Solving Right Triangles

Watch the video explainer

Learning goals

Choose a useful ratio or theorem from the given information
Find missing sides by solving a trigonometric ratio equation
Recover acute angles with valid inverse operations in degree mode
Preserve calculator precision and check lengths, angles, and ratios
Decide whether the supplied data determine a right triangle

Match a method to the data

The ratios from Sine, Cosine, and Tangent connect an acute reference angle to two sides. For that chosen angle, write the opposite leg as $\text{opp}$ , the adjacent leg as $\text{adj}$ , and the hypotenuse as $\text{hyp}$ . The adjacent leg excludes the hypotenuse.

For example, with a $30^\circ$ angle and hypotenuse $18$ units, the opposite leg is $9$ units:

\begin{aligned} \sin30^\circ&=\frac9{18},\\ 9&=18\sin30^\circ. \end{aligned}

The known angle supplies a ratio; algebra then supplies a length.

The given angle connects the opposite leg with the hypotenuse.

Choose a ratio containing the known side and the side you want. Writing an equation first keeps multiplying and dividing from becoming a guess.

Side pair relative to the selected acute angle	Ratio
Opposite and hypotenuse	$\sin\theta=\text{opp}/\text{hyp}$
Adjacent leg and hypotenuse	$\cos\theta=\text{adj}/\text{hyp}$
Opposite and adjacent legs	$\tan\theta=\text{opp}/\text{adj}$

With two known sides and a missing third side, Pythagoras gives a direct route. With two known sides and a missing acute angle, reverse a ratio using an inverse operation, introduced below. The other acute angle follows by subtracting from $90^\circ$ .

Check your understanding

A right triangle has a known hypotenuse and a known acute angle $\theta$ . Write the opposite leg as $\text{opp}$ and the hypotenuse as $\text{hyp}$ . Which equation directly finds the opposite leg?

\cos\theta=\text{opp}/\text{hyp}

\tan\theta=\text{opp}/\text{hyp}

\sin\theta=\text{opp}/\text{hyp}

\text{opp}=\text{hyp}+\theta

Find sides from a known angle

Use degree mode for every calculator calculation in this lesson. A quick check is that entering the sine of $30^\circ$ gives $0.5$ . Enter the angle as the input to the trigonometric function, then use the resulting ratio in the equation.

Worked example 1 A given hypotenuse

Problem. Triangle $ABC$ is right at $C$ . Its hypotenuse $AB$ is $14$ cm, and angle $A$ measures $37^\circ$ . Find both legs to the nearest tenth of a centimeter and find angle $B$ .

The marked angle at A determines the side roles.

Solution. Relative to $A$ , side $BC$ is opposite and $CA$ is the adjacent leg. Use each with the original known hypotenuse:

\begin{aligned} \sin37^\circ&=\frac{BC}{14},\\ BC&=14\sin37^\circ,\\ BC&\approx8.4\text{ cm}. \end{aligned}

For the other leg,

\begin{aligned} \cos37^\circ&=\frac{CA}{14},\\ CA&=14\cos37^\circ,\\ CA&\approx11.2\text{ cm}. \end{aligned}

The acute angles add to $90^\circ$ , so

m\angle B=53^\circ.

Both legs are shorter than $14$ cm. Using the original data for both calculations avoids passing a rounded leg into a second calculation.

Worked example 2 The unknown is below the fraction bar

Problem. Triangle $QRT$ is right at $R$ . Angle $T$ measures $64^\circ$ , and leg $RT$ is $7.5$ m. Find hypotenuse $QT$ to the nearest tenth of a meter.

Read the right-angle and reference-angle markings in this orientation.

Solution. Relative to $T$ , the given side is the adjacent leg. Cosine uses this side and the unknown hypotenuse:

\cos64^\circ=\frac{7.5}{QT}.

Multiply by $QT$ , then divide by the positive cosine value:

\begin{aligned} QT\cos64^\circ&=7.5,\\ QT&=\frac{7.5}{\cos64^\circ},\\ QT&\approx17.1\text{ m}. \end{aligned}

The hypotenuse exceeds the given leg. Multiplying $7.5$ by the cosine would instead make the supposed hypotenuse shorter than that leg.

Check your understanding

A right triangle has adjacent leg 9 units and unknown hypotenuse h, relative to an acute angle of 40 degrees. Which expression gives h?

9\cos40^\circ

\cos40^\circ/9

9/\cos40^\circ

\cos^{-1}(9/40)

Reverse a ratio to find an acute angle

A trigonometric function takes an angle and returns a ratio. An inverse trigonometric operation takes a valid ratio and returns the angle that produces it. Here we use these operations only to recover acute angles of nondegenerate right triangles.

For example, a right triangle with adjacent leg $7$ and hypotenuse $10$ has

\cos\theta=0.7.

The inverse cosine operation in degree mode gives

\begin{aligned}\theta&=\cos^{-1}(0.7),\\\theta&\approx45.6^\circ.\end{aligned}

The two labeled side lengths determine the acute angle.

There is only one acute angle with that ratio. Fixing the adjacent leg at $7$ and the hypotenuse at $10$ fixes the remaining leg by Pythagoras, and SSS then fixes the triangle’s angles. Scaling both lengths preserves the angle. The same argument works for any valid adjacent-to-hypotenuse ratio. Fixing the opposite leg and hypotenuse establishes the corresponding fact for sine. For tangent, choosing an adjacent leg of one unit fixes the opposite leg from its ratio, so SAS fixes the right triangle’s shape.

The notation has a specific meaning: $\sin^{-1}$ , $\cos^{-1}$ , and $\tan^{-1}$ ask for an angle. They are also called arcsine, arccosine, and arctangent. An inverse-function mark after the function name is not a reciprocal of its value. For example,

\sin^{-1}(0.5)=30^\circ.

Taking a reciprocal instead would give

\frac1{0.5}=2.

That number is not the requested angle. Calculator keys may require a shift or second-function key; check the displayed operation and degree mode.

For the acute-angle problems here, the allowed inputs are strict:

Operation	Input that can come from a nondegenerate right triangle’s acute angle
Inverse sine or inverse cosine	A ratio strictly between $0$ and $1$
Inverse tangent	Any positive ratio

The hypotenuse exceeds each leg, but one leg may exceed the other. Thus tangent may exceed $1$ . A sine or cosine ratio of $1.2$ cannot describe an acute angle. Inputs $0$ and $1$ for sine or cosine produce boundary angles rather than an acute angle. A zero or negative tangent ratio also fails the positive-leg condition.

Worked example 3 Two given legs

Problem. A right triangle has legs $13$ and $19$ cm. Let $\theta$ be the angle opposite the $13$ cm leg. Find both acute angles to the nearest tenth of a degree.

The unknown angle faces the 13 cm leg.

Solution. Both known sides are legs. Relative to $\theta$ , their opposite-to-adjacent ratio gives

\tan\theta=\frac{13}{19}.

The ratio is positive, so it gives an acute angle. Use inverse tangent:

\theta=\tan^{-1}\!\left(\frac{13}{19}\right).

In degree mode,

\theta\approx34.4^\circ.

Keep the unrounded inverse value to find the other acute angle $\phi$ :

\begin{aligned} \phi&=90^\circ-\theta,\\ \phi&\approx55.6^\circ. \end{aligned}

The angle opposite the longer leg is larger, matching side-angle ordering.

Check your understanding

An acute angle $\theta$ in a right triangle has cosine 0.35. Which operation finds $\theta$ in degrees?

\theta=1/0.35

\theta=\cos(0.35^\circ)

\theta=\sin^{-1}(0.35)

\theta=\cos^{-1}(0.35)

Complete a triangle from two sides

Two positive leg lengths determine a right triangle by SAS. A positive leg and a longer hypotenuse determine the missing leg by Pythagoras, then determine the triangle by SSS. To find an acute angle, select the inverse operation that uses the original known pair directly.

Worked example 4 A leg and the hypotenuse

Problem. A right triangle has hypotenuse $23$ m and one leg $9$ m. Let $\alpha$ be the acute angle opposite that leg. Find the missing leg to the nearest tenth of a meter and both acute angles to the nearest tenth of a degree.

The known leg is across from the selected acute angle.

Solution. Subtract squares for the missing leg $x$ :

\begin{aligned} x^2&=23^2-9^2,\\ x^2&=448,\\ x&=8\sqrt7,\\ x&\approx21.2\text{ m}. \end{aligned}

The known $9$ m leg is opposite $\alpha$ . Its ratio to the hypotenuse is between zero and one:

\alpha=\sin^{-1}\!\left(\frac9{23}\right).

Thus,

\alpha\approx23.0^\circ.

Subtract the unrounded angle from $90^\circ$ to obtain the other acute angle, approximately $67.0^\circ$ . The smaller leg faces the smaller angle. The exact leg $8\sqrt7$ also checks the side calculation:

9^2+(8\sqrt7)^2=23^2.

Keep precision, then check

An expression such as $14\sin37^\circ$ can be left exact until its final decimal is needed. Keep the calculator’s stored values or re-enter the original expression, rather than copying a rounded ratio or angle into the next step.

Worked example 5 Two calculator records

Problem. A right triangle has an acute angle of $17.3^\circ$ and an adjacent leg of $32$ cm. Find the opposite leg to the nearest tenth of a centimeter. Compare the result with rounding the tangent value to $0.31$ before multiplying.

The original angle and side length are the calculator inputs.

Solution. Use the opposite-to-adjacent ratio, keeping the whole expression:

\begin{aligned} x&=32\tan17.3^\circ,\\ x&\approx9.966890\text{ cm},\\ x&\approx10.0\text{ cm}. \end{aligned}

The middle line shows extra digits for comparison, not a new exact value. Rounding the ratio first gives a different computation:

32(0.31)=9.92.

That altered computation rounds to $9.9$ cm. The required result from the original data is $10.0$ cm. Rounding belongs at the requested final result.

Check results against the geometry and the original data. Each leg must be positive and shorter than the hypotenuse. The acute angles must lie strictly between $0^\circ$ and $90^\circ$ and add to $90^\circ$ . Substitute into the original ratio, and use Pythagoras as a side check when all three sides are available. Rounded values usually satisfy these equations only approximately.

Check your understanding

Two known legs are 6 and 13 units. You recover their acute angle $\theta$ opposite 6, then use the adjacent leg to find the hypotenuse h. Which process preserves the original data until final rounding?

A Evaluate

h=\frac{13}{\cos(\tan^{-1}(\frac6{13}))}

in degree mode, then round h at the end B Round

\theta

to a whole degree before using cosine C Round 6/13 to a whole number before taking the inverse D Replace both leg lengths by their average before calculating

Decide whether there is enough information

The right angle is essential. These methods are for nondegenerate right triangles, not an arbitrary triangle with a familiar-looking sketch.

An acute angle fixes the shape, but it does not fix the size. A right triangle with angles $28^\circ$ , $62^\circ$ , and $90^\circ$ can have a hypotenuse of $10$ cm or $20$ cm. The second triangle is a dilation of the first. Angles alone cannot determine its side lengths.

The angle data agree, while the side lengths differ by a positive scale factor.

One side alone is also insufficient: with a leg fixed, the other leg can change, changing the acute angles. In contrast, a valid pair of sides, or one acute angle and a side whose role relative to that angle is specified, fixes both the shape and size.

A different size measurement can sometimes replace a side length. For example, if the perimeter is $P$ and an acute angle is $\theta$ , write the two legs using the hypotenuse $h$ . Their sum with the hypotenuse gives

P=h\sin\theta+h\cos\theta+h.

Factoring $h$ lets you recover the size. This uses the same ratio algebra, not a new triangle theorem.

Check consistency too. A stated hypotenuse of $6$ cm with a leg of $8$ cm is impossible. So is an acute-angle cosine ratio of $1$ . Getting a calculator output does not make the geometric data valid.

Check your understanding

Which data allow more than one noncongruent right triangle? Each set includes a right angle.

A Legs of 6 and 7 units B An acute angle of

38^\circ

and a 9-unit leg opposite that angle C A 9-unit hypotenuse and a 4-unit leg D Acute angles of

29^\circ

and

61^\circ

, with no lengths

Solving Right Triangles

Learning goals

Match a method to the data

Find sides from a known angle

Reverse a ratio to find an acute angle

Complete a triangle from two sides

Keep precision, then check

Decide whether there is enough information

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress