Question

Construct $△ABC$ Such that $\angle B={100}^{\circ },$
$BC=6.4cm,\angle C={50}^{\circ }$ and constant its incircle

Answer 1

$\left(i\right)$ Drawn BC = 6.4 cm
$\left(ii\right)\angle B={100}^{\circ }$ and $\angle C={50}^{\circ }$ using the protector. Where the rays $YB$ and $XC$ meet, name the point as $A$
$△ABC$ is thus obtained
$\left(iii\right)$ Construct the angle bisectors of$\angle B$ and $\angle C$ and let them meet at point O.
$\left(iv\right)$ Through point O, draw a perpendicular to line. Let the perpendicular meet the line at point P.
$\left(v\right)$ With O as centre and OP as radius , draw a circle inside the $△ABC$.
This is the required incircle.