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____ __

In ∆*STU*, 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° + 2*x *= 180°

2*x *= 180°– 70°

2*x *= 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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7th Maths : Term 2 Unit 4 : Geometry : Application of Angle Sum Property of Triangle | Geometry | Term 2 Chapter 4 | 7th Maths

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