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Sine, Cosine, and Tangent Functions

Trigonometric ratios based on sides of right triangles in relation to an angle.

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Sine, Cosine, Tangent

An isoceles right triangle has leg lengths of 4 units each. What is the sine of each of the triangle's acute angles?


The trigonometric ratios sine, cosine and tangent refer to the known ratios between particular sides in a right triangle based on an acute angle measure.

In this right triangle, side \begin{align*}c\end{align*}c is the hypotenuse.

If we consider the angle \begin{align*}B\end{align*}B, then we can describe each of the legs by its position relative to angle \begin{align*}B\end{align*}B: side \begin{align*}a\end{align*}a is adjacent to \begin{align*}B\end{align*}B; side \begin{align*}b\end{align*}b is opposite \begin{align*}B\end{align*}B

If we consider the angle \begin{align*}A\end{align*}A, then we can describe each of the legs by its position relative to angle \begin{align*}A\end{align*}A: side \begin{align*}b\end{align*}b is adjacent to \begin{align*}A\end{align*}A; side \begin{align*}a\end{align*}a is opposite \begin{align*}A\end{align*}A

Now we can define the trigonometry ratios as follows:

\begin{align*}\text{{\color{red}S}ine is} \ \frac{{\color{red}o}pposite}{{\color{red}h}ypotenuse} \quad \text{{\color{red}C}osine is} \ \frac{{\color{red}a}djacent}{{\color{red}h}ypotenuse} \quad \text{{\color{red}T}angent is} \ \frac{{\color{red}o}pposite}{{\color{red}a}djacent}\end{align*}

{\color{red}S}ine is oppositehypotenuse{\color{red}C}osine is adjacenthypotenuse{\color{red}T}angent is oppositeadjacent

A shorthand way to remember these ratios is to take the letters in red above and write the phrase:

\begin{align*}\text{{\color{red}SOH \ CAH \ TOA}}\end{align*}

{\color{red}SOH \ CAH \ TOA}

Now we can find the trigonometric ratios for each of the acute angles in the triangle above.

\begin{align*}\sin A &=\frac{a}{c} \qquad \sin B=\frac{b}{c} \\ \cos A &=\frac{b}{c} \qquad \cos B=\frac{a}{c} \\ \tan A &=\frac{a}{b} \qquad \tan B=\frac{b}{a}\end{align*}


It is important to understand that given a particular (acute) angle measure in a right triangle, these ratios are constant no matter how big or small the triangle. For example; if the measure of the angle is \begin{align*}25^\circ\end{align*}25, then \begin{align*}\sin 25^\circ \approx 0.4226\end{align*}sin250.4226 and ratio of the opposite side to the hypotenuse is always 0.4226 no matter how big or small the triangle.

Example A

Find the trig ratios for the acute angles \begin{align*}R\end{align*}R and \begin{align*}P\end{align*}P in \begin{align*}\Delta PQR\end{align*}ΔPQR.

Solution: From angle \begin{align*}R\end{align*}R, \begin{align*}O=8\end{align*}O=8; \begin{align*}A=15\end{align*}A=15; and \begin{align*}H=17\end{align*}H=17. Now the trig ratios are:

\begin{align*}\sin R=\frac{8}{17}; \ \cos R=\frac{15}{17}; \ \tan R=\frac{8}{15}\end{align*}

sinR=817; cosR=1517; tanR=815

From angle \begin{align*}P\end{align*}P, \begin{align*}O=15\end{align*}O=15; \begin{align*}A=8\end{align*}A=8; and \begin{align*}H=17\end{align*}H=17. Now the trig ratios are:

\begin{align*}\sin P=\frac{15}{17}; \ \cos P=\frac{8}{17}; \ \tan P=\frac{15}{8}\end{align*}

sinP=1517; cosP=817; tanP=158

Do you notice any patterns or similarities between the trigonometric ratios? The opposite and adjacent sides are switched and the hypotenuse is the same. Notice how this switch affects the ratios:

\begin{align*}\sin R=\cos P \quad \cos R=\sin P \quad \tan R=\frac{1}{\tan P}\end{align*}


Example B

Use trigonometric ratios to find the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

Solution: First identify or label the sides with respect to the given acute angle. So, \begin{align*}x\end{align*} is opposite, \begin{align*}y\end{align*} is hypotenuse (note that it is the hypotenuse because it is the side opposite the right angle, it may be adjacent to the given angle but the hypotenuse cannot be the adjacent side) and 6 is the adjacent side.

To find \begin{align*}x\end{align*}, we must use the given length of 6 in our ratio too. So we are using opposite and adjacent. Since tangent is the ratio of opposite over adjacent we get:

\begin{align*}\tan 35^\circ &=\frac{x}{6} \\ x &=6 \tan 35^\circ \quad \text{multiply both sides by 6}\\ x & \approx 4.20 \quad \quad \quad \text{Use the calculator to evaluate-type in 6TAN(35) ENTER}\end{align*}

NOTE: make sure that your calculator is in DEGREE mode. To check, press the MODE button and verify that DEGREE is highlighted (as opposed to RADIAN). If it is not, use the arrow buttons to go to DEGREE and press ENTER. The default mode is radian, so if your calculator is reset or the memory is cleared it will go back to radian mode until you change it.

