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# Application of Angle Sum Property of Triangle

We are familiar with the angle sum property of a triangle which can be stated as the sum of all angles in a triangle is 180°. We can verify this by doing the following activity.

Application of Angle Sum Property of Triangle

We are familiar with the angle sum property of a triangle which can be stated as the sum of all angles in a triangle is 180°. We can verify this by doing the following activity.

Activity

Draw any triangle and colour the angles. Check the property as shown below: From the above we have verified that the sum of three angles of any triangle is 180°

From the above activity, we get the result as, the sum of three angles of any triangle is 180°.

Now we prove the result in a formal method.

Given: A triangle ABC, where A = x , B = y and C = z .

Now let us prove, x + y + z = 180°.

To show this we need to extend BC to D and draw a line CE, parallel to AB.

Now CE makes two angles, ACE and ECD . Let it be u and v respectively.

Now z, u, v are the angles formed at a point on a straight line.

Therefore z + u + v = 180°   . ... (1)

Since AB and CE are parallel and DB is a transversal,

v = y (corresponding angles).

Again AB and CE are parallel lines and AC is a transversal,

u = x (alternate angles). Also z + u + v = 180° [by (1)]

Hence, by replacing u as x and v as y, we get x + y + z = 180°.

Hence the sum of all three angles in a triangle is 180°.

Example 4.1

Can the following angles form a triangle?

(i) 80°, 70°, 50°

(ii) 56°, 64°, 60°

Solution

(i) Given angles 80°, 70°, 50°

Sum of the angles = 80°+70°+ 50° = 200° ≠ 180°

The given angles cannot form a triangle.

(ii) Given angles 56°, 64°, 60° Sum of the angles = 56°+ 64°+ 60° = 180°

The given angles can form a triangle.

Example 4.2

Find the measure of the missing angle in the given triangle ABC. Solution

Let A = x

We know that,

A+ B+C=180° (angle sum property)

x + 44° + 31° = 180°

x + 75° = 180°

x = 180°– 75°

x = 105°

Example 4.3

InSTU, if SU = UT, SUT = 70°, STU = x, find the value of x.

Solution

Given, SUT = 70° UST = STU = x [Angles opposite to equal sides]

SUT + UST +STU = 180°

70° + x + x = 180°

70° + 2x = 180°

2x = 180°– 70°

2x = 110°

x = 110º / 2 = 55°

Example 4.4

If two angles of a triangle having measures 65° and 35°, find the measure of the third angle.

Solution

Given angles are 65° and 35°. Let the third angle be x

65° + 35°+ x = 180°

100° + x = 180°

x= 180°–100°

x = 80°

Tags : Geometry | Term 2 Chapter 4 | 7th Maths , 7th Maths : Term 2 Unit 4 : Geometry
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