Comparison of Sides and Angles in a Triangle

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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°.

9th Grade Math

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