Question

Construct a ∆PQR where PQ = 5 cm, ∠P = 75° and ∠R = 55°.

Answer 1

Steps of construction :

1. Draw a line segment PQ = 5 cm.

2. Make ∠75° at point P with the help of compass.

3. Sum of all angles of a triangle is 180°.

∠P + ∠Q + ∠R = 180°
∠5° + ∠Q + 55° = 180°
130° + ∠Q = 180°
∠Q = 180°- 130° = 50°

4. Make ∠50° at point Q which intersect the line making ∠75° at point R.

5. Join RQ.

Thus ∆PQR is required triangle.