Solve problems involving exact value right triangles

When we find angles using trigonometric ratios, we will often get long decimal answers. If, for example, we put $\cos30^\circ$cos30° into the calculator, the result looks like $0.86602$0.86602$\dots$… which we might then want to round. However, taking the $\cos$cos, $\sin$sin or $\tan$tan of some special angles, we can express the answer as an exact value, 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 look at some of those important angles.

Exact value triangles

45 degree angles

Below is a right-angle isosceles triangle, with the equal sides of $1$1 unit. Using Pythagoras' theorem, we can work out that the hypotenuse is $\sqrt{2}$√2 units. Further, because the angles in a triangle add up to $180^\circ$180° and the base angles in an isosceles triangle are equal, we can deduce that the other two unknown angles are both $45^\circ$45°.

Using our trigonmetric ratios, we can see that:

• $\sin45^\circ=\frac{1}{\sqrt{2}}$sin45°=1√2​
• $\cos45^\circ=\frac{1}{\sqrt{2}}$cos45°=1√2​
• $\tan45^\circ=\frac{1}{1}$tan45°=11​$=$=$1$1

 

30 and 60 degree angles

To find the exact ratios of $30$30 and $60$60 degree angles, we can start with a equilateral triangle with side lengths of $2$2 units. Remember that all the angles in an equilateral triangle are $60^\circ$60°.

Then we are going to draw a line that cuts the triangle in half into two congruent triangles. The base line is cut into two $1$1 unit pieces and the length of this centre line is then found using Pythagoras' theorem.

Now let's just focus on one half of this triangle.

Using our trig ratios, we can see that:

• $\sin30^\circ=\frac{1}{2}$sin30°=12​
• $\cos30^\circ=\frac{\sqrt{3}}{2}$cos30°=√32​
• $\sin60^\circ=\frac{\sqrt{3}}{2}$sin60°=√32​
• $\cos60^\circ=\frac{1}{2}$cos60°=12​

 

Rationalising the denominator

In most cases it is preferred to give fractional answers with a rational denominator. We can do so by multiplying the numerator and denominator by the surd in the denominator.

For example, $\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$1√3​=1√3​×√3√3​$=$=$\frac{\sqrt{3}}{3}$√33​.

Exact value summary table

This table is another way to display the information in the exact value triangles. Think about which method you prefer to help you remember these exact ratios.

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

 

Practice questions

QUESTION 1

Question 2

Question 3