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8.01 Trigonometric functions as right triangle ratios


Trigonometric ratios

Trigonometric ratios describe the relationships between the different sides of a right triangle. These sides are described in terms of where they are compared to one of the two acute angles.

So a trigonometric ratio tells us which two sides are in a given ratio, as well as the angle the ratio is being measured from.

In the triangle below we’ve chosen one of the acute angles, labeled $x$x. The sides of the triangle can then be referred to as the side opposite $x$x, the side adjacent to $x$x and the hypotenuse.


Three trigonometric ratios are:

The sine ratio: $\sin\left(x\right)=\frac{\text{Opposite}}{\text{Hypotenuse}}$sin(x)=OppositeHypotenuse

The cosine ratio: $\cos\left(x\right)=\frac{\text{Adjacent}}{\text{Hypotenuse}}$cos(x)=AdjacentHypotenuse

The tangent ratio: $\tan\left(x\right)=\frac{\text{Opposite}}{\text{Adjacent}}$tan(x)=OppositeAdjacent

There are also what we call the reciprocal trigonometric ratios, each of which are a reciprocal of one of the three ratios above.

The cosecant ratio: $\csc\left(x\right)=\frac{1}{\sin\left(x\right)}=\frac{\text{Hypotenuse}}{\text{Opposite}}$csc(x)=1sin(x)=HypotenuseOpposite

The secant ratio: $\sec\left(x\right)=\frac{1}{\cos\left(x\right)}=\frac{\text{Hypotenuse}}{\text{Adjacent}}$sec(x)=1cos(x)=HypotenuseAdjacent

The cotangent ratio: $\cot\left(x\right)=\frac{1}{\tan\left(x\right)}=\frac{\text{Adjacent}}{\text{Opposite}}$cot(x)=1tan(x)=AdjacentOpposite

Notice similar triangles will have the same the trigonometric ratio of a particular angle, because enlarging or shrinking a triangle doesn't change the size of its angles or the ratio of its sides. 
For example, if $\tan x=\frac{2}{1}$tanx=21 we can tell that the side opposite to $x$x is twice as big as the side adjacent to $x$x. The sides might have lengths $6$6 and $3$3, not necessarily $2$2 and $1$1.

Even though we don’t know the exact length of the sides just from knowing these ratios, we can still use one ratio to calculate another ratio.


Using one ratio to find another

Given one trigonometric ratio, we can construct a right triangle that has an angle and sides that match the information described by the ratio.

We can then use the Pythagorean Theorem to find the length of the other side.

Once we have all three side lengths, we can find any other trigonometric ratio of the acute angle by using the appropriate formula.


Worked example

Question 1

A right triangle has $\sin\theta=\frac{13}{85}$sinθ=1385, find the value of $\csc\theta$cscθ and $\sec\theta$secθ.

Think: If $\sin\theta=\frac{13}{85}$sinθ=1385, then this means we can make the length of the hypotenuse is $85$85 and the length of the opposite side $13$13. We need to find the length of the adjacent side for $\sec\theta$secθ.


Finding $\csc\theta$cscθ:

$\csc\theta$cscθ $=$= $\frac{1}{\sin\theta}$1sinθ
  $=$= $\frac{1}{\frac{13}{85}}$11385
  $=$= $\frac{85}{13}$8513

Finding $\sec\theta$secθ, by first finding the adjacent side:

$a^2+b^2$a2+b2 $=$= $c^2$c2

State the Pythagorean Theorem

$a^2+13^2$a2+132 $=$= $85^2$852

Fill in the given information

$a^2+169$a2+169 $=$= $7225$7225

Evaluate the squares

$a^2$a2 $=$= $7056$7056

Subtract $169$169 from both sides

$a$a $=$= $\sqrt{7056}$7056

Take the square root of both sides

$a$a $=$= $84$84

Simplify the square root


We get the triangle below

Now to find $\sec\theta$secθ

$\sec\theta$secθ $=$= $\frac{1}{\cos\theta}$1cosθ
  $=$= $\frac{\text{hypotenuse }}{\text{adjacent }}$hypotenuse adjacent
  $=$= $\frac{85}{84}$8584


Practice questions

Question 2

Question 3

Consider the following triangle.

A right triangle with vertices labeled $A$A, $B$B, and $C$C. Each of the non-right angles at vertices $A$A and $B$B is marked with a small arc near the vertex. Vertex $C$C has a right angle, indicated by a small square corner. An angle $\theta$θ is indicated at vertex $B$B. Side $BC$BC is labeled 5 and is adjacent the angle theta. Side $AC$AC is labeled 12 and is opposite the angle $\theta$θ. Side $AB$AB is the hypotenuse and labeled x

  1. Find the value of $x$x.

  2. Hence find the value of $\sin\theta$sinθ. Express your answer as a simplified fraction.

  3. Hence find the value of $\cos\theta$cosθ. Express your answer as a simplified fraction.

Question 4

Consider the triangle below. Given that $\sin x=\frac{4}{5}$sinx=45 , find the value of $\csc x$cscx.

Question 5

Consider the triangle below.

  1. Calculate the length of the missing side.

  2. Find each of the following trigonometric ratios.

    $\sin x$sinx $=$= $\editable{}$
    $\cos x$cosx $=$= $\editable{}$
    $\tan x$tanx $=$= $\editable{}$
    $\sec x$secx $=$= $\editable{}$
    $\operatorname{cosec}x$cosecx $=$= $\editable{}$
    $\cot x$cotx $=$= $\editable{}$

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