$Construct a triangle whose perimeters 10.4 cm and the base angles are 45^{\circ} a n d 120^{\circ}$

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Solution

Steps of construction:
1. Draw a line segment PQ = 10.4 cm.
2. Construct an angle of 45° and bisect it to get $∠ Q P X$ .
3. Construct an angle of 120° and bisect it to get $∠ P Q Y$ .
4. The ray XP and YQ intersect at A.
5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.
6. Join AB and AC.
Thus, △ ABC is the required triangle.

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