**Examples on ****Theorem 1**

Published on ginnyent on 16th May, 2019.

Example 1

The angles of a triangle are x, 2x and 3x. Find the value of x in degrees.

**Solution**

Recall what theorem 1 says.

It now follows that x + 2x + 3x = 180^{0}

Collecting like terms we have:

6x = 180^{0}

Therefore x = 180^{0}/6

= 30^{0}

x = 30^{0}, 2x = 60^{0} and 3x = 90^{0}. You can see that the triangle is a right-angled triangle.

**Example 2**

An isosceles triangle is such that each of the base angles is twice the vertical angle. Find the angles of the triangle.

**Solution**

Let’s make a sketch. See the diagram below.

From the diagram, we have assumed the vertical angle to be *a,* and the two base angles to be *2a*.

We know from theorem 1 that the sum of the angles in a triangle is 180^{0}.

Therefore, we will sum the three angles and equate it to 180^{0 }

a + 2a + 2a = 180^{0}

Collecting like terms, we have:

5a = 180^{0}

a = 180^{0}/5

a = 36^{0}

2a =

The angles of the triangle are therefore: 36^{0} ,^{ }72^{0} and 72^{0}

Example 3

In a right-angled triangle, one of the acute angles is 200 greater than the other. Find the angles of the triangle.

**Solution**

Let’s look at the diagram below.

From the diagram, we have the angles of the triangle to be: 90^{0}, a, and a+20^{0}

We already know from theorem 1 that the sum of the angles in a triangle is 180^{0}

Therefore, we will equate the sum of the angles to 180^{0}

90^{0}+ a + (a+ 20^{0}) = 180^{0}

Collecting like terms, we have:

110^{0} + 2a = 180^{0}

2a = 180^{0} – 110^{0}

2a = 70^{0}

a = 70^{0}/2

= 35^{0}

a = 35^{0}

a + 20^{0} = 55^{0}

Therefore the angles of the triangle are: 90^{0}, 55^{0} and 35^{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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