To find \begin{align*}y\end{align*} using trig ratios and the given length of 6, we have adjacent and hypotenuse so we’ll use cosine:

\begin{align*}\cos 35^\circ &=\frac{6}{y} \\ \frac{\cos 35^\circ}{1} &=\frac{6}{y} \quad \quad \quad \quad \text{set up a proportion to solve for} \ y \\ 6 &= y \cos 35^\circ \quad \text{cross multiply} \\ y &=\frac{6}{\cos 35^\circ} \quad \ \text{divide by} \cos 35^\circ \\ y &=7.32 \quad \quad \ \ \text{Use the calculator to evaluate-type in 6/TAN(35) ENTER}\end{align*}

Alternatively, we could find \begin{align*}y\end{align*} using the value we found for \begin{align*}x\end{align*} and the Pythagorean theorem:

\begin{align*}4.20^2+6^2 &=y^2 \\ 53.64 &=y^2 \\ y & \approx 7.32 \end{align*}

The downside of this method is that if we miscalculated our \begin{align*}x \end{align*} value, we will double down on our mistake and guarantee an incorrect \begin{align*}y \end{align*} value. In general you will help avoid this kind of mistake if you use the given information whenever possible.

Example C

Given \begin{align*}\Delta ABC\end{align*}, with \begin{align*}m \angle A=90^\circ, m\angle C=20^\circ\end{align*} and \begin{align*}c=9\end{align*}, find \begin{align*}a\end{align*} and \begin{align*}b\end{align*}.

Solution: Visual learners may find it particularly useful to make a sketch of this triangle and label it with the given information:

To find \begin{align*}a\end{align*} (the hypotenuse) we can use the opposite side and the sine ratio: \begin{align*}\sin 20^\circ=\frac{9}{a}\end{align*}, solving as we did in Example B we get \begin{align*}a=\frac{9}{\sin 20^\circ} \approx 26.31\end{align*} To find \begin{align*}b\end{align*} (the adjacent side) we can use the opposite side and the tangent ratio: \begin{align*}\tan 20^\circ=\frac{9}{b}\end{align*}, solving for \begin{align*}b\end{align*} we get \begin{align*}b=\frac{9}{\tan 20^\circ} \approx 24.73\end{align*}.

Concept Problem Revisit If you draw the triangle described in this problem, you will see that the sine \begin{align*}\frac {opposite}{hypotenuse}\end{align*} of each of the acute angles in the same. It is \begin{align*}\frac{4}{hypotenuse}\end{align*}. So we need to find the hypotenuse.

Let's use the Pythagorean Theorem.

\begin{align*}4^2 + 4^2 = c^2\\ 16 + 16 = c^2\\ 32 = c^2\\ c = 4\sqrt{2}\end{align*}

Therefore, the sine of both of the acute angles is \begin{align*}\frac{4}{4\sqrt{2}}\end{align*} or \begin{align*}\frac{\sqrt{2}}{2}\end{align*}.

Guided Practice

1. Use trig ratios to find \begin{align*}x\end{align*} and \begin{align*}y\end{align*}:

2. Given \begin{align*}\Delta ABC\end{align*} with \begin{align*}m \angle B=90^\circ, m \angle A=43^\circ\end{align*} and \begin{align*}a=7\end{align*}, find \begin{align*}b\end{align*} and \begin{align*}c\end{align*}.

3. The base of a playground slide is 6 ft from the base of the platform and the slide makes a \begin{align*}60^\circ\end{align*} angle with the ground. To the nearest tenth of a foot, how high is the platform at the top of the slide?


1. For \begin{align*}x\end{align*}: \begin{align*}\cos 62^\circ &=\frac{5}{x} \\ x &=\frac{5}{\cos 62^\circ} \approx 10.65\end{align*}

For \begin{align*}y\end{align*}: \begin{align*}\tan 62^\circ &=\frac{y}{5} \\ y &=5 \tan 62^\circ \approx 9.40\end{align*}

2. For \begin{align*}b\end{align*}: \begin{align*}\sin 43^\circ &=\frac{7}{b} \\ b &=\frac{7}{\sin 43^\circ} \approx 10.26\end{align*}

For \begin{align*}c\end{align*}: \begin{align*}\tan 43^\circ &=\frac{7}{c} \\ c &=\frac{7}{\tan 43^\circ} \approx 7.51 \end{align*}

3. \begin{align*}\tan 60^\circ &=\frac{h}{6}\\ h &=6 \tan 60^\circ \approx 10.39\end{align*}

, so the height of the platform is 10.4 ft

Explore More

Use you calculator to find the following trigonometric ratios. Give answers to four decimal places.

  1. \begin{align*}\sin 35^\circ\end{align*}
  2. \begin{align*}\tan 72^\circ\end{align*}
  3. \begin{align*}\cos 48^\circ\end{align*}
  4. \begin{align*}\tan 45^\circ\end{align*}
  5. \begin{align*}\sin 30^\circ\end{align*}
  6. \begin{align*}\cos 88^\circ\end{align*}
  7. Write the three trigonometric ratios of each of the acute angles in the triangle below.

Use trigonometric ratios to find the unknown side lengths in the triangles below. Round your answers to the nearest hundredth.

For problems 11-13 use the given information about \begin{align*}\Delta ABC\end{align*} with right angle \begin{align*}B\end{align*} to find the unknown side lengths. Round your answer to the nearest hundredth.

  1. \begin{align*}a=12\end{align*} and \begin{align*}m \angle A=43^\circ\end{align*}
  2. \begin{align*}m \angle C=75^\circ\end{align*} and \begin{align*}b=24\end{align*}
  3. \begin{align*}c=7\end{align*} and \begin{align*}m \angle A=65^\circ\end{align*}
  4. A ramp needs to have an angle of elevation no greater than 10 degrees. If the door is 3 ft above the sidewalk level, what is the minimum possible ramp length to the nearest tenth of a foot?
  5. A ship, Sea Dancer, is 10 km due East of a lighthouse. A second ship, Nelly, is due north of the lighthouse. A spotter on the Sea Dancer measures the angle between the Nelly and the lighthouse to be \begin{align*}38^\circ\end{align*}. How far apart are the two ships to the nearest tenth of a kilometer?

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