How to Solve Special Right Triangle Problems (With Examples!)

Special right triangle problems are a regular feature in studies of geometry and trigonometry. These triangles have unique properties that simplify the process of solving for unknown sides and angles.

Using these triangles allows us to solve problems without using the pythagorean theorem or trigonometric functions. Instead, we are able to rely on the known ratios between the sides of the triangle.

In this guide, I will walk you through the 30-60-90 triangle and the 45-45-90 triangle so that you can gain a better understanding of how to apply them to solve various trigonometry problems without needing to rely on trigonometric ratios or ​advanced methods.

We’ll explore each special triangle, their side ratios, and the simple strategies we can use to find missing sides in special triangle problems.

What Are Special Right Triangles?

The two most common special right triangles are the 30-60-90 triangle and the 45-45-90 triangle. While both triangles include a right angle, their acute angles (either 30°, 60°, or 45°) make them unique. 

Each special triangle is shown in the image below:

We can derive the 30-60-90 triangle by dividing an equilateral triangle into two pieces. The result is a triangle with interior angles that are 30°, 60°, and 90°, and the lengths of the sides that follow a specific ratio: \(1:\sqrt{3}:2\). 

Note that the shortest side is always half the length of the hypotenuse, and the longer leg is \(\sqrt{3}\) times the shortest side.

The 45-45-90 triangle is also known as an isosceles right triangle because two sides are equal, and the two acute angles are both 45°. The ratio of the sides of length is \(1:1:\sqrt{2}\), where the two equal legs are the same, and the length of the hypotenuse is the square root of 2 times the length of either leg. 

This type of triangle often appears in geometry problems that involve squares or an isosceles triangle.

How To Solve Special Right Triangles Examples

Let’s take a look at a few special right triangles problems so that you can get used to applying each triangle!

30-60-90 Right Triangle Examples

Example 1: Finding a Missing Side Length

You are given a 30-60-90 triangle with a hypotenuse of 10 units. Find the lengths of the other two sides.

Solution:

In a 30-60-90 triangle, the side lengths follow the ratio \(1:\sqrt{3}:2\). The hypotenuse (the longest side of a right triangle) is twice the length of the shortest side. So, the shortest side (opposite the 30° angle) is 10 ÷ 2 = 5 units.

The longer leg (opposite the 60° angle) is \(\sqrt{3}\) times the shortest side. Thus, the longer leg = \(5\sqrt{3}\) ≈ 8.66 units.

So, the side lengths are:

• Shortest side = 5 units
• Longer leg = 8.66 units
• Hypotenuse = 10 units

Example 2: Application with a Ladder

An 18-foot ladder is leaning against a wall, forming a 30° angle with the ground. How high up the wall does the ladder reach?

Solution:

In a 30-60-90 triangle, the height of the ladder (opposite the 60° angle) is \(\sqrt{3}\) times the shortest side.

The ladder’s length (18 feet) is the hypotenuse. The height is \(\frac{\sqrt{3}}{2}\) of the hypotenuse.

Therefore, height = \(18\times\frac{\sqrt{3}}{2}\) ≈ 15.59 feet.

So, the ladder reaches approximately 15.59 feet up the wall.

45-45-90 Right Triangle Examples

Example 3: Find the Length of the Hypotenuse

You have a 45-45-90 triangle with legs of length 7 units each. Find the length of the hypotenuse.

Solution:

In a 45-45-90 triangle, the ratio of the side lengths is \(1:1:\sqrt{2}\), meaning the hypotenuse is \(\sqrt{2}\) times the length of the legs.

Therefore, the hypotenuse = \(7 \times \sqrt{2}\) ≈ 9.9 units.

So, the hypotenuse is approximately 9.9 units long.

Example 4: Cutting a Square in Half

A square has sides of 10 units. If you cut the square in half along its diagonal, what is the length of the diagonal?

Solution:

Cutting the square along its diagonal forms two 45-45-90 triangles. The diagonal is the hypotenuse of the triangle.

Since the sides of the square are 10 units, each leg of the triangle is also 10 units. The hypotenuse (diagonal) = \(10 \times \sqrt{2}\) ≈ 14.14 units.

So, the diagonal of the square is approximately 14.14 units long.

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