Question

Construct a triangle ABC, where $\angle BAC=\angle ACB$. Let D be any point on the line AB, such that the $\angle CBD={120}^{\circ }$, AB = 3 m. [2 MARKS]

Answer 1

Steps: 1 Mark
Construction: 1 Mark

Draw a line segment AB = 3 m and extend it.

Sum of two interior opposite angles is equal to the exterior angle.

$\angle BAC+\angle ACB={120}^{\circ }$

Hence, $\angle ABC={60}^{\circ }$

Hence, the given triangle is an equilateral triangle.

Draw AB = 3 m

Draw a line AC of length 3 m making an angle of 60 degrees with AB. Join BC.