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.

**Solution:**

∠XZY = 180° - (∠XYZ + ∠YXZ)

= 180° - (35° + 63°)

= 180° - 98°

= 82°

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.

**Solution:**

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°.

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