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?

Diagram of Sine Ratios

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.

Although the side lengths of the right triangles 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(37) is always .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

Want to try some word problems on this topic? Click here

Problem 1

Step 1

In the triangle above, what side is adjacent to $$\angle MLN$$?

Remember: When you are naming angles, the middle 'letter' or 'L' in this case is the angle in question.

The side adjacent to $$ \angle M\color{Red}{L}N = ML $$

Step 2

What side is the hypotenuse?

LN

Step 3

Calculate $$cos(\angle MLN)$$

$$ cos(\angle M\color{Red}{L}N) = \frac{adjacent}{hypotenuse}= \frac{8}{10} = .8 $$

Step 4

Calculate $$ cos(\angle MNL)$$

Remember: When you are naming angles, the middle 'letter' or 'L' in this case is the angle in question.

$$ cos(\angle M\color{Red}{N}L) = \frac{6}{10} = .6 $$

Problem 4

What is the sine ratio of $$\angle ACB$$ in the triangle on the right?

Remember that the sine ratio is the

$$ \frac{\text{Opposite } }{ hypotenuse}. \\ sin(ACB) = \frac{opposite}{hypotenuse} = \frac{6}{10} = .6 $$

Problem 5

What is the cosine ratio of $$\angle ACB$$ the triangle on the right?

Remember that the cosine ratio is the side adjacent to the angle (x) / hypotenuse.

$$ cos(ACB) = \frac{adjacent}{hypotenuse}= \frac{4}{5} = .8 $$

Problem 6

What is the tangent ratio of $$\angle BAC$$ in the triangle on the right?

Remember that the tangent ratio is the side opposite /adjacent sides.

Therefore
$$ tan(BAC) = \frac{opposite}{adjacent} = \frac{24}{7} = 3.4285714285714284 $$

Problem 7

Find the sine, cosine and tangent of $$\angle BRT$$?

$$ \text{ for } \angle B\color{Red}{R}T \\ sin(\color{Red}{R})= \frac{opp}{hyp} = \frac{12}{13} = .923 \\ cos( \color{Red}{R})= \frac{adj}{hyp} = \frac{9}{13} = .69 \\ tan( \color{Red}{R})= \frac{opp}{adj} = \frac{12}{9} = 1.3 $$

Problem 8

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

$$ sin(Y\color{red}{X}Z ) = \frac{opposite}{hypotenuse} = \frac{24}{25} = .96 $$

Problem 9

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

$$ sin(Y\color{red}{X}Z) = \frac{adjacent}{hypotenuse} = \frac{7}{25} = .28 $$

Problem 10

What is $$sin(\angle GHI), cos(\angle GHI), tan(\angle GHI)$$

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

Problem 11

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

$$ tangent = \frac{opposite}{adjacent} $$

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

$$ \angle L \\ tan(L) = \frac{9}{12} $$

Problem 12

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

$$ cosine = \frac{adjacent}{hypotenuse} $$

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

$$ \angle K \\ cos(k) = \frac{9}{15} $$

Problem 13

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 $$\angle C $$

$$ tan( \angle A) = \frac{48}{14} \approx 3.42857 \\ \color{Red}{ tan( \angle C) = \frac{14}{48} \approx.29167 } $$

Problem 14

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<

$$ tan( \angle P) = \frac{5}{12} \approx 0. 41666 \\ \color{Red}{ tan( \angle R) = \frac{12}{5} = 2.4 } $$

Challenge Problem challenge problem

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


pick angle

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)

Word Problems

Problem 15

In $$\triangle JKL$$, sin(k) = $$\frac{3}{5} $$, what is tan(k)?

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

sine picture

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 2

Use the Pythagorean theorem!

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

sine picture 3
Problem 16

In $$\triangle ABC , cos(b) = \frac{7}{25} $$, what is sin(b)?

Step 1

Draw this triangle and label the sides: cosine word problem

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 2

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

cosine
back to SOCHATOA Side lengths next to SOHCAHTOA to find side lengths