In triangle XYZ below on the left, the exterior angle is “a” and the two opposite interior angles are x and y.
To prove that a = p + q, draw a line at Z (i.e. line ZW) to divide the exterior angle “a” into angles p and q and let this line be parallel to line XY (see the diagram above).
From the diagram on the right,
x = p (corresponding angles ZW//XY)
y = q (alternate angles ZW//XY)
Hence: a = x + y = p + q
Therefore, the exterior angle of a triangle is equal to the sum of the two opposite interior angles.
Example 1
Find the size of angle x in this triangle
Solution
x + 53º + 87º = 180º (sum of angles of a Triangle)
x + 140º = 180º
x = 180º – 140º
x = 40º
Example 2
a. In the diagram below, find the value of x
b. Hence, use the value of x to find the actual values of the interior angles
Solution
a. <ABC = 2x (vertically opposite angles)
2x + 3x + 4x = 180º (sum of angles of a triangle)
9x = 180º
Divide both sides by 9
\( \frac{9x}{9} = \frac{180}{9} \)x = \( \frac{180}{9} \)
x = 20º
b. If a = 20º, then 2x = 2 x 20º = 40º
3x = 3 x 20º = 60º
4x = 4 x 20º = 80º
The angles are 40º, 60º, 80º
Example 3
State the sizes of the lettered angles and give reasons
Solution
140º + a = 180º (angles on a straight line is 180º)
a = 180º – 140º
a = 40º
a + b + c = 180 (sum of angles in a triangle)
40 + b + c = 180
b + c = 180 – 40
b + c = 140
b = 70, c = 70 (base angles in an isosceles triangle are equal)
b = c {sum of angles of an isosceles triangle base}
OR
140 = b + c (the exterior angle of a triangle is equal to the sum of the two opposite interior angles)
b = 70, c = 70 (base angles in an isosceles triangle are equal)
Example 4
Find angles x, y, z
Solution
Consider Δ ABC
x + 32º + 90º = 180º (sum of angles in a triangle)
x = 180º – 90º – 32º
x = 180º – 122º
x = 58º
x + y = 180 (angles on a straight line)
58 + y = 180
y = 180 – 58
y = 122º
Consider Δ ACD
z + 30 + 122 = 180 (sum of angles in a triangle)
z + 152 = 180
z = 180 – 152
z = 28º
Therefore: x = 58º, y = 122º, z = 28º
Easy subject