Right Triangle Trigonometry

18 min read

A surveyor cannot stretch a tape measure to the top of a flagpole, and a sailor cannot walk across the water to a lighthouse. Both can measure two things from where they stand: one distance along the ground and one angle. Right triangle trigonometry turns that distance and that angle into the height, the range, or whichever side cannot be reached directly. The reason it works is one fact about similar triangles: in a right triangle, fixing one acute angle fixes the shape, so the ratio of any two sides depends only on that angle and not on how big the triangle is. Six named ratios (sine, cosine, tangent, and their reciprocals) record those side ratios, and the mnemonic SOH-CAH-TOA recalls the three primary ones on demand.

By the end of this page you can:

• Label the opposite side, the adjacent side, and the hypotenuse relative to a chosen acute angle.
• State and apply SOH-CAH-TOA to write sine, cosine, and tangent as side ratios.
• Find a missing side of a right triangle from one side and one acute angle.
• Find a missing acute angle from two sides using an inverse trigonometric function.
• Solve a right triangle: find all six parts (three sides and three angles).
• Set up and solve an angle-of-elevation or angle-of-depression problem.

The floor to take away: in a right triangle, $\sin\theta$ is the opposite side over the hypotenuse, $\cos\theta$ is the adjacent side over the hypotenuse, and $\tan\theta$ is the opposite side over the adjacent side, all measured relative to the acute angle $\theta$. Knowing one angle and one side is enough to find any other side, because the ratio is fixed by the angle alone.

Quick Reference

Before You Start

Check each box you can do from memory. A box you cannot check yet is not a problem; it points to a quick refresher.

• I can find the third side of a right triangle from the other two using $a^2 + b^2 = c^2$. (Pythagorean Theorem)
• I can state the side lengths of the 30-60-90 triangle ($1$, $\sqrt{3}$, $2$) and the 45-45-90 triangle ($1$, $1$, $\sqrt{2}$). (Special Right Triangles)
• I can recognize that $\sin\theta$ means the sine function applied to the angle $\theta$, not a product.

A 60-second self-check. Pick an answer, then reveal the reasoning.

Try This First

Before any formula, draw a right triangle and pick one of its acute angles. Call it $\theta$.

Measure (or estimate) two things: the length of the side directly across the triangle from $\theta$, and the length of the longest side (the one across from the square corner). Write the first length over the second as a fraction.

Now draw a second right triangle that has the same angle $\theta$ but is twice as large, and form the same fraction.

Predict: will the two fractions be equal, or will the larger triangle give a larger fraction?

The Idea Behind Right Triangle Trigonometry

In a right triangle, fixing one acute angle fixes the shape, so the ratios of the sides depend only on that angle. A trigonometric function takes an angle as input and returns a side ratio as output.

Picture two right triangles with the same acute angle but different sizes, like a photo and an enlargement. They are similar, so corresponding sides are in the same proportion. The ratio “opposite side over hypotenuse” is identical in both triangles. Sine, cosine, and tangent are names for three of these angle-determined ratios, and cosecant, secant, and cotangent are their reciprocals.

Because the ratio is set by the angle, two questions open up. Given the angle and one side, any other side follows: multiply or divide by the known ratio. Given two sides, the angle follows: an inverse trigonometric function reads the angle back from the ratio. Those two moves, together with the Pythagorean theorem, are enough to find every part of a right triangle.

Prerequisite Hub

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    click D "right-triangle-trig.html"

Builds on (review these first if needed):

Unlocks (comes after this):

Scope note. Oblique triangles (no right angle), handled by the Law of Sines and the Law of Cosines, are a separate skill. Everything here assumes one angle is exactly $90^\circ$.

Official Definitions

The six trigonometric ratios of an acute angle

For an acute angle $\theta$ in a right triangle, with sides labeled relative to $\theta$: \[ \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}, \] \[ \csc\theta = \frac{\text{hypotenuse}}{\text{opposite}}, \quad \sec\theta = \frac{\text{hypotenuse}}{\text{adjacent}}, \quad \cot\theta = \frac{\text{adjacent}}{\text{opposite}}. \] The mnemonic SOH-CAH-TOA records the first three: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.

Source: Stewart 7e, Section 6.2 Trigonometry of Right Triangles. The same six ratios appear in OpenStax Precalculus 2e, Section 5.4.

Labeling the sides relative to $\theta$

Angle of elevation and angle of depression

The angle of elevation is the angle measured up from a horizontal line to the line of sight of an object above the horizontal. The angle of depression is the angle measured down from a horizontal line to the line of sight of an object below the horizontal.

Source: Stewart 7e, Section 6.2 Trigonometry of Right Triangles.

Quick Reference

The three primary ratios (SOH-CAH-TOA). For an acute angle $\theta$: \[ \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}. \]

The three reciprocal ratios. \[ \csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{1}{\tan\theta}. \] The reciprocal pairing crosses the “co”: cosecant pairs with sine, secant pairs with cosine.

Exact values to know (from the special triangles).

