# Properties of Angles of a Triangle

We will discuss about some of the properties of angles of a triangle.

1. The three angles of a triangle are together equal to two right angles.

ABC is a triangle.

Then ∠ZXY + ∠XYZ + ∠YZX = 180°

Using this property, let us solve some of the examples.

Solved examples:

(i) In ∆XYZ, ∠X = 55° and ∠Y = 75°. Find ∠Z.

Solution:

∠X + ∠Y + ∠Z = 180°

or, 55° + 75° + ∠Z = 180°

or, 130° + ∠Z = 180°

or, 130° - 130° + ∠Z = 180° - 130°

Therefore, ∠Z = 50°

(ii) In the ∆XYZ, ∠Y = 5∠Z and ∠X= 3∠Z. Find the angles of the triangle.

Solution:

∠X + ∠Y + ∠Z = 180°

or, 3∠Z + 5∠Z + ∠Z = 180°

or, 9∠Z = 180°

or, $$\frac{9∠Z}{9}$$ = $$\frac{180°}{9}$$

Therefore, ∠Z = 20°

We know, ∠X= 3∠Z

Now, plug-in the value of ∠Z

∠X= 3 × 20°

Therefore, ∠X= 60°

Again we know, ∠Y= 5∠Z

Now, plug-in the value of ∠Z

∠Y= 5 × 20°

Therefore, ∠Y= 100°

Hence, the angles of the triangle are ∠X = 60°, ∠Y = 100° and ∠Z = 20°.

2. If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

The side QR of the ∆PQR is produced to S.

Then ∠PRS = ∠RPQ + ∠PQR

Corollary 1: An exterior angle of a triangle is greater than either of the interior opposite angles.

In ∆PQR, QR is produced to S.

Therefore, ∠PRS > ∠RPQ and ∠PRS ∠PQR

Corollary 2: A triangle can have only one right angle.

Corollary 3: A triangle can have only one obtuse angle.

Corollary 4: A triangle must have at least two acute angles.

Corollary 5: In a right-angled triangle, the acute angles are complementary.

Now, using this property, let us solve some of the following examples.

Solved examples:

(i) Find ∠Q from the given figure.

Solution:

∠P + ∠Q = ∠PRS

Given, ∠P = 50° and ∠PRS = 120°

or, 50° + ∠Q = 120°

or, 50° - 50° + ∠Q = 120° - 50°

or, ∠Q = 120° - 50°

Therefore, ∠Q = 70°

(ii) From the given figure find all the angles of ∆ABC, given that ∠B = ∠C.

Solution:

Given, ∠B = ∠C

We know, ∠DAC = 150°

∠DAC + ∠CAB = 180°, as they form a linear pair

or, 150° + ∠CAB = 180°

or, 150° - 150° + ∠CAB = 180° - 150°

or, ∠CAB = 30°

Let ∠B = ∠C = x°

Therefore, x° + x° = 150°, as the exterior angle of a triangle is equal to the sum of the interior opposite angles.

or, 2x° = 150°

or, $$\frac{2x°}{2}$$ = $$\frac{150°}{2}$$

or, x° = 75°

Therefore, ∠B = ∠C = 75°.