**The triangle inequality theorem**

The sum of the lengths of any two sides in a triangle is greater than or equal to the length of the remaining side. This is valid for any triangle.

In other words, if $a, b, c$ are lengths of sides in a triangle $ABC$, then:

$$a+b\geq c,$$

$$b+c\geq a,$$

$$c+a\geq b.$$

*Example 1*.

Let’s construct a triangle $ABC$ whose lengths of sides are $ c = 6$, $ b = 2$, and $ a = 3$.

We can see that it’s not possible to construct a triangle with the given lengths of sides, because $a+b<c$.

In comparison, triangle whose lengths of sides are $4, 5, 6$ it’s possible construct. We have:

$ 4 + 5 = 9 > 6,$

$ 5 + 6 = 11 > 4,$

$ 4 + 6 = 10 > 5.$

*Example 2*.

Consider a triangle $ABC$ whose default lengths of sides are $a=5$ and $b=7$.

We don’t know the length of side $c$, however, we can use the triangle inequality theorem to find in which interval the length of side $c$ is.

Now, we have:

$$ c < a + b \Rightarrow c < 12$$

$$ c + a < b \Rightarrow c > b – a \Rightarrow c > 2$$

$$ c + b < a \Rightarrow c > a – b \Rightarrow c > – 2$$

Now we must observe the intersection of these intervals: $ c < 12$, $ c > 2$ $\Rightarrow $ $ c \in < 2, 12 >$.

In a triangle the longest side is across the greatest angle, and in reverse, the greatest angle is across the longest side. This is valid for any triangle.

We will construct any random triangle and measure its sides and angles.

Observe this triangle for instance. The largest side is obviously $\overline{BC}$ with a length of $12.96$, across of $\overline{BC}$ is an angle $\angle{BAC}$ with a measure of $ 111.05^{\circ}$ which is the angle of greatest measure for this triangle.

Remember, exterior angles are angles that are supplementary to interior angles (they add up to $ 180^{\circ}$).

Let $\alpha, \beta, \gamma$ be interior angles and $\alpha ‘, \beta ‘, \gamma ‘$ be exterior angles.

What is interesting about these angles is that every exterior angle’s measure is equal to the sum of the measures of other two interior angles (those who are not supplementary to that exterior angle).

This means that:

$\alpha ‘ = \beta + \gamma$,

$\beta ‘ = \alpha + \gamma$,

$\gamma ‘= \alpha + \beta$.

From this we can conclude:

$$\alpha’ > \beta \quad and \quad \alpha ‘ > \gamma$$

$$\beta’ > \alpha \quad and \quad \beta ‘ > \gamma,$$

$$\gamma ‘ > \alpha \quad and \quad \gamma’ > \beta.$$

## Triangle inequality theorem worksheets

**Angle - Integers** (541.1 KiB, 511 hits)

**Angle - Decimals** (554.1 KiB, 376 hits)

**Angle - Fractions/Mixed numbers** (691.2 KiB, 382 hits)

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