8.02 Special right triangles and exact value trigonometric ratios

$45^\circ-45^\circ-90^\circ$45°−45°−90° Triangles

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. 

The first kind of triangle that we want to look at is the $45^\circ-45^\circ-90^\circ$45°−45°−90° triangle. 

The first thing to notice about this kind of triangle is that it has two congruent angles and is, therefore, an isosceles triangle. This, in turn, means that the two sides of the triangle that are not the hypotenuse are equal in length. That is, $a=b$a=b in the diagram above.

Knowing this fact means that we can also use the Pythagorean Theorem to find out what the length of the hypotenuse should be. For example, let's look at the case where $a=b=1$a=b=1.

We can then find the value of $c$c in the following way.

$1^2+1^2$12+12	$=$=	$c^2$c2	Using the Pythagorean Theorem
$1+1$1+1	$=$=	$c^2$c2	Simplifying the powers
$c^2$c2	$=$=	$2$2	Evaluating the addition
$c$c	$=$=	$\sqrt{2}$√2	Taking the positive square root of both sides

So, here we have a $45^\circ-45^\circ-90^\circ$45°−45°−90° with all side lengths filled in:

Note that any triangle with angle measures of $45^\circ$45°, $45^\circ$45° and $90^\circ$90° is necessarily similar to this triangle. Using what we know about similar triangles, this means that the sides of any $45^\circ-45^\circ-90^\circ$45°−45°−90° triangle are in the ratio $1:1:\sqrt{2}$1:1:√2.

 

 

Practice questions

Question 1

Question 2

Question 3

 

$30^\circ-60^\circ-90^\circ$30°−60°−90° Triangles

Just like the $45^\circ$45°-$45^\circ$45°-$90^\circ$90° triangle, the $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle is another special type of right triangle. We can construct a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle by starting with an equilateral triangle and cutting it into two halves.

Here is an equilateral triangle:

To form a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle, we draw in an altitude from any vertex. In fact, we get two congruent $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangles by doing so:

In order to look at the relationships between the side lengths of such a triangle, let's suppose that the initial equilateral triangle $\triangle ABC$△ABC had side lengths of $2$2 units each. Since the altitude $\overline{AD}$AD bisects the side $\overline{BC}$BC, this means that the length of the short side $\overline{BD}$BD will be $1$1 unit. Here is this information on the $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle $\triangle ABD$△ABD:

There is only one unknown side length left on this triangle, $AD$AD, which has been labeled $x$x on the diagram above. Since $\triangle ABD$△ABD is a right triangle, we can use the Pythagorean Theorem to find this length:

$1^2+x^2$12+x2	$=$=	$2^2$22	Using the Pythagorean Theorem
$1+x^2$1+x2	$=$=	$4$4	Simplifying
$x^2$x2	$=$=	$3$3	Subtracting $1$1 from both sides
$x$x	$=$=	$\sqrt{3}$√3	Taking the positive square root of both sides

Here is a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° with all side lengths filled in:

Note that any triangle with angle measures of $30^\circ$30°, $60^\circ$60° and $90^\circ$90° is necessarily similar to this triangle. Using what we know about similar triangles, this means that the sides of any $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle are in the ratio $1:\sqrt{3}:2$1:√3:2.

 

Practice questions

Question 4

Question 5

Question 6

 

Trigonometric ratios for special triangles

You'll have noticed by now that when you find angles using trigonometric ratios, you often get long decimal answers. If, for example, you put $\cos30^\circ$cos30° into the calculator, you will see an answer of $0.86602$0.86602... which we'd have to round. However, when you take cos, sin or tan of some angles, you can express the answer as an exact number, rather than a decimal. It just may include irrational numbers. We often use these exact ratios in relation to $30^\circ$30°, $45^\circ$45° and $60^\circ$60°.

Let's use our special triangles to find the exact value trig ratios for $30^\circ$30°, $45^\circ$45° and $60^\circ$60°.

 

Using what we know about trigonometric ratios, we can find the exact values for sine, cosine and tangent.

For example, using the second triangle: 

$\sin\theta$sinθ	$=$=	$\frac{\text{opposite }}{\text{hypotenuse }}$opposite hypotenuse ​
$\sin30^\circ$sin30°	$=$=	$\frac{1}{2}$12​

 

This table is one way to display the information in the exact value triangles. You can choose which method you prefer to help you remember these exact ratios.

	sin	cos	tan
$30^\circ$30°	$\frac{1}{2}$12​	$\frac{\sqrt{3}}{2}$√32​	$\frac{1}{\sqrt{3}}$1√3​
$45^\circ$45°	$\frac{1}{\sqrt{2}}$1√2​	$\frac{1}{\sqrt{2}}$1√2​	$1$1
$60^\circ$60°	$\frac{\sqrt{3}}{2}$√32​	$\frac{1}{2}$12​	$\sqrt{3}$√3

You may be asked to rationalize the denominators as we generally don't like to have radicals in the denominator. In that case, the table becomes:

	sin	cos	tan
$30^\circ$30°	$\frac{1}{2}$12​	$\frac{\sqrt{3}}{2}$√32​	$\frac{\sqrt{3}}{3}$√33​
$45^\circ$45°	$\frac{\sqrt{2}}{2}$√22​	$\frac{\sqrt{2}}{2}$√22​	$1$1
$60^\circ$60°	$\frac{\sqrt{3}}{2}$√32​	$\frac{1}{2}$12​	$\sqrt{3}$√3

 

Practice questions

QUESTION 7

Question 8

Question 9