Construct $\vartriangle ABC$ in which $BC=7\ cm,\ \angle B=75^{o}$ and $AB+AC=12\ cm$.

Given: Base $BC = 7\ cm,\ \angle B=75^o$ and sum of two sides $AB+AC=12\ cm$

To do: To Construct $\vartriangle ABC$.

Follow the steps-

$1.$ Draw a ray $BX$ and cut off a line segment $BC=7\ cm$ from it.

$2.$ Construct $\angle YBX=75^o$ at $B$.

$3.$ With $B$ as centre and radius $=12\ cm\ ( \because AB+AC=12\ cm)$ draw an arc to meet $BY$ at $D$.

$4.$ Let's join $CD$.

$5.$ Draw Perpendicular bisector $PQ$ of $CD$ intersecting $BD$ at $A$.

$6.$ Join $AC$.

Then $\vartriangle ABC$ is the required triangle.