Finding a missing angle. If $\sin\theta = k$, then $\theta = \sin^{-1}k$ (read “inverse sine of $k$”). The functions $\cos^{-1}$ and $\tan^{-1}$ work the same way for cosine and tangent.

Labeling the Sides

The labels depend on which acute angle is in view. The hypotenuse is always across from the right angle. The opposite side is across from the chosen angle $\theta$. The adjacent side is the remaining leg, the one touching $\theta$.

Example 1. A right triangle has the right angle at $C$ and acute angle $\theta$ at $A$. Relative to $\theta$, name the three sides.

Predict first: which single side keeps its name no matter which acute angle is chosen?

The hypotenuse keeps its name; it is the side opposite the right angle $C$ and the longest side. The opposite side is the leg across from $A$. The adjacent side is the other leg, which touches $A$. Check: the prediction holds. Only the two legs swap roles when the chosen angle changes; the hypotenuse is fixed by the right angle, which never moves.

The Three Primary Ratios

Sine, cosine, and tangent are the three ratios captured by SOH-CAH-TOA.

Example 2. A right triangle has the side opposite $\theta$ equal to $3$, the adjacent side equal to $4$, and hypotenuse $5$. Find $\sin\theta$, $\cos\theta$, and $\tan\theta$.

Predict first: the opposite side ($3$) is shorter than the hypotenuse ($5$), so $\sin\theta$ should be less than $1$.

\[ \sin\theta = \frac{3}{5} = 0.6, \qquad \cos\theta = \frac{4}{5} = 0.8, \qquad \tan\theta = \frac{3}{4} = 0.75. \]

Check: $0.6$ is less than $1$, as predicted. Each value is a ratio of two specific sides, read off the labels.

Finding a Missing Side

When two legs are not both given, the same ratios produce a missing side. If one acute angle and one side are known, choose the ratio that links the known side to the wanted side.

Example 3. A surveyor stands $40$ feet from the base of a flagpole. The angle of elevation to the top of the pole is $30^\circ$. Find the height $h$ of the pole.

Predict first: the horizontal distance is $40$ feet and the angle is shallow, so the height should be noticeably less than $40$ feet.

The height is opposite the $30^\circ$ angle and the $40$-foot distance is adjacent, so tangent (opposite over adjacent) links them: \[ \tan(30^\circ) = \frac{h}{40} \implies h = 40\tan(30^\circ) = 40 \cdot \frac{\sqrt{3}}{3} = \frac{40\sqrt{3}}{3} \approx 23.1 \text{ feet}. \]

Check: $23.1$ feet is less than $40$ feet, as predicted.

Finding a Missing Angle

When two sides are known but no acute angle is, the angle is recovered with an inverse trigonometric function. If $\tan\theta = k$, then $\theta = \tan^{-1}k$, the angle whose tangent is $k$. The functions $\sin^{-1}$ and $\cos^{-1}$ do the same for sine and cosine.

Example 4. A right triangle has the side opposite $\theta$ equal to $3$ and the adjacent side equal to $4$. Find $\theta$ to the nearest degree.

Predict first: the opposite side ($3$) is a little shorter than the adjacent side ($4$), so $\theta$ should be a little less than $45^\circ$ (at exactly $45^\circ$ the two legs would be equal).

The two known sides are the opposite and the adjacent, so tangent links them: \[ \tan\theta = \frac{3}{4} = 0.75 \implies \theta = \tan^{-1}(0.75) \approx 36.87^\circ \approx 37^\circ. \]

Check: $37^\circ$ is a little less than $45^\circ$, as predicted.

Solving a Right Triangle

To solve a triangle is to find all six parts: the three sides and the three angles. In a right triangle one angle is already known ($90^\circ$), so any two more pieces of information (one side plus one acute angle, or two sides) determine the rest. The tools are the three ratios for missing sides, the inverse ratios for missing angles, and the fact that the two acute angles add to $90^\circ$.

Example 5. A right triangle has the right angle at $C$, the side opposite angle $A$ equal to $7$, and the side opposite angle $B$ equal to $24$. Solve the triangle.

Predict first: side $24$ is much longer than side $7$, so angle $B$ (opposite the long side) should be much larger than angle $A$, and the hypotenuse should be a little more than $24$.

Step 1, the hypotenuse. By the Pythagorean theorem, \[ c = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25. \]

Step 2, angle $A$. Side $7$ is opposite $A$ and the hypotenuse is $25$, so \[ \sin A = \frac{7}{25} = 0.28 \implies A = \sin^{-1}(0.28) \approx 16.26^\circ. \]

Step 3, angle $B$. The two acute angles add to $90^\circ$, so \[ B = 90^\circ - 16.26^\circ \approx 73.74^\circ. \]

Solution. Sides $7$, $24$, $25$; angles $A \approx 16.26^\circ$, $B \approx 73.74^\circ$, $C = 90^\circ$.

Check: the hypotenuse $25$ is a little more than $24$, and angle $B \approx 73.74^\circ$ is much larger than angle $A \approx 16.26^\circ$, both as predicted. A further check: $\sin B = \tfrac{24}{25} = 0.96$ and $\sin^{-1}(0.96) \approx 73.74^\circ$, which agrees with Step 3.

