Here we will solve different types of problems on comparison of sides and angles in a triangle.
1. In ∆XYZ, ∠XYZ = 35° and ∠YXZ = 63°.
Arrange the sides of the triangle in the descending order of their lengths.
∠XZY = 180° - (∠XYZ + ∠YXZ)
= 180° - (35° + 63°)
= 180° - 98°
Therefore, ∠XZY > ∠YXZ > ∠XYZ
⟹ XY > YZ > XZ, as the greater angle has the greater side opposite to it.
2. In a ∆PQR, QR = 6 cm, PQ = 6.7 cm and PR = 5 cm. Arrange x°, y° and z and z° in the ascending order.
Given that PQ > QR > PR. Therefore, ∠PRQ > ∠QPR >∠PQR, as the greater side has the greater angle opposite to it.
Therefore, (180° - ∠PRQ) < (180° - ∠QPR) < (180° - ∠PQR)
⟹ z° < x° < y°.