Answer 1

Step 1: Draw a line segment $PQ=11.8cm=(AB+BC+CA)$

Step 2: At $P$, construct an angle ${60}^o$ and at $Q={45}^o$.

Step 3: Bisect three angles

Step 4: Draw perpendicular bisector $DE$ of $AP$ to intersect $PQ$ at $B$ and $FG$ of $AQ$ to intersect $PQ$ at $C$.

Step 5: Join $AB$ and $AC$