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
  1. the angle ray from $B$
  2. the base $BC$
  3. perpendicular to $BC$
  4. 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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