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# Construct $âˆ†ABC$, given $m\angle A = 60^{\circ}$, $m\angle B = 30^{\circ}$ and $AB = 5.8\ cm$.

**To do:** To construct $∆ABC$, given $m\angle A = 60^{\circ}$, $m\angle B = 30^{\circ}$ and $AB = 5.8\ cm$.

**Steps of construction :**

- Let us draw a line segment $AB=5.8\ cm$.
- At A, draw ray AY making an angle of $60^{\circ}$ with AB such that $\angle YAB=60^{\circ}$.
- At $B$, let us draw ray $BX$ making an angle of $30^{\circ}$ with AB such that $\angle ABX=30^{\circ}$.
- $C$ is the point where rays $BX$ and $AY$ intersect.

$\triangle ABC$ is the required triangle.

Hence constructed.

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