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8. Right Triangles & Trigonometry
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Geometry

8.02 Special right triangles

Lesson
Practice

Special right triangles

Exploration

Drag the points to change the segments on the applet.

Loading interactive...

  1. Drag one of the sliders. How would you classify the triangle that forms?

  2. What are the angle measures of the triangle?

  3. If the length of the base is 1 unit, what are the lengths of the hypotenuse and the other leg?

  4. Write out the measure of each side length and the opposite angle's measure. What do you notice?

  5. Repeat questions 1-3 with the other triangle.

While the Pythagorean theorem can apply to any kind of right triangle, there are particular types of right triangles whose side lengths and angles have helpful properties.

We will look at two special right triangles.

45 \degree- 45 \degree- 90\degree triangle theorem

In a 45 \degree- 45 \degree- 90\degree triangle, given \text{leg length}=x, we have: \text{hypotenuse}=x\cdot \sqrt{2}

This means that the ratio of sides is 1:1: \sqrt{2}

Right triangle ABC.Angle A and B are both 45°, angle C is 90°. Side CA and BC are labeled 's' and side AB (the hypotenuse) is labeled 's square root of 2'

The 45 \degree- 45 \degree- 90\degree triangle is an isosceles right triangle.

30 \degree- 60 \degree- 90\degree triangle theorem

In a 30 \degree- 60 \degree- 90\degree triangle, given \text{short leg}=x, we have: \text{hypotenuse}= x\cdot 2 \\ \text{longer leg}= x\cdot \sqrt{3}

This means that the ratio of sides 1:\sqrt{3}: 2

Right triangle ABC/ Angle A is 30°, angle B is 60° and angle C is 90°.Side AB (Hypotenuse )is 2s, side BC is 's', and side CA is 's square root of 3'

The 30 \degree- 60 \degree- 90\degree triangle can be created by starting with an equilateral triangle and constructing an altitude.

When solving problems involving special right triangles, radicals are often involved. When dividing by a radical, answers should be given with a rationalized denominator. A rationalized denominator is a denominator that does not contain a radical.

To rationalize the denominator that contains a radical, we multiply the numerator and denominator by the radical term from the denominator.

\displaystyle \dfrac{6}{\sqrt{2}}\displaystyle =\displaystyle \dfrac{6}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}Use that \dfrac{\sqrt{2}}{\sqrt{2}}=1 to make an equivalent fraction
\displaystyle =\displaystyle \dfrac{6\sqrt{2}}{2}Use that \sqrt{2}\cdot\sqrt{2}=2 to make the denominator rational
\displaystyle =\displaystyle \dfrac{\cancel{6}\sqrt{2}}{\cancel{2}}Divide out the common factor of 2
\displaystyle =\displaystyle 3\sqrt{2}Write in simplest form

We can check on the calculator that \dfrac{6}{\sqrt{2}}=3\sqrt{2}.

Examples

Example 1

Consider the triangle below:

Right triangle with both legs of length 1, and two angles measuring 45 degrees.
a

Find the length of the hypotenuse.

Worked Solution
Create a strategy

Call the hypotenuse c. Use the Pythagorean theorem to find the missing side of the triangle.

Apply the idea
\displaystyle a^{2}+b^{2}\displaystyle =\displaystyle c^{2}Pythagorean theorem
\displaystyle 1^{2}+1^{2}\displaystyle =\displaystyle c^{2}Substitution
\displaystyle 2\displaystyle =\displaystyle c^{2}Evaluate the exponents and addition
\displaystyle \sqrt{2}\displaystyle =\displaystyle cEvaluate the square root of both sides of the equation
b

Find and justify the ratio of proportionality between the side lengths of any 45 \degree- 45 \degree- 90\degree triangle.

Worked Solution
Create a strategy

Draw a triangle with the given angle measures.

Right triangle with two angles measuring 45 degrees.
Apply the idea

All 45 \degree- 45 \degree- 90\degree triangles are similar because they have three corresponding congruent angles. We also know a 45 \degree- 45 \degree- 90\degree triangle is an isosceles triangle by the converse of the base angles theorem. Therefore, its legs are congruent.

Let the legs of the triangle be x. Call the hypotenuse of the triangle y.

Right triangle with both legs of length x, hypotenuse of length y, and two angle measuring 45 degrees.

We can use the Pythagorean theorem to show that:

\displaystyle a^{2}+b^{2}\displaystyle =\displaystyle c^{2}Pythagorean theorem
\displaystyle x^{2}+x^{2}\displaystyle =\displaystyle y^{2}Substitution
\displaystyle 2x^{2}\displaystyle =\displaystyle y^{2}Combine like terms
\displaystyle x \sqrt{2}\displaystyle =\displaystyle yEvaluate the square root of both sides of the equation

Using this, we can say that the ratio of the side lengths of any 45 \degree- 45 \degree- 90\degree triangle is x : x : x \sqrt{2}, or 1:1: \sqrt{2}.

