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# Trigonometric Ratios

"Trigon" is the Greek word for triangle and "metric" is the Greek word for measurement. Bearing this in mind, it comes as no surprise that trigonometry deals with the measurements of triangles.

The first trig topic students encounter is trigonometric ratios: special measurements about right triangles. Recall that all right triangles have one angle measuring exactly 90 degrees, the two sides of a right triangle that form the 90° angle are called the legs, and the third side (opposite the right angle) is called the hypotenuse.

In this article, we'll explore the six most common trigonometric ratios. Let's get started!

## Three basic trigonometric ratios: sine, cosine, tangent

The three basic trigonometric ratios are called sine, cosine, and tangent. Each of these represents a proportion between the lengths of two of a triangle's sides as shown below:

• Sine is the length of the leg opposite to the angle divided by the length of the hypotenuse. Abbreviated "sin."
• Cosine is the length of the leg adjacent to the angle divided by the length of the hypotenuse. Abbreviated "cos."
• Tangent is the length of the leg opposite to the angle divided by the length of the leg adjacent to the angle. Abbreviated "tan."

You will need to memorize all three of these, and generations of math teachers have used "SOHCAHTOA" as a mnemonic to help with that. The acronym SOHCAHTOA means Sine Opposite over Hypotenuse, Cosine Adjacent over Hypotenuse, Tangent Opposite over Adjacent (capitalized words are in the acronym).

If we're given a right triangle, we can find the sine, cosine, and tangent of either of the non-right angles. Let's look at an example:

We can see that angle C is a right angle, making this a right triangle. The length of the leg opposite ∠ A is a, the adjacent leg is b, and c is the hypotenuse. Using the trigonometric ratios above, we get:

$\mathrm{sin}\angle a=\frac{a}{c}$ (opposite over hypotenuse)

$\mathrm{cos}\angle a=\frac{b}{c}$ (adjacent over hypotenuse)

$\mathrm{tan}\angle a=\frac{a}{b}$ (opposite over adjacent)

You'll frequently see numbers instead of letters, but otherwise, this is what working with trigonometric ratios is like.

## Three more trigonometric ratios: secant, cosecant, and cotangent

The other three trigonometric ratios are secant, cosecant, and cotangent. You can think of them as the opposite of the ratios above since they flip-flop the proportion's numerator and denominator:

Secant = length of the hypotenuse/length of the leg adjacent to the angle. Abbreviated "sec"

$\mathrm{sec}\left(x\right)=\frac{1}{\mathrm{cos}\left(x\right)}$

Cosecant = length of the hypotenuse/length of the leg opposite to the angle. Abbreviated "csc"

$\mathrm{csc}\left(x\right)=\frac{1}{\mathrm{sin}\left(x\right)}$

Cotangent = length of the leg adjacent to the angle/length of the leg opposite to the angle. Abbreviated "cot."

$\mathrm{cot}\left(x\right)=\frac{1}{\mathrm{tan}\left(x\right)}$

These need to be memorized too, but there's no equivalent to SOHCAHTOA. Instead, most students use SOHCAHTOA and then reverse them with secant corresponding to cosine, cosecant to sine, and cotangent to tangent.

Let's take another look at our diagram:

The length of the leg opposite ∠A is a, the adjacent leg is b, and the hypotenuse is c. Using this, we can write expressions for each of the trigonometric ratios above:

$\mathrm{sec}\angle A=\frac{c}{b}$ (hypotenuse over adjacent)

$\mathrm{csc}\angle A=\frac{c}{a}$ (hypotenuse over opposite)

$\mathrm{cot}\angle A=\frac{b}{a}$ (adjacent over opposite)

## Practice questions on trigonometric ratios

a. The triangle below is a right triangle. If the tangent of angle C is $\frac{3}{7}$ , what is the length of segment BC?

The tangent is the opposite over adjacent and segment BC is adjacent to angle C. Therefore, the length of the segment BC must be 7 units with the given tangent.

b. If the sine of an angle is 2/3 and the cosine of the angle is 5/12, what is the tangent of the angle?

${\mathrm{tan}}_{\left(\theta \right)}=\frac{\mathrm{sin}\left(\theta \right)}{\mathrm{cos}\left(\theta \right)}$

$\mathrm{tan}\left(\theta \right)=\frac{\frac{2}{3}}{\frac{5}{12}}$

Now, dividing by a fraction is the same as multiplying by its reciprocal:

$\mathrm{tan}\left(\theta \right)=\frac{2}{3}×\frac{12}{5}$

Multiply the numerators and denominators:

${\mathrm{tan}}_{\left(\theta \right)}=\frac{2×12}{3×5}$

${\mathrm{tan}}_{\left(\theta \right)}=\frac{24}{15}$

Finally, simplify the fraction:

${\mathrm{tan}}_{\left(\theta \right)}=\frac{8}{5}$

So, the tangent of the angle is $\frac{8}{5}$ .

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