Construct a triangle $ABC$ in which $BC = 5 \, cm$,$\angle B = 60^{\circ}$ and $AC + AB = 7.5 \, cm$.

(N/A) Given: In $\triangle ABC$,$BC = 5 \, cm$,$AC + AB = 7.5 \, cm$ and $\angle B = 60^{\circ}$.
Required: To construct $\triangle ABC$.
Steps of construction:
$1.$ Draw a ray $BX$ and cut off a line segment $BC = 5 \, cm$ from it.
$2.$ At $B$,construct $\angle XBY = 60^{\circ}$.
$3.$ With $B$ as centre and radius $= 7.5 \, cm$,draw an arc to meet $BY$ at $D$.
$4.$ Join $CD$.
$5.$ Draw the perpendicular bisector of $CD$,intersecting $BD$ at $A$.
$6.$ Join $AC$. Then,$ABC$ is the required triangle.