Reflect and check

Notice that this means if the legs were length 5, then the hypotenuse would have length 5\sqrt{2}.

Example 2

Consider the triangle below:

Right triangle with both legs of length 2, and both base angles measuring 60 degrees. A vertical segment is drawn from the apex to the base of the triangle. This segment divides the remaining angle of the triangle into two, both measuting 30 degrees.
a

Find the height of the triangle.

Worked Solution
Create a strategy

Equiangular triangles are also equilateral, so the base is also 2 units. The altitude of an equilateral triangle is also the median, so half of the base must be 1 unit. We can use the Pythagorean Theorem to find the missing side of the triangle, which we will call b.

Right triangle with one leg of length 1, hypotenuse length of 2. The angle opposite the side of length 1 has a measure of 30 degrees. The adjacent angle has a measure of 60 degrees.
Apply the idea
\displaystyle a^2+b^2\displaystyle =\displaystyle c^2Pythagorean theorem
\displaystyle 1^2+b^2\displaystyle =\displaystyle 2^2Substitution
\displaystyle 1 + b^{2}\displaystyle =\displaystyle 4Evaluate the exponents
\displaystyle b^{2}\displaystyle =\displaystyle 3Subtract 1 from both sides of the equation
\displaystyle b\displaystyle =\displaystyle \sqrt{3}Evaluate the square root of both sides of the equation
Reflect and check

Notice that the smallest angle is opposite the shortest side and the largest angle is opposite the longest side.

A right triangle with angles 90-60-30. The side opposite of the 90° angle is labeled '2', the side opposite the 30° angle is labeled '1' and the side opposite the 60° angle is labeled 'square root of 3'.

  • Side with length 1 is opposite the 30\degree angle

  • Side with length \sqrt{3}\approx 1.732 is opposite the 60\degree angle

  • Side with length 2 is opposite the 90\degree angle

b

Find and justify the ratio of proportionality between the side lengths of any 30 \degree- 60 \degree- 90\degree triangle.

Worked Solution
Create a strategy

Draw and label an equilateral triangle with a 30 \degree- 60 \degree- 90\degree triangle drawn.

An equilateral triangle with an angle bisector drawn. The angle bisector is also the height of the triangle.
Apply the idea

All 30 \degree- 60 \degree- 90\degree triangles are similar because they have three corresponding congruent angles. We also know a 30 \degree- 60 \degree- 90\degree triangle is half of an equilateral triangle.

Let the short leg of the triangle be x. Then the hypotenuse of the triangle must be 2x. Call the missing leg of the triangle y.

An equilateral triangle with a bisector 'y' from the vertex to the base. Half of the base is labeled 'x'.Angle measurements are written. One side of the triangle is labeled '2x'.

We can use the Pythagorean theorem to show that:

\displaystyle a^{2}+b^{2}\displaystyle =\displaystyle c^{2}Pythagorean theorem
\displaystyle x^{2}+y^{2}\displaystyle =\displaystyle 4x^{2}Substitution and evaluate the exponent
\displaystyle y^{2}\displaystyle =\displaystyle 3x^{2}Subtract x^{2} from both sides of the equation
\displaystyle y\displaystyle =\displaystyle x \sqrt{3}Evaluate the square root of both sides of the equation and apply the commutative property of multiplication

Using this, we can say that the ratio of the side lengths of any 30 \degree- 60 \degree- 90\degree triangle is x : x \sqrt{3} : 2x, or 1:\sqrt{3}:2.

Example 3

Consider the triangle below:

A right triangle with legs of length a and 15, and a hypotenuse of length c. The angle opposite the leg with length a has a measure of 45 degrees.
a

Find the exact value of a.

Worked Solution
Create a strategy

The triangle has a right angle and a 45\degree angle shown. This means that this triangle is a 45 \degree- 45 \degree- 90\degree triangle because the sum of interior angles in a triangle is equal to 180 \degree.

Apply the idea

Legs in 45 \degree- 45 \degree- 90\degree special right triangles are congruent, and the ratio of the sides is 1:1: \sqrt{2}. We are given that 15 is the length of one leg from the diagram.

a is the other leg of the triangle as a is opposite a 45 \degree angle.

So, a=15.

b

Find the exact value of c.

Worked Solution
Create a strategy

Using the 45 \degree- 45 \degree- 90\degree triangle theorem, the hypotenuse is \sqrt{2} times the length of the leg. We want to use this property to find c.

Apply the idea

We know the length of the legs is 15 and c is the hypotenuse as it is opposite the right angle.

So, c=15\sqrt{2}.

Reflect and check

We could have also used the Pythagorean theorem to solve for c.

