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Right Triangle Trigonometry

Use right triangles to evaluate trigonometric functions.

Using Right Triangles to Evaluate Trigonometric Functions

In earlier sections, we used a unit circle to define the trigonometric functions. In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of \(t\) is its value at \(t\) radians. First, we need to create our right triangle. shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point \((x,y)\) to the x-axis, we have a right triangle whose vertical side has length \(y\) and whose horizontal side has length \(x.\) We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

We know

\[\cos \ t=\frac{x}{1}=x\]

Likewise, we know

\[\sin \ t=\frac{y}{1}=y\]

These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using \((x,y)\) coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of \(x,\) we will call the side between the given angle and the right angle the adjacent side to angle \(t.\) (Adjacent means “next to.”) Instead of \(y,\) we will call the side most distant from the given angle the opposite side from angle \(t.\) And instead of \(1,\) we will call the side of a right triangle opposite the right angle the hypotenuse. These sides are labeled in .

Given a right triangle with an acute angle of \(t,\)

\[\begin{array}{l}\sin (t)=\frac{\text{opposite}}{\text{hypotenuse}} \\ \cos (t)=\frac{\text{adjacent}}{\text{hypotenuse}} \\ \tan (t)=\frac{\text{opposite}}{\text{adjacent}}\end{array}\]

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.”

Example

Try it.

Given the triangle shown in , find the value of \(\cos \ \alpha .\)

Solution

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so:

\[\begin{array}{l}\cos (\alpha )=\frac{\text{adjacent}}{\text{hypotenuse}} \\ =\frac{15}{17}\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Key Equations

Cofunction Identities \(\begin{array}{l}\begin{array}{l} \\ \cos \ t=\sin (\frac{\pi }{2}-t)\end{array} \\ \sin \ t=\cos (\frac{\pi }{2}-t) \\ \tan \ t=\text{cot}(\frac{\pi }{2}-t) \\ \text{cot}\ t=\tan (\frac{\pi }{2}-t) \\ \text{sec}\ t=\text{csc}(\frac{\pi }{2}-t) \\ \text{csc}\ t=\text{sec}(\frac{\pi }{2}-t)\end{array}\)

Key Concepts

  • We can define trigonometric functions as ratios of the side lengths of a right triangle. See .
  • The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See .
  • We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See .
  • Any two complementary angles could be the two acute angles of a right triangle.
  • If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See .
  • We can use trigonometric functions of an angle to find unknown side lengths.
  • Select the trigonometric function representing the ratio of the unknown side to the known side. See .
  • Right-triangle trigonometry permits the measurement of inaccessible heights and distances.
  • The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Given the triangle shown in , find the value of \(\cos \ \alpha .\)

    Fi àwọn àgbèwọlé hàn

    The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so:

    \[\begin{array}{l}\cos (\alpha )=\frac{\text{adjacent}}{\text{hypotenuse}} \\ =\frac{15}{17}\end{array}\]

  2. Given the triangle shown in , find the value of \(\text{sin}\ t.\)

    Fi àwọn àgbèwọlé hàn

    \(\frac{7}{25}\)

  3. Using the triangle shown in , evaluate \(\sin \ \alpha ,\) \(\cos \ \alpha ,\) \(\tan \ \alpha ,\) \(\text{sec}\ \alpha ,\) \(\text{csc}\ \alpha ,\) and \(\text{cot}\ \alpha .\)

    Fi àwọn àgbèwọlé hàn
    \[\begin{array}{l}\sin \ \alpha =\frac{\text{opposite }\alpha }{\text{hypotenuse}}=\frac{4}{5} \\ \cos \ \alpha =\frac{\text{adjacent to }\alpha }{\text{hypotenuse}}=\frac{3}{5} \\ \tan \ \alpha =\frac{\text{opposite }\alpha }{\text{adjacent to }\alpha }=\frac{4}{3} \\ \text{sec}\ \alpha =\frac{\text{hypotenuse}}{\text{adjacent to }\alpha }=\frac{5}{3} \\ \text{csc}\ \alpha =\frac{\text{hypotenuse}}{\text{opposite }\alpha }=\frac{5}{4} \\ \text{cot}\ \alpha =\frac{\text{adjacent to }\alpha }{\text{opposite }\alpha }=\frac{3}{4}\end{array}\]

