Relative Magnitudes of Sides and Angles

Relative Magnitudes of Sides and Angles

We have seen that in a triangle, the angles opposite two equal sides are equal. Conversely, if two angles are equal in a triangle, then the sides opposite these angles are equal.

What if in a triangle, two sides (or angles) are unequal? Can we then talk about the relative magnitudes of the angles (or sides) opposite these?

Theorem: If two sides of a triangle are unequal, then the greater side has the greater angle opposite to it.

Take a look at the following figure of a triangle ABC, in which AC is larger than AB. We have to show that ∠B is greater than ∠C.

Greater side opposite greater angle

Proof: Since AC is larger than AB, we can take a point D on AC such that AB = AD, as shown below:

Greater side opposite greater angle example

In ∆ABD, since AB = AD, the angles opposite these equal sides must be equal: ∠ABD = ∠ADB. Now, note that ∠ADB is an exterior angle for ∆BDC, which means that ∠ADB > ∠C. Thus, ∠ABD > ∠C.

Finally, since ∠B must be greater than ∠ABD, we have: ∠B > ∠C.

The converse of this theorem also holds.

Theorem: If two angles of a triangle are unequal, then the greater angle has the greater side opposite to it. Consider the following triangle, for which it is given that ∠B > ∠C. We have to show that AC > AB.

Smaller angle opposite smaller side

Proof: Note that AC cannot be equal to AB, since then ∠B would have been equal to ∠C. Similarly, AC cannot be less than AB, since then ∠B would have been less than ∠C (we have just proven this result). Thus, the only possibility is that AC is greater than AB.

Download SOLVED Practice Questions of Relative Magnitudes of Sides and Angles for FREE
Triangles and Quadrilaterals
grade 9 | Questions Set 1
Triangles and Quadrilaterals
grade 9 | Answers Set 1
Triangles and Quadrilaterals
grade 9 | Questions Set 2
Triangles and Quadrilaterals
grade 9 | Answers Set 2