\displaystyle a^{2} + b^{2}\displaystyle =\displaystyle c^{2}Pythagorean theorem
\displaystyle 15^{2}+15^{2}\displaystyle =\displaystyle c^{2}Substitute the side lengths a=15 and b=15
\displaystyle 450\displaystyle =\displaystyle c^{2}Evaluate the exponents and addition
\displaystyle 21.21\displaystyle =\displaystyle cEvaluate the square root of both sides of the equation

15\sqrt{2} \approx 21.21.

Example 4

Find the value of \theta.

A right triangle. The angle between the hypotenuse and the shortest side is labeled 'θ'. The shortest side is labeled '5 square root of 3', the 2nd longest side, which is opposite of the angle labeled 'θ' is labeled '15', and there's no label on the hypotenuse.
Worked Solution
Create a strategy

We can see that this is a right triangle where the ratio between the legs is 15:5\sqrt{3}. Since the legs are not congruent, we know it cannot be a 45 \degree- 45 \degree- 90\degree triangle.

To confirm it is a 30 \degree- 60 \degree- 90\degree triangle, we should simplify the ratio between the sides. If it is a 30 \degree- 60 \degree- 90\degree triangle, then we can use the side-angle relationship to determine the angle measure.

Apply the idea

We first need to simplify the ratio or show that \text{longer leg}= \text{shorter leg}\cdot \sqrt{3} to confirm it a 30 \degree- 60 \degree- 90\degree triangle.

\displaystyle \text{longer leg}\displaystyle =\displaystyle \text{shorter leg}\cdot \sqrt{3}Use 30 \degree- 60 \degree- 90\degree triangle theorem
\displaystyle \text{longer leg}\displaystyle =\displaystyle \left(5\sqrt{3}\right)\cdot \sqrt{3}Substitute length of shorter leg
\displaystyle \text{longer leg}\displaystyle =\displaystyle 5\cdot 3Use that \sqrt{3}\cdot\sqrt{3}=3
\displaystyle \text{longer leg}\displaystyle =\displaystyle 15Evaluate the product

Now that we know that this is a 30 \degree- 60 \degree- 90\degree triangle, we can see that \theta is the angle opposite the longer leg, so must be the larger angle that isn't the right angle. \theta = 60\degree

Reflect and check

If we weren't sure which leg was longer, we could evaluate 5\sqrt{3}\approx 8.66 on a calculator, to see that 5\sqrt{3} is the shorter leg.

Example 5

Norman is building a backyard playground for this dog and wants to include the King of the Hill obstacle. He wants one side of the ramp to be 6 feet long and for the ramp to make a 45 \degree angle with the ground.

An obstacle in the shape of a triangle and a dog running above. A line is drawn from the vertex to the base at 90º representing the height labeled 'x'. The angle opposite of this side is 45°. The side opposite the 90º angle is 6ft.

Determine x, the maximum height of the ramp.

Worked Solution
Create a strategy

Looking at one half of the ramp, we can see that it forms a right triangle with the ground. One acute angle is 45\degree, so using the interior angle sum of a triangle the other acute angle is 180\degree - 90\degree -45 \degree=45 \degree. This means that half of the ramp makes a 45 \degree- 45 \degree- 90\degree triangle.

This means that we can use that if a leg has length x, then the hypotenuse has length x\sqrt{2}.

Apply the idea
\displaystyle \text{leg}\cdot \sqrt{2}\displaystyle =\displaystyle \text{hypotenuse}45 \degree- 45 \degree- 90\degree triangle theorem
\displaystyle x\cdot \sqrt{2}\displaystyle =\displaystyle 6Substitute given information
\displaystyle x\displaystyle =\displaystyle \dfrac{6}{\sqrt{2}}Divide both sides by \sqrt{2}
\displaystyle x\displaystyle =\displaystyle \dfrac{6}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}Rationalize the denominator
\displaystyle x\displaystyle =\displaystyle \dfrac{6\sqrt{2}}{2}Evaluate the product
\displaystyle x\displaystyle =\displaystyle 3\sqrt{2}Write fraction in simplest form
Reflect and check

We can check our solution using that the 45 \degree- 45 \degree- 90\degree triangle theorem says:

If \text{leg}=x, \text{hypotenuse}=x\sqrt{2}

So since \text{leg}=3\sqrt{2}, \text{hypotenuse}=\left(3\sqrt{2}\right)\sqrt{2}

This can be simplified to \text{hypotenuse}=3\left(\sqrt{2}\right)^{2}=3\cdot 2 = 6, which is what we were given.

Example 6

Find the exact value of each variable in the diagram shown.

