In the previous section we saw median and altitude. In this section, we will discuss about exterior angle.
This ∠RQS is an exterior angle of triangle PQR. Let us see the details of how this exterior ∠RQS is formed:
• We take the two sides PQ and QR.
These two are adjacent sides of the triangle. But it may be noted that, as there are only 3 sides in a triangle, which ever two sides we take, they will naturally be adjacent to each other.
• The two sides PQ and QR intersect at the vertex Q. This point of intersection is of importance as far as the to the ∠RQS is concerned
• The opposite side of this vertex Q, which is the side PR has no role to play here
• An extension is given to PQ at Q, and so PQ becomes PS
• The angle between this extended line PS and the adjacent side at the point of intersection Q is marked. This is the exterior angle at Q
In short, we can say:
■ An exterior angle can be marked at any of the three vertices of a triangle
■ At a vertex, one of the sides is extended
■ The angle is marked between this extended line, and the other side of the triangle.
■ This angle is the exterior angle of the triangle at that vertex.
According to the above defenition, another exterior angle can be marked at the same vertex Q. This is shown in fig.(b). This time, RQ is extended, and the angle is marked between RT and PQ.
So we get two exterior angles at Q. But interestingly, these two angles have the same value. This can be explained using fig.(c). From this fig., we can see that ∠RQS and ∠PQT are opposite angles. So they will be equal. Fig(d) shows an exterior angle at another vertex R.
Let us now see the method for calculating the value of any exterior angle:
Consider the fig.9.5(a) below. Triangle PQR and the exterior angle is drawn on a card board. Then, the angle at P (shown in green colour) and the angle at R (shown in magenta colour) are cut out. These two cut out angles are placed together at Q.
We can see that they fit exactly into the exterior angle at Q. This means that, the exterior angle at Q is equal to the sum of the interior angle at P and the interior angle at Q. As far as the vertex Q is concerned, P and Q are opposite interior angles. Because, they are opposite to Q. They are also called remote interior angles because they are situated remotely from Q. So we can write a general rule:
■ The exterior angle is equal to the sum of the remote interior angles.
Let us see the theoretical proof for this general rule:
• In the fig9.5(b), a line UQV is drawn through Q
• This line is parallel to the side PR
• This line UV splits the exterior angle at Q into ∠x and ∠y
• So we have two parallel lines PR and UV, cut by a transversal QR
• Then ∠R = ∠x (∵ they are alternate interior angles)
• We also have two parallel lines PR and UV cut by another transversal PQ
• Then ∠P = ∠y (∵ they are corresponding angles)
• So we get ∠P + ∠Q = ∠x + ∠y
• But P and Q are the remote interior angles. And ∠x + ∠y = the exterior angle at Q.
• So we get ∠P + ∠Q = sum of remote interior angles = ∠x + ∠y = the exterior angle at Q
We will now see some solved examples:
Solved example 9.1
An exterior angle of a triangle is 75o. One of it’s remote interior angle is 30o. What is the other remote interior angle?
Solution:
• We have: Exterior angle = Sum of the remote interior angles
• The exterior angle = 75o
• One remote interior angle = 30o
• Let the unknown remote interior angle be x
• So we can write: 75 = 30 + x
∴ x = 75 – 30 = 45o
Solved example 9.2
Find the value of x in the following figs:
Solution:
(a) x = sum or remote interior angles = 56 + 66 = 122o
(b) x = sum of remote interior angles = 39 + 122 = 161o
In the remaining two problems, we have to work in a 'reverse order'.
(c) 95 = x + 51 ∴ x = 95 – 51 = 44o
(d) 151 = 95 + x therefore x = 151- 95 = 56o
In the next section we will discuss about the sum of all the 3 interior angles of a triangle.
Exterior angle of a triangle
Consider the triangle PQR in the fig 9.4(a) below. PQ is extended towards S. Then an angle RQS is formed between PS and QR. |
| Fig.9.4 |
• We take the two sides PQ and QR.
These two are adjacent sides of the triangle. But it may be noted that, as there are only 3 sides in a triangle, which ever two sides we take, they will naturally be adjacent to each other.
• The two sides PQ and QR intersect at the vertex Q. This point of intersection is of importance as far as the to the ∠RQS is concerned
• The opposite side of this vertex Q, which is the side PR has no role to play here
• An extension is given to PQ at Q, and so PQ becomes PS
• The angle between this extended line PS and the adjacent side at the point of intersection Q is marked. This is the exterior angle at Q
In short, we can say:
■ An exterior angle can be marked at any of the three vertices of a triangle
■ At a vertex, one of the sides is extended
■ The angle is marked between this extended line, and the other side of the triangle.
■ This angle is the exterior angle of the triangle at that vertex.
According to the above defenition, another exterior angle can be marked at the same vertex Q. This is shown in fig.(b). This time, RQ is extended, and the angle is marked between RT and PQ.
So we get two exterior angles at Q. But interestingly, these two angles have the same value. This can be explained using fig.(c). From this fig., we can see that ∠RQS and ∠PQT are opposite angles. So they will be equal. Fig(d) shows an exterior angle at another vertex R.
Let us now see the method for calculating the value of any exterior angle:
Consider the fig.9.5(a) below. Triangle PQR and the exterior angle is drawn on a card board. Then, the angle at P (shown in green colour) and the angle at R (shown in magenta colour) are cut out. These two cut out angles are placed together at Q.
 |
| Fig.9.5 |
■ The exterior angle is equal to the sum of the remote interior angles.
Let us see the theoretical proof for this general rule:
• In the fig9.5(b), a line UQV is drawn through Q
• This line is parallel to the side PR
• This line UV splits the exterior angle at Q into ∠x and ∠y
• So we have two parallel lines PR and UV, cut by a transversal QR
• Then ∠R = ∠x (∵ they are alternate interior angles)
• We also have two parallel lines PR and UV cut by another transversal PQ
• Then ∠P = ∠y (∵ they are corresponding angles)
• So we get ∠P + ∠Q = ∠x + ∠y
• But P and Q are the remote interior angles. And ∠x + ∠y = the exterior angle at Q.
• So we get ∠P + ∠Q = sum of remote interior angles = ∠x + ∠y = the exterior angle at Q
We will now see some solved examples:
Solved example 9.1
An exterior angle of a triangle is 75o. One of it’s remote interior angle is 30o. What is the other remote interior angle?
Solution:
• We have: Exterior angle = Sum of the remote interior angles
• The exterior angle = 75o
• One remote interior angle = 30o
• Let the unknown remote interior angle be x
• So we can write: 75 = 30 + x
∴ x = 75 – 30 = 45o
Solved example 9.2
Find the value of x in the following figs:
 |
| Fig.9.6 |
(a) x = sum or remote interior angles = 56 + 66 = 122o
(b) x = sum of remote interior angles = 39 + 122 = 161o
In the remaining two problems, we have to work in a 'reverse order'.
(c) 95 = x + 51 ∴ x = 95 – 51 = 44o
(d) 151 = 95 + x therefore x = 151- 95 = 56o
In the next section we will discuss about the sum of all the 3 interior angles of a triangle.
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