Let's remind ourselves of the names of the sides of a right triangle relative to an angle \theta.

The three fundamental trigonometric ratios, sine, cosine, and tangent, represent the ratios of the sides of a right triangle. The ratios are often shortened to \sin , \cos, and \tan, and they are defined as

\eqalign{ \sin(\theta) &= \dfrac {\text{opposite}} {\text{hypotenuse}}, \\[5pt] \cos(\theta) &= \dfrac {\text{adjacent}} {\text{hypotenuse}}, \\[5pt] \tan(\theta) &= \dfrac {\text{opposite}} {\text{adjacent}}. \\ }

For a given angle \theta , these ratios remain the same no matter what the size of the triangle is. For example, if we were to double the length of each side of a given triangle, the values of these ratios would remain unchanged.

An easy way of remembering how to compute the different trigonometric ratios is by using the three-syllable acronym SOH-CAH-TOA. It stands for:

  • SOH \rightarrow Sine = Opposite / Hypotenuse,

  • CAH \rightarrow Cosine = Adjacent / Hypotenuse,

  • TOA \rightarrow Tangent = Opposite / Adjacent.

FLAG

Find \sin{\theta}, \cos{\theta}, and \tan{\theta} for the triangle below.

EXPLANATION

Remember the acronym "SOH-CAH-TOA."

  • The sine of an angle is equal to the opposite divided by the hypotenuse (S-O-H). Relative to the angle \theta, the opposite side is \overline{AC}, and the hypotenuse is \overline{BC}. Therefore, \sin{\theta} = \dfrac{AC}{BC} = \dfrac{3}{5}.

  • The cosine of an angle is equal to the adjacent divided by the hypotenuse (C-A-H). Relative to the angle \theta, the adjacent side is \overline{AB}, and the hypotenuse is \overline{BC}. Therefore, \cos{\theta} = \dfrac{AB}{BC} = \dfrac{4}{5}.

  • The tangent of an angle is equal to the opposite divided by the adjacent (T-O-A). Relative to the angle \theta, the opposite side is \overline{AC}, and the adjacent side is \overline{AB}. Therefore, \tan{\theta} = \dfrac{AC}{AB} = \dfrac{3}{4}.

FLAG

Use the diagram above to determine $\cos{Q}.$

a
b
c
d
e

In the right triangle $\triangle KLM,$ $\angle K$ is a right angle, $KL=24,$ $KM=32$ and $LM=40.$ Find $\sin M.$

a
$\dfrac{2}{5}$
b
$\dfrac{3}{5}$
c
$\dfrac{4}{5}$
d
$\dfrac{1}{4}$
e
$\dfrac{3}{4}$

For the right triangle $PQR$ above, which of the following statements are true?

  1. $\cos{Q} = \dfrac{8}{17}$
  2. $\tan{Q} = \dfrac{8}{15}$
  3. $\sin{R} = \dfrac{15}{17}$
a
I only
b
II only
c
III only
d
I and II only
e
II and III only

In the right triangle \triangle ABC, the angle \angle A is a right angle, and the side lengths are AB=2, AC=\sqrt{5}, and BC=3. Find \tan B.

EXPLANATION

Let's illustrate the given triangle.

The tangent of an angle is equal to the opposite divided by the adjacent (T-O-A). Relative to \angle B, the opposite side is \overline{AC}, and the adjacent side is \overline{AB}. Therefore, \eqalign{ \tan{B} &= \dfrac{AC}{AB} = \dfrac{\sqrt{5}}{2}. }

FLAG

Use the diagram above to determine $\sin{C}.$

a
b
c
d
e

In the right triangle $\triangle ABC,$ $\angle A$ is a right angle, $AB=2\sqrt{7},$ $AC=6$ and $BC=8.$ Find $\sin B.$

a
$\dfrac{\sqrt 7}{3}$
b
$\dfrac{4}{3}$
c
$\dfrac{\sqrt{21}}{7}$
d
$\dfrac{\sqrt 7}{4}$
e
$\dfrac{3}{4}$

In the right triangle $\triangle{STU},$ $\angle S$ is a right angle, $ST=2\sqrt{10},$ $SU=3,$ and $TU=7.$ Find $\tan T.$

a
b
c
d
e

Given that \tan\theta = \dfrac{1}{4}, find the value of x.

EXPLANATION

The tangent of an angle is equal to the opposite divided by the adjacent (T-O-A).

Relative to the angle \theta, the length of the opposite side is x-1, and the length of the adjacent side is 2x. Therefore,

\begin{align*} \tan\theta &= \dfrac{\text{opposite}}{\text{adjacent}} \\[5pt] \frac{1}{4} &= \frac{x-1}{2x}\\[5pt] 2x &= 4(x-1)\\[5pt] 2x &= 4x - 4\\[5pt] 2x &= 4\\[5pt] x &= 2. \end{align*}

FLAG

Given that $\cos\theta =\dfrac{11}{61}$, the value of $x$ is

a
$11$
b
$2$
c
$6$
d
$13$
e
$7$

If $\sin\theta = \dfrac{12}{13},$ then $x = $

a
b
c
d
e

Calculators usually have a special key that allows us to compute \sin\theta, \cos\theta, and \tan\theta for some specific value of \theta.

For example, using a calculator, we can compute the trigonometric ratios for \theta = 30^\circ and get the following results:

\begin{align*} \sin \left(30^\circ\right) &= 0.5\\ \cos \left(30^\circ\right) &= 0.866...\\ \tan \left(30^\circ\right) &= 0.577... \end{align*}

Watch out! Make sure that the calculator is in degrees mode (instead of radians mode) before you start the computation.

FLAG

Evaluate \cos\left(50^\circ\right). Round the answer to two decimal places.

EXPLANATION

Using a calculator, we find

\begin{align*} \cos \left(50^\circ\right) &= 0.64278...\approx 0.64, \end{align*}

rounded to two decimal places.

FLAG

Rounded to $3$ decimal places, $\sin\left(72^\circ\right)\approx$

a
b
c
d
e

Rounded to $3$ decimal places, $\tan\left(55^\circ\right)=$

a
$0.574$
b
$1.181$
c
$0.819$
d
$1.574$
e
$1.428$

Which of the following is true for $\angle B?$

  1. $\cos{B} \approx 0.82$
  2. $\sin{B} \approx 0.44$
  3. $\tan{B} \approx 0.70$
a
I and III only
b
III only
c
I only
d
II only
e
II and III only
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