A composite shape of two right triangles. The hypotenuse of the first triangle is the 2nd longest side of the other. The 30° angles of the triangles are adjacent.The first triangle has side labels of 'r' for the shortest side, 's' for the 2nd longest side and '6' for the longest side or hypotenuse. The second triangle has side labels of 'p' for the shortest side and 'q' for the 2nd longest side.
Worked Solution
Create a strategy

Both triangles are special 30 \degree- 60 \degree- 90\degree triangles. We can use that if \text{short leg}=x, then \text{hypotenuse}=2\cdot x and \text{longer leg} = x\cdot \sqrt{3}.

In other words, that the side lengths of a 30 \degree- 60 \degree- 90\degree triangle have a ratio of 1:\sqrt{3}: 2.

Apply the idea

For the bottom triangle, we have that \text{hypotenuse}=6, so we get:

\displaystyle \text{hypotenuse}\displaystyle =\displaystyle 2 \cdot \text{short leg}Using 30 \degree- 60 \degree- 90\degree triangle theorem
\displaystyle 6\displaystyle =\displaystyle 2 \cdot rSide opposite 30\degree angle is the shorter leg
\displaystyle 3\displaystyle =\displaystyle rDivide both sides by 2

And then:

\displaystyle \text{longer leg}\displaystyle =\displaystyle \sqrt{3} \cdot \text{short leg}Using 30 \degree- 60 \degree- 90\degree triangle theorem
\displaystyle s\displaystyle =\displaystyle \sqrt{3} \cdot 3Using that r=3 is the shorter leg
\displaystyle s\displaystyle =\displaystyle 3\sqrt{3}

For the top triangle, we have that \text{longer leg}=6, so we get:

\displaystyle \text{longer leg}\displaystyle =\displaystyle \sqrt{3} \cdot \text{short leg}Using 30 \degree- 60 \degree- 90\degree triangle theorem
\displaystyle 6\displaystyle =\displaystyle \sqrt{3} \cdot pSide opposite 30\degree angle is the shorter leg
\displaystyle \dfrac{6}{\sqrt{3}}\displaystyle =\displaystyle pDivide both sides by \sqrt{3}
\displaystyle \dfrac{6}{\sqrt{3}}\cdot \dfrac{\sqrt{3}}{\sqrt{3}}\displaystyle =\displaystyle pRationalize the denominator
\displaystyle \dfrac{6\sqrt{3}}{3}\displaystyle =\displaystyle pEvaluate products
\displaystyle 2\sqrt{3}\displaystyle =\displaystyle pWrite fraction in simplest form

And then:

\displaystyle \text{hypotenuse}\displaystyle =\displaystyle 2 \cdot \text{short leg}Using 30 \degree- 60 \degree- 90\degree triangle theorem
\displaystyle q\displaystyle =\displaystyle 2 \cdot 2\sqrt{3}Using p=2\sqrt{3} is the shorter leg
\displaystyle q\displaystyle =\displaystyle 4\sqrt{3}Evaluate the product

So we have that:

  • p=4\sqrt{3}
  • q=4\sqrt{3}
  • r=3
  • s=3\sqrt{3}
Reflect and check

We could also have found one length in each triangle using the 30 \degree- 60 \degree- 90\degree triangle theorem, and then found the third side using the Pythagorean theorem.

Idea summary

We can use special right triangles to find missing side lengths or angles in particular right triangles.

The 45 \degree- 45 \degree- 90\degree triangle can be drawn by creating an isosceles right triangle with leg lengths of 1 or x.

Right triangle ABC.Angle A and B are both 45°, angle C is 90°. Side CA and BC are labeled 's' and side AB (the hypotenuse) is labeled 's square root of 2'

The 45 \degree- 45 \degree- 90\degree triangle theorem says that, given \text{leg length}=x, we have: \text{hypotenuse}=x\cdot \sqrt{2}So the ratio of sides is 1:1: \sqrt{2}

The 30 \degree- 60 \degree- 90\degree triangle can be drawn by creating an equilateral triangle with side lengths of 2 or 2x and then bisecting the top vertex with an altitude and looking only at one half.

Right triangle ABC/ Angle A is 30°, angle B is 60° and angle C is 90°.Side AB (Hypotenuse )is 2s, side BC is 's', and side CA is 's square root of 3'

The 30 \degree- 60 \degree- 90\degree triangle theorem says that, given \text{short leg}=x, we have: \text{hypotenuse}= x\cdot 2 \\ \text{longer leg}= x\cdot \sqrt{3}So the ratio of sides 1:\sqrt{3}: 2

Outcomes

G.TR.4

The student will model and solve problems, including those in context, involving trigonometry in right triangles and applications of the Pythagorean Theorem.

G.TR.4d

Solve problems using the properties of special right triangles.

G.TR.4e

Solve for missing lengths in geometric figures, using properties of 45°-45°-90° triangles, where rationalizing denominators may be necessary.

G.TR.4f

Solve for missing lengths in geometric figures, using properties of 30°-60°-90° triangles, where rationalizing denominators may be necessary.

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