Worked Example: An Angle-of-Depression Application

Example 6. From the top of a $60$-foot lighthouse, the angle of depression to a boat is $30^\circ$. How far is the boat from the base of the lighthouse?

Predict first: the angle is shallow ($30^\circ$), so the boat should be farther from the base than the lighthouse is tall, that is, more than $60$ feet out.

The angle of depression from the top equals the angle of elevation from the boat, because they are alternate interior angles formed by two parallel horizontal lines cut by the line of sight. So the angle at the boat is $30^\circ$. The lighthouse height $60$ is opposite that angle and the ground distance $d$ is adjacent, so tangent links them: \[ \tan(30^\circ) = \frac{60}{d} \implies d = \frac{60}{\tan(30^\circ)} = \frac{60}{\frac{\sqrt{3}}{3}} = 60 \cdot \frac{3}{\sqrt{3}} = 60\sqrt{3} \approx 103.9 \text{ feet}. \]

Check: $103.9$ feet is more than $60$ feet, as predicted. A shallow angle places the boat far out, which the answer reflects.

Common Misconceptions

Leveled Practice

Level 1 - Direct Application

Problem 1. A right triangle has opposite side $6$, adjacent side $8$, hypotenuse $10$. Find $\sin\theta$ and $\tan\theta$.

Problem 2. State $\sin(45^\circ)$, $\cos(45^\circ)$, and $\tan(45^\circ)$.

Level 2 - Find the Missing Side

Problem 3. A right triangle has adjacent side $9$ and hypotenuse $15$. Find the opposite side, then $\sin\theta$.

Problem 4. A guy wire runs from the top of a $30$-foot pole to a stake on the ground, making a $50^\circ$ angle with the ground. Find the length of the wire to the nearest tenth of a foot.

Level 3 - Find the Missing Angle

Problem 5. A right triangle has the side opposite an acute angle equal to $5$ and the side adjacent equal to $5$. Find the angle.

Problem 6. A ladder $30$ feet long leans against a wall and reaches $24$ feet up the wall. Find the angle of elevation the ladder makes with the ground, to the nearest degree.

Level 4 - Solve the Triangle

Problem 7. A right triangle has the right angle at $C$, hypotenuse $c = 20$, and one acute angle $A = 35^\circ$. Solve the triangle (find $B$, side $a$ opposite $A$, and side $b$ opposite $B$), rounding sides to the nearest tenth.

Level 5 - Applied Modeling

Problem 8. From a window $30$ feet above the street, the angle of elevation to the top of the building across the street is $50^\circ$, and the angle of depression to the base of that building is $20^\circ$. Find the height of the building across the street. Predict whether the height is more or less than $30$ feet before computing.

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes.

Novice (Levels 1-2):

• Label the opposite, adjacent, and hypotenuse relative to a chosen acute angle.
• State and apply SOH-CAH-TOA for sine, cosine, and tangent.
• Find a missing side from one acute angle and one known side.

Competent (Levels 3-4):

• Recall the exact values of sine, cosine, and tangent at $30^\circ$, $45^\circ$, and $60^\circ$.
• Find a missing acute angle from two sides with $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$.
• Solve a right triangle: find all three sides and all three angles.

Proficient (Level 5):

• Set up and solve an angle-of-elevation or angle-of-depression problem, including the two-triangle case with a shared horizontal distance.
• Explain why corresponding ratios depend only on the angle, not the triangle’s size.

Mental Model

Think of a trig function as a dial that reads a side ratio off an angle. Set the dial to an acute angle, and a right triangle of any size at that angle hands back the same three ratios, because all such triangles are scaled copies of one another. Sine reads opposite over hypotenuse, cosine reads adjacent over hypotenuse, tangent reads opposite over adjacent, captured by SOH-CAH-TOA. Turning the dial forward (angle in, ratio out) finds a missing side; turning it backward with an inverse function (ratio in, angle out) finds a missing angle. With those two turns plus the Pythagorean theorem, every part of a right triangle is reachable from two known parts.

Check Yourself

Close the notes and answer each from memory, then reveal it. Pulling an idea back from memory is one of the strongest ways to make it stick.

Connections

Looking back:

• Special Right Triangles supplies the exact $30^\circ$, $45^\circ$, $60^\circ$ ratios used here without a calculator.
• Pythagorean Theorem provides the third side whenever two sides are known.
• The Unit Circle: Sine and Cosine shows these acute-angle ratios as first-quadrant coordinates of a point on a circle of radius $1$.

Looking ahead:

• Area of a Triangle: the formula $A = \tfrac{1}{2}ab\sin C$ measures area from two sides and the included angle, using the sine introduced here.
• Oblique triangles, with no right angle, are handled by the Law of Sines and the Law of Cosines in a later skill.

Real-world connections:

• Surveying and navigation turn one measured distance and one measured angle into an unreachable height or range.
• Physics resolves a force or velocity into horizontal and vertical components with sine and cosine.
• Computer graphics rotate and project points using exactly these ratios.

Resources

Last updated: 2026-06-24