  4. Using the triangle shown in , evaluate \(\sin \ t,\) \(\cos \ t,\) \(\tan \ t,\) \(\text{sec}\ t,\) \(\text{csc}\ t,\) and \(\text{cot}\ t.\)

    Fi àwọn àgbèwọlé hàn

    \(\begin{array}{l}sin\ t=\frac{33}{65},\cos \ t=\frac{56}{65},tan\ t=\frac{33}{56}, \\ \text{sec}\ t=\frac{65}{56},\text{csc}\ t=\frac{65}{33},\text{cot}\ t=\frac{56}{33}\end{array}\)

  5. Find the exact value of the trigonometric functions of \(\frac{\pi }{3},\) using side lengths.

    Fi àwọn àgbèwọlé hàn
    \[\begin{array}{l}\sin (\frac{\pi }{3})=\frac{\text{opp}}{\text{hyp}}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2} \\ \cos (\frac{\pi }{3})=\frac{\text{adj}}{\text{hyp}}=\frac{s}{2s}=\frac{1}{2} \\ \tan (\frac{\pi }{3})=\frac{\text{opp}}{\text{adj}}=\frac{\sqrt{3}s}{s}=\sqrt{3} \\ \text{sec}(\frac{\pi }{3})=\frac{\text{hyp}}{\text{adj}}=\frac{2s}{s}=2 \\ \text{csc}(\frac{\pi }{3})=\frac{\text{hyp}}{\text{opp}}=\frac{2s}{\sqrt{3}s}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3} \\ \text{cot}(\frac{\pi }{3})=\frac{\text{adj}}{\text{opp}}=\frac{s}{\sqrt{3}s}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\end{array}\]

  6. Find the exact value of the trigonometric functions of \(\frac{\pi }{4},\) using side lengths.

    Fi àwọn àgbèwọlé hàn

    \(\sin (\frac{\pi }{4})=\frac{1}{\sqrt{2}},\cos (\frac{\pi }{4})=\frac{1}{\sqrt{2}},\tan (\frac{\pi }{4})=1,\)
    \(\text{sec}(\frac{\pi }{4})=\sqrt{2},csc(\frac{\pi }{4})=\sqrt{2},\text{cot}(\frac{\pi }{4})=1\)

  7. If \(\sin \ t=\frac{5}{12},\) find \(\cos (\frac{\pi }{2}-t).\)

    Fi àwọn àgbèwọlé hàn

    According to the cofunction identities for sine and cosine,

    \[\sin \ t=\cos (\frac{\pi }{2}-t).\]

    So

    \[\cos (\frac{\pi }{2}-t)=\frac{5}{12}.\]

  8. If \(\text{csc}(\frac{\pi }{6})=2,\) find \(\text{sec}(\frac{\pi }{3}).\)

    Fi àwọn àgbèwọlé hàn

    2

  9. Find the unknown sides of the triangle in .

    Fi àwọn àgbèwọlé hàn

    We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

    \[\tan (30^{\circ})=\frac{7}{a}\]

    We rearrange to solve for \(a.\)

    \[\begin{array}{l}a=\frac{7}{\tan (30^{\circ})} \\ \approx 12.1\end{array}\]

    We can use the sine to find the hypotenuse.

    \[\sin (30^{\circ})=\frac{7}{c}\]

    Again, we rearrange to solve for \(c.\)

    \[\begin{array}{l}c=\frac{7}{\sin (30^{\circ})} \\ \approx 14\end{array}\]

  10. A right triangle has one angle of \(\frac{\pi }{3}\) and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

    Fi àwọn àgbèwọlé hàn

    \(\text{adjacent}=10;\) \(\text{opposite}=10\sqrt{3}\); missing angle is \(\frac{\pi }{6}\)

  11. To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of \(57^{\circ}\) between a line of sight to the top of the tree and the ground, as shown in . Find the height of the tree.

    Fi àwọn àgbèwọlé hàn

    We know that the angle of elevation is \(57^{\circ}\) and the adjacent side is 30 ft long. The opposite side is the unknown height.

    The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of \(57^{\circ},\) letting \(h\) be the unknown height.

    \[\begin{array}{ll}\ \tan \ \theta =\frac{\text{opposite}}{\text{adjacent}} & \\ \text{tan}(57^{\circ})=\frac{h}{30} & \text{Solve for }h. \\ h=30\tan (57^{\circ}) & \text{Multiply}. \\ h\approx 46.2 & \text{Use a calculator}.\end{array}\]

    The tree is approximately 46 feet tall.

  12. How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of \(\frac{5\pi }{12}\) with the ground? Round to the nearest foot.

    Fi àwọn àgbèwọlé hàn

    About 52 ft

  13. For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle.

    Fi àwọn àgbèwọlé hàn

  14. When a right triangle with a hypotenuse of 1 is placed in the unit circle, which sides of the triangle correspond to the x- and y-coordinates?

  15. The tangent of an angle compares which sides of the right triangle?

    Fi àwọn àgbèwọlé hàn

    The tangent of an angle is the ratio of the opposite side to the adjacent side.

  16. What is the relationship between the two acute angles in a right triangle?

  17. Explain the cofunction identity.

    Fi àwọn àgbèwọlé hàn

    For example, the sine of an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement.

  18. \(\cos (\text{34^{\circ}})=\sin (\text{__^{\circ}})\)

  19. \(\cos (\frac{\pi }{3})=\sin \text{(___)}\)

    Fi àwọn àgbèwọlé hàn

    \(\frac{\pi }{6}\)

  20. \(\text{csc}(\text{21^{\circ}})=\text{sec}(\text{___^{\circ}})\)

  21. \(\tan (\frac{\pi }{4})=\text{cot}(\text{__})\)

    Fi àwọn àgbèwọlé hàn

    \(\frac{\pi }{4}\)

  22. \(\cos \ B=\frac{4}{5},a=10\)

  23. \(\sin \ B=\frac{1}{2},\ a=20\)

    Fi àwọn àgbèwọlé hàn

    \(b=\frac{20\sqrt{3}}{3},c=\frac{40\sqrt{3}}{3}\)

  24. \(\tan \ A=\frac{5}{12},b=6\)

  25. \(\tan \ A=100,b=100\)

    Fi àwọn àgbèwọlé hàn

    \(a=10,000,c=10,000.5\)

  26. \(\sin \ B=\frac{1}{\sqrt{3}},\ a=2\)

  27. \(a=5,\ ∡\ A={60}^{∘}\)

    Fi àwọn àgbèwọlé hàn

    \(b=\frac{5\sqrt{3}}{3},c=\frac{10\sqrt{3}}{3}\)

  28. \(c=12,\ ∡\ A={45}^{∘}\)

  29. \(\sin \ A\)

    Fi àwọn àgbèwọlé hàn

    \(\frac{5\sqrt{29}}{29}\)

  30. \(\cos \ A\)

  31. \(\tan \ A\)

    Fi àwọn àgbèwọlé hàn

    \(\frac{5}{2}\)

  32. \(\text{csc}\ A\)

  33. \(\text{sec}\ A\)

    Fi àwọn àgbèwọlé hàn

    \(\frac{\sqrt{29}}{2}\)

  34. \(\text{cot}\ A\)

  35. \(\sin \ A\)

    Fi àwọn àgbèwọlé hàn

    \(\frac{5\sqrt{41}}{41}\)

  36. \(\cos \ A\)

  37. \(\tan \ A\)

    Fi àwọn àgbèwọlé hàn

    \(\frac{5}{4}\)

  38. \(\text{csc}\ A\)

  39. \(\text{sec}\ A\)

    Fi àwọn àgbèwọlé hàn

    \(\frac{\sqrt{41}}{4}\)

  40. \(\text{cot}\ A\)

Symbols used here

\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
\sin,\ \cos,\ \tan
sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.
^\circ
degrees
1/360 of a full turn. 180° = π radians.
\theta
theta
The usual name for an angle.
\arcsin,\ \sin^{-1}
inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

How to: Right Triangle Trigonometry

  1. Use right triangles to evaluate trigonometric functions.
  2. Find function values for
  3. Use cofunctions of complementary angles.
  4. Use the definitions of trigonometric functions of any angle.
  5. Use right triangle trigonometry to solve applied problems.
  6. Find the sine as the ratio of the opposite side to the hypotenuse.
  7. Find the cosine as the ratio of the adjacent side to the hypotenuse.
  8. Find the tangent as the ratio of the opposite side to the adjacent side.

Questions people ask

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

Wárá

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Diẹ̀ nínú Trigonometry

The unit circleTrigonometric equationsTrigonometric identitiesDegrees and radiansRight-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosinesGraphs of sine, cosine and tangentInverse trigonometric functions