Evaluating Trigonometric Ratios Using Reference Triangles

Hello! The previous lesson developed two coordinate-geometry tools: the distance formula from Pythagoras’ theorem, and the section formula for dividing a segment in a ratio. Pythagoras’ theorem returns again here, but now it helps build two right triangles whose side ratios give exact trigonometric values.

This begins the trigonometry-and-circles module. First, you will establish the values of sine, cosine, and tangent at $30^{\circ }$, $45^{\circ }$, and $60^{\circ }$. Those exact values will be essential in the next lesson, where you turn heights-and-distances diagrams into trigonometric equations.

By the end of this lesson, you should be able to draw the correct reference triangle, identify the opposite, adjacent, and hypotenuse sides relative to a chosen angle, and evaluate the three trig ratios without a calculator.

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1. Trigonometric ratios depend on the chosen angle

For an acute angle $\theta$ in a right triangle:

$$\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}$$ $$\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$$

The mnemonic SOH CAH TOA helps you recall these definitions.

The hypotenuse is always the side opposite the right angle, and it is the longest side. The labels opposite and adjacent, however, depend on which acute angle you are using:

• the opposite side is across from the chosen angle;
• the adjacent side touches the chosen angle but is not the hypotenuse.

This last point is especially important in a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle. The same two legs are present whether you are evaluating $30^{\circ }$ or $60^{\circ }$, but their roles swap.

The image is useful as a final check, but aim to understand where the side lengths come from. If you can rebuild the triangles, you do not need to rely on a memorised table.

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2. The two reference triangles

The $45^{\circ }$-$45^{\circ }$-$90^{\circ }$ triangle

A right triangle with two equal acute angles has angles:

$$45^{\circ },45^{\circ },90^{\circ }.$$

Because equal angles have equal opposite sides, its two legs are equal. Choose each leg to have length $1$. Pythagoras gives the hypotenuse $h$:

$$1^2+1^2=h^2$$ $$2=h^2$$ $$h=\sqrt{2}.$$

So the side ratio is:

$$\boxed{1:1:\sqrt{2}}$$

where $\sqrt{2}$ is the hypotenuse.

The $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle

This triangle can be made by drawing an altitude through the middle of an equilateral triangle of side length $2$. The altitude divides the equilateral triangle into two congruent right triangles.

Each right triangle has:

• hypotenuse $2$, inherited from one side of the equilateral triangle;
• short leg $1$, because the base has been split into two equal halves;
• one angle of $60^{\circ }$ and one of $30^{\circ }$.

Use Pythagoras to calculate the remaining leg $l$:

$$1^2+l^2=2^2$$ $$1+l^2=4$$ $$l^2=3$$ $$l=\sqrt{3}.$$

So the side ratio is:

$$\boxed{1:\sqrt{3}:2}$$

The short leg, length $1$, is opposite $30^{\circ }$. The longer leg, length $\sqrt{3}$, is opposite $60^{\circ }$. The side of length $2$ is the hypotenuse.

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3. Evaluating the ratios

Values for $30^{\circ }$

Use the $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle. Relative to the $30^{\circ }$ angle:

• opposite side: $1$;
• adjacent side: $\sqrt{3}$;
• hypotenuse: $2$.

Therefore,

$$\sin 30^{\circ }=\frac{1}{2}$$ $$\cos 30^{\circ }=\frac{\sqrt{3}}{2}$$ $$\tan 30^{\circ }=\frac{1}{\sqrt{3}}.$$

It is conventional to rationalise the denominator:

$$\frac{1}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}.$$

Thus,

$$\boxed{\tan 30^{\circ }=\frac{\sqrt{3}}{3}}$$

Values for $60^{\circ }$

Stay with the same triangle, but now view it from the $60^{\circ }$ angle. The hypotenuse is still $2$, but the two legs exchange roles:

• opposite side: $\sqrt{3}$;
• adjacent side: $1$;
• hypotenuse: $2$.

So,

$$\boxed{\sin 60^{\circ }=\frac{\sqrt{3}}{2}}$$ $$\boxed{\cos 60^{\circ }=\frac{1}{2}}$$ $$\boxed{\tan 60^{\circ }=\sqrt{3}}$$

Notice the pattern:

$$\sin 30^{\circ }=\cos 60^{\circ }$$

and

$$\cos 30^{\circ }=\sin 60^{\circ }.$$

This happens because $30^{\circ }$ and $60^{\circ }$ are complementary angles: together they make $90^{\circ }$.

Values for $45^{\circ }$

In a $45^{\circ }$-$45^{\circ }$-$90^{\circ }$ triangle, either leg can be opposite or adjacent because the legs are both $1$. The hypotenuse is $\sqrt{2}$.

$$\sin 45^{\circ }=\frac{1}{\sqrt{2}}$$ $$\cos 45^{\circ }=\frac{1}{\sqrt{2}}$$

Rationalising each gives:

$$\boxed{\sin 45^{\circ }=\cos 45^{\circ }=\frac{\sqrt{2}}{2}}$$

For tangent:

$$\boxed{\tan 45^{\circ }=\frac{1}{1}=1}$$

Here is the full table in exact form:

Angle	$\sin \theta$	$\cos \theta$	$\tan \theta$
$30^{\circ }$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$
$45^{\circ }$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$
$60^{\circ }$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$

A useful way to rebuild the sine values is:

$$\sin 30^{\circ }=\frac{\sqrt{1}}{2}$$ $$\sin 45^{\circ }=\frac{\sqrt{2}}{2}$$ $$\sin 60^{\circ }=\frac{\sqrt{3}}{2}.$$

Cosine uses exactly the same values in reverse order:

$$\cos 30^{\circ }=\frac{\sqrt{3}}{2},\cos 45^{\circ }=\frac{\sqrt{2}}{2},\cos 60^{\circ }=\frac{\sqrt{1}}{2}.$$

Finally, if you forget a tangent value, use:

$$\tan \theta =\frac{\sin \theta }{\cos \theta }.$$

For example:

$$\tan 60^{\circ }=\frac{\sin 60^{\circ }}{\cos 60^{\circ }}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}.$$

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4. Using the values accurately

In a question asking you to evaluate, give the exact value unless it explicitly asks for a decimal approximation.

Example 1: A single ratio

Evaluate:

$$\cos 30^{\circ }.$$

From the table, or from the $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle:

$$\boxed{\cos 30^{\circ }=\frac{\sqrt{3}}{2}}$$

A decimal such as $0.866$ is only an approximation, while $\frac{\sqrt{3}}{2}$ is exact.

Example 2: Combining two standard values

Evaluate:

$$2\sin 60^{\circ }-\cos 30^{\circ }.$$

Substitute the standard values:

$$2\left(\frac{\sqrt{3}}{2}\right)-\frac{\sqrt{3}}{2}$$ $$=\sqrt{3}-\frac{\sqrt{3}}{2}$$ $$=\frac{\sqrt{3}}{2}.$$

Therefore,

$$\boxed{2\sin 60^{\circ }-\cos 30^{\circ }=\frac{\sqrt{3}}{2}}$$

Example 3: A product involving tangent

Evaluate:

$$\tan 30^{\circ }\tan 60^{\circ }.$$ $$=\left(\frac{\sqrt{3}}{3}\right)\left(\sqrt{3}\right)$$ $$=\frac{3}{3}$$ $$\boxed{=1}$$

This result also makes sense from the unreduced forms:

$$\tan 30^{\circ }=\frac{1}{\sqrt{3}}$$

and

$$\tan 60^{\circ }=\sqrt{3}.$$

Their product is $1$.

Common errors to avoid

• Using the hypotenuse as the adjacent side. The hypotenuse is never called adjacent in SOH CAH TOA.
• Putting $\sqrt{3}$ opposite $30^{\circ }$. In the $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle, the shortest side $1$ is opposite the smallest angle, $30^{\circ }$.
• Forgetting that side labels depend on the chosen angle. When switching from $30^{\circ }$ to $60^{\circ }$, opposite and adjacent switch.
• Giving approximations when exact values are expected. Prefer $\frac{\sqrt{3}}{2}$, not $0.866$.
• Writing $\tan 30^{\circ }=\frac{1}{3}$. The correct exact value is $\frac{\sqrt{3}}{3}$.

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Key takeaways

The two reference triangles are enough to evaluate the main standard acute angles:

$$45^{\circ }\text{-}45^{\circ }\text{-}90^{\circ }\text{has side ratio}1:1:\sqrt{2}$$ $$30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }\text{has side ratio}1:\sqrt{3}:2.$$

Use SOH CAH TOA with respect to the selected angle:

$$\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}},\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}},\tan \theta =\frac{\text{opposite}}{\text{adjacent}}.$$

Rather than trying to memorise nine disconnected answers, rebuild them from the reference triangles. In the next lesson, you will use these ratios in labelled heights-and-distances diagrams, where deciding which sides are opposite and adjacent becomes the key first step.