To construct $\triangle ABC$ with $BC = 5, \angle B = 60^{\circ}, AB - AC = 2$, $X$ is marked on
To construct $\triangle ABC$ with $BC = 5, \angle B = 60^{\circ}, AB - AC = 2$, $X$ is marked on
- the angle ray from $B$
- the base $BC$
- perpendicular to $BC$
- angle bisector at $B$
Solution
$X$ is on the ray of angle $\angle B$ at distance $AB - AC$ from $B$.
Asked in: MH-SSC-9
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