## TRIANGLES: Finding Unknown Angles

#### Properties of Triangles

Following are the important properties of triangles:

We will prove some of these properties in the questions below.

- The sum of angles of a triangle is 180°.
- Isosceles triangles have two sides that are equal.
- The angles opposite the equal sides in an isosceles triangle are equal.
- All sides of an equilateral triangle are equal.
- All angles of an equilateral triangle are equal. Each angle is 60°.
- A right-angled triangle has one angle that is equal to 90°.
- The sum of the other two angles of a right-angled triangle is 90°.

We will prove some of these properties in the questions below.

#### 1. Angles of an isosceles triangle

Make an isosceles triangle as below.

Now, using a protractor, measure the 2 angles opposite its equal sides. What do you notice?

In the figure above,

∠a = 70°

∠b = 70°

Now, using a protractor, measure the 2 angles opposite its equal sides. What do you notice?

In the figure above,

∠a = 70°

∠b = 70°

**The angles opposite the equal sides of the isosceles triangle are equal.**#### 2. Angles of an equilateral triangle

Make an equilateral triangle as below with equal sides.

Now, using a protractor, measure all the angles of the triangle. What do you notice?

In the figure above,

∠x = 60°

∠y = 60°

∠z = 60°

Now, using a protractor, measure all the angles of the triangle. What do you notice?

In the figure above,

∠x = 60°

∠y = 60°

∠z = 60°

**All 3 angles of an equilateral triangle are equal. Each angle is 60°.**#### 3. Angles of a right-angled triangle

Consider the right-angled triangle above.

∠p + ∠r + ∠q = 180°

∠p + ∠r + 90° = 180°

∠p + ∠r = 90°

**The sum of the other two angles of a right-angled triangle is 90°.**

#### 4. In the triangle below, AB = BC and ∠ABC = 80°. Find the other two angles.

180° − 80° = 100°

100° ÷ 2 = 50°

∠a = 50°

∠c = 50°

100° ÷ 2 = 50°

∠a = 50°

∠c = 50°

The sum of angles of a triangle is 180°.

Since AB = BC, ΔABC is an isosceles triangle.

So, ∠a = ∠c.

So, ∠a = ∠c.

#### 5. In triangle PQR below, PQ = QR and ∠QPR = 45°. Is ΔPQR a right-angled triangle?

∠p = ∠r

∠r = 45°

∠q = 180° − 45° − 45° = 90°

∠r = 45°

∠q = 180° − 45° − 45° = 90°

Since, PQ = QR, ΔPQR is an isosceles triangle, and ∠p = ∠r.

Since, one of the angles of triangle PQR is 90°, ΔPQR is a right-angled triangle.

It is, in fact, a right-angled isosceles triangle (as one of its angles is 90° and two of its sides are equal).

It is, in fact, a right-angled isosceles triangle (as one of its angles is 90° and two of its sides are equal).