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Solving Right Triangles: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Triangle ABC

Triangle $ABC$ is right at $C$ . The hypotenuse $AB$ is $18$ cm, and angle $A$ measures $41^\circ$ .

Find both legs to the nearest tenth of a centimeter and angle $B$ exactly.

The side labels and marked angle are the givens.

Text description of this figure

Triangle ABC is right at C. Side CB is labeled BC, side AC is labeled CA, and hypotenuse AB is labeled 18 cm. The marked angle at A is labeled 41°.
1. Hint 1
  Identify each leg relative to the given angle at $A$ .
2. Hint 2
  Use the original hypotenuse with sine for the opposite leg and cosine for the adjacent leg.
3. Hint 3
  The acute angles add to a right angle.
Answer

$\begin{aligned}BC&\approx11.8\text{ cm},\\CA&\approx13.6\text{ cm},\\m\angle B&=49^\circ.\end{aligned}$

Full solution

Relative to $A$ , side $BC$ is opposite and $CA$ is the adjacent leg.
Both use the known hypotenuse directly.

The sine equation gives
$\begin{aligned}\sin41^\circ&=\frac{BC}{18},\\BC&=18\sin41^\circ,\\BC&\approx11.8\text{ cm}.\end{aligned}$

The cosine equation gives
$\begin{aligned}\cos41^\circ&=\frac{CA}{18},\\CA&=18\cos41^\circ,\\CA&\approx13.6\text{ cm}.\end{aligned}$

Subtract the given acute angle from a right angle.
$m\angle B=90^\circ-41^\circ$
Thus angle $B$ is exactly $49^\circ$ .

Each leg is shorter than $18$ cm.
The leg opposite the smaller angle is shorter, as expected.

Answer

$\begin{aligned}BC&\approx11.8\text{ cm},\\CA&\approx13.6\text{ cm},\\m\angle B&=49^\circ.\end{aligned}$

Key idea

Calculate each requested leg from the original given side to avoid carrying a rounded intermediate length.
Problem 2 A rotated configuration

Triangle $PQR$ is right at $Q$ . Angle $P$ measures $58^\circ$ , and the opposite leg $QR$ is $6.8$ m.

Find $PR$ and $PQ$ to the nearest tenth of a meter.

The marked reference angle is at P, and the right angle is at Q.

Text description of this figure

Triangle PRQ is right at Q. Side QR is labeled 6.8 m, side PQ is labeled PQ, and hypotenuse PR is labeled PR. The marked angle at P is labeled 58°.
1. Hint 1
  The known side is opposite the selected angle; decide which ratio contains it and each unknown.
2. Hint 2
  Sine places the hypotenuse in the denominator. Tangent places the adjacent leg in the denominator.
3. Hint 3
  Solve