**Examples on Theorem Two**

Published on ginnyent.net on 21st May, 2019

Look at the two diagrams below.

In Fig 1, find α and β.

In Fig 2, find, a and b.

**Solution for Fig. 1**

Recall what theorem 1, says. You can refresh your memory by visiting theorem 1.

In triangle XYZ, 70^{0 }+ 30^{0} + α = 180^{0}(Sum of the angles in a triangle)

100^{0} + α = 180^{0}

α = 180^{0 }-100^{0}

= 80^{0}

The exterior angle of triangle XYZ is represented by β

α + β = 180^{0} (sum of the angles on a straight line)

Since we have got α, 80^{0} + β = 180^{0}

β** = **180^{0}-80^{0}

= 100^{0}

**Solution for Fig. 2**

In triangle RST, a + 50^{0} + b = 180^{0} (sum of the angles in a triangle), but we don’t know a and b; that is we have two **unknowns. **

The exterior angle of triangle RST is 120^{0}, and we know that;

a + 50^{0 }= 120^{0} (exterior angle of a triangle = sum of two opposite interior angles)

Therefore a = 120^{0} – 50^{0}

= 70^{0}.

Now that we have found a, let’s find b.

50^{0 }+ a + b = 180^{0} (sum of the angles in a triangle)

50^{0 }+ 70^{0} + b = 180^{0}

b= 180^{0} – (50^{0} + 70^{0})

= 180^{0} – 120^{0}

b = 60^{0 }

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CHG Ofokansi is an author, medical laboratory scientist, a teacher who has several books for children and adults.
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