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°.

**From ****Comparison of Sides and Angles in a Triangle**** to HOME PAGE**

**Didn’t find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**