In the previous section we saw Exterior angle. In this section, we will discuss about the sum of all the 3 Interior angles of a triangle.
Let us do an experiment: Consider the triangle PQR shown in fig.9.7(a). Draw it on a cardboard and cut out the three angles at the 3 vertices. Arrange the three angles side by side as shown in fig(b).
We can see that the three angles add up in such a way that the base of the resulting shape is a line. That is., the three angles add up to give 180o.
Let us do one more experiment: Draw a triangle on a cardboard and make a cut out. Make two more copies of the same triangle. Thus we will get three identical triangles as shown in fig.9.8(a) Arrange them side by side as shown in fig(b).
Here also the base of the resulting shape is a line. So the sum of the angles is 180o. Thus we can say that the sum of the interior angles of a triangle is 180o.
We can prove it theoretically as follows: Consider the triangle in fig.9.9 below. We know that exterior angle will be equal to the sum of the remote interior angles.
• So ∠1 + ∠2 = ∠x
• Adding ∠3 on both sides we get ∠1 + ∠2 + ∠3 = ∠x + ∠3
• But x and 3 form a linear pair. That is., ∠x + ∠3 = 180o
• So we an write ∠1 + ∠2 + ∠3 = 180o
■ Thus we prove that the sum of the three interior angles of a triangle is 180o
We will see some solved examples based on the above discussion:
Solved example 9.3
Find the value of x in the following figs.
Solution: The given problems can be solved by using the fact that the sum of the three interior angles of a triangle will always be equal to 180o.
(a) 52 + 55 + x = 180 ⇒ 107 + x = 180 ∴ x = 180 - 107 = 73o
(b) 28 + 118 + x = 180 ⇒ 146 + x = 180 ∴ x = 180 - 146 = 34o
(c) 3x + x + 80 = 180 ⇒ 4x + 80 = 180 ⇒ 4x = 180 - 80 = 100 ∴ x = 100/4 =25o
(d) 2x + x + 90 = 180 ⇒ 3x + 90 = 180 ⇒ 3x = 180 - 90 = 90 ∴ x = 90/3 = 30o
Solved example 9.4
Find the value of x and y in the following figs.:
Solution:
(a) • 65 + 55 + x = 180 ⇒ 120 + x = 180 ∴ x = 180 - 120 = 60o
• x and y form a linear pair. So x + y = 180o ⇒ 60 + y = 180 ∴ y = 180 - 60 = 120
(b) • 50 + x + y = 180
• But x and 70 are opposite angles. So x = 70
• So we can write: 50 + 70 + y = 180 ⇒ 120 + y = 180 ∴ y = 180 - 120 = 60o
In the next section we will discuss about Equilateral and Isosceles triangles.
Let us do an experiment: Consider the triangle PQR shown in fig.9.7(a). Draw it on a cardboard and cut out the three angles at the 3 vertices. Arrange the three angles side by side as shown in fig(b).
 |
| Fig.9.7 |
Let us do one more experiment: Draw a triangle on a cardboard and make a cut out. Make two more copies of the same triangle. Thus we will get three identical triangles as shown in fig.9.8(a) Arrange them side by side as shown in fig(b).
 |
| Fig.9.8 |
We can prove it theoretically as follows: Consider the triangle in fig.9.9 below. We know that exterior angle will be equal to the sum of the remote interior angles.
 |
| Fig.9.9 |
• Adding ∠3 on both sides we get ∠1 + ∠2 + ∠3 = ∠x + ∠3
• But x and 3 form a linear pair. That is., ∠x + ∠3 = 180o
• So we an write ∠1 + ∠2 + ∠3 = 180o
■ Thus we prove that the sum of the three interior angles of a triangle is 180o
We will see some solved examples based on the above discussion:
Solved example 9.3
Find the value of x in the following figs.
 |
| Fig.9.10 |
(a) 52 + 55 + x = 180 ⇒ 107 + x = 180 ∴ x = 180 - 107 = 73o
(b) 28 + 118 + x = 180 ⇒ 146 + x = 180 ∴ x = 180 - 146 = 34o
(c) 3x + x + 80 = 180 ⇒ 4x + 80 = 180 ⇒ 4x = 180 - 80 = 100 ∴ x = 100/4 =25o
(d) 2x + x + 90 = 180 ⇒ 3x + 90 = 180 ⇒ 3x = 180 - 90 = 90 ∴ x = 90/3 = 30o
Solved example 9.4
Find the value of x and y in the following figs.:
Solution:
(a) • 65 + 55 + x = 180 ⇒ 120 + x = 180 ∴ x = 180 - 120 = 60o
• x and y form a linear pair. So x + y = 180o ⇒ 60 + y = 180 ∴ y = 180 - 60 = 120
(b) • 50 + x + y = 180
• But x and 70 are opposite angles. So x = 70
• So we can write: 50 + 70 + y = 180 ⇒ 120 + y = 180 ∴ y = 180 - 120 = 60o
In the next section we will discuss about Equilateral and Isosceles triangles.
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