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  • How will you construct a 150° angle?

How will you construct a 150° angle?

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Take a line segment, say, AB.

Choose a point C on it.

With center C, and radius BC, draw an arc.

With center B and radius BC, cut the previous arc at say DD.

(\angle DCB=60^{\circ}\ because\ we\ just\ made\ CD=BC=BD )

With center D, and radius BC draw an arc.

With center B and radius BC, cut this arc at, say, EE.

Then EC is the bisector of ∠BCD, and

Hence, \angle BCE=30^{\circ}

Then, \angle ACE=180^{\circ} - \angle BCE=150^{\circ}

 

Posted by

Divya Prakash Singh

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