﻿ Sine, Cosine and Tangent ratios of a triangle. How to write the trig ratios of right triangles ## Please disable adblock in order to continue browsing our website.

Unfortunately, in the last year, adblock has now begun disabling almost all images from loading on our site, which has lead to mathwarehouse becoming unusable for adlbock users.

# Sine, Cosine, Tangent Ratios

Practice writing the ratios

### Video Tutorial How to write sohcahtoa ratios, given side lengths

#### What do we mean by 'ratio' of sides?

Sine, cosine and tangent of an angle represent the ratios that are always true for given angles. Remember these ratios only apply to right triangles.

The 3 triangles pictured below illustrate this.

Diagram 1

Although the side lengths are different , each one has a 37° angle, and as you can see, the sine of 37 is always the same!

In other words, $$sin(\red { 37^{\circ} } )$$ is always $$\red {.6 }$$ .

(Note: I rounded to the nearest tenth) That's, of course, why we can use a calculator to find these sine, cosine and tangent ratios.

### Practice Problems

##### Problem 1
Step 1

In the diagram 2, what side is adjacent to $$\angle L$$?

$$\overline{ML}$$.

Step 2

What side is the hypotenuse?

$$\overline{ LN}$$

Step 3

Calculate $$cos(\angle L)$$.

Use sochcahtoa to help remember the ratio.

$cos(\angle \red L) = \frac{adjacent}{hypot} \\ cos(\angle \red L) = \frac{8}{10} \\ = .8$

Step 4

Calculate $$cos(\angle N)$$ (a different angle from prior question, look carefully at letters).

$cos(\angle \red N) = \frac{6}{10} \\ = .6$

##### Problem 2

What is the sine ratio of $$\angle C$$ ?

Remember the sine ratio

$sin(\angle C) =\frac{\text{opp } }{ hypot} \\ sin(\angle C) = \frac{6}{10} \\ sin(\angle C) = \frac{6}{10} \\ sin(\angle C) = .6$

##### Problem 3

What is the cosine ratio of $$\angle C$$ in $$\triangle ABC$$ ?

Remember the cosine ratio

$cos(\angle C) =\frac{\text{adj } }{ hypot} \\ cos(C) = \frac{4}{5} \\ cos(C) = .8$

##### Problem 4

What is the tangent ratio of $$\angle A$$ ?

Use sochcahtoa to help remember the ratio.

$tan(A) = \frac{opposite}{adjacent} \\ = \frac{24}{7} \\= 3.42857142$

##### Problem 5

Find the sine, cosine and tangent of $$\angle R$$.

Use sochcahtoa to help remember the ratios.

$$\text{ for } \angle \red R \\ \boxed { Sine } \\ sin(\red R )= \frac{opp}{hyp} \\ sin(\red R )= \frac{12}{13} \\ sin(\red R )= .923 \\ \boxed{ cosine } \\ cos(\red R)= \frac{adj}{hyp} \\ cos(\red R)= \frac{9}{13} = \\ cos(\red R).69 \\ \boxed{ tangent } \\ tan(\red R)= \frac{opp}{adj} \\ tan(\red R) = \frac{12}{9} \\ tan(\red R)= 1.3$$

##### Problem 6

What is $$sin(\angle X)$$?

$$sin(\red X ) = \frac{opp}{hyp} \\ sin(\red X ) = \frac{24}{25} \\ sin(\red X ) = .96$$

##### Problem 7

What is $$cos(\angle X)$$?

$$sin(\red X) = \frac{adj}{hyp} \\ sin(\red X) = \frac{7}{25} \\ sin(\red X) = .28$$

##### Problem 8

Calculate : $$\text{ a) } sin(\angle H) \\ \text{ b) }cos(\angle H) \\ \text{ c) } tan(\angle H)$$

$$sin(\angle \red H) = \frac{3}{5} \\ sin( \angle \red H) = .6 \\ cos(\angle \red H) =\frac{ 4}{5} \\ cos (\angle \red H)= .8 \\ tan(\angle \red H) = \frac{3}{4} \\ tan(\angle \red H)= .75$$

### Practice Problems II

##### Problem 9

Which angle below has a tangent ratio of $$\frac{3}{4}$$?

Diagram A
Diagram B

Use sochcahtoa to help remember the ratio for tangent.

$$tangent = \frac{opp}{adj}$$

So which angle has a tangent that is equivalent to $$\frac{3}{4}$$?

You only have 2 options. Either $$\angle L$$ or $$\angle K$$. So let's calculate the tangent for each angle.

$tan( \angle K) = \frac{12}{9} \\ \frac{12}{9} \red {\ne} \frac 3 4$

$tan( \angle L) = \frac{9}{12} \\ \frac{9}{12} = \frac 3 4$
##### Problem 10

Which angle below has a cosine of $$\frac{3}{5}$$?

Diagram A
Diagram B

Use sochcahtoa to help remember the ratio for tangent.

$$cosine = \frac{adj}{hyp}$$

So which angle has a cosine that is equivalent to $$\frac 3 5$$?

You only have 2 options. Either $$\angle L$$ or $$\angle K$$. So let's calculate the tangent for each angle.

$cos( \angle K) = \frac{9}{15} \\ \frac{9}{15} \red {\ne} \frac 3 4$

$cos(L) = \frac{12}{15} \\ \frac{12}{15} = \frac 3 5$
##### Problem 11

Which angle below has a tangent $$\approx$$ .29167?

This is a little trickier because you are given the ratio as a decimal; however, you only have two options. Either $$\angle A$$ or $$\angle C$$.

$$tan(\angle A) = \frac{48}{14} \\ tan(\angle A)= 3.42857$$

$$tan(\angle C) = \frac{14}{48} \\ tan(\angle C) = .29167$$
##### Problem 12

Which angle below has a tangent of 2.4?

Again, there are two options (angles R or P), but since your ratio is greater than 1 you might quickly be able to notice that it must be R.

$$tan(\angle P) = \frac{5}{12} \\ tan(\angle P) = 0. 41666$$

$$tan(\angle R) = \frac{12}{5} \\ tan(\angle R) = 2.4$$

### Word Problems

##### Problem 13
Step 1

It's easiest to do a word problem like this one, by first drawing the triangle and labelling the sides. We know the opposite side of $$\angle K$$ and we know the hypotenuse.

To get the tangent ratio we need to know the length of the adjacent side.

How can we find the length of the adjacent side?

Step 2

Use the Pythagorean theorem!

$$a^2 + b^2 = c^2 \\ 3^2 + b^2 = 5^2 \\ b^2 = 5^2- 3^2 = 25-9 = 16 \\ b = 4$$

Now, use the tangent ratio!

Step 3

Use the Pythagorean theorem!

$$tan(k) = \frac{opposite}{adjacent} \\ tan(k) = \frac{3}{4}$$

##### Problem 14
Step 1

Draw this triangle and label the sides:

Remember that the cosine ratio = $$\frac{adjacent}{hypotenuse}$$.

How can we find the length of the opposite side?
(Remember that sine involves the opposite so we need to find that somehow).

Step 2

Use the Pythagorean theorem!

$$a^2 + b^2 = c^2 \\ 7^2 + b^2 = 25^2 \\ b^2 = 25^2- 7^2 = 625 - 49 = 576 \\ b = \sqrt{576} =24$$

Now, use the sine ratio!

Step 3

$$sin(b) = \frac{opposite}{hypotenuse}$$

top | Practice I | Practice II | Word Problems | #Challenge Problem
##### Challenge Problem

Be careful!
In the triangle below, which angle has a sine ratio of 2.6?

There is no angle that has a sine of 2.6.

Remember that the maximum value of the sine ratio is 1.

The same, is true, by the way of cosine ratio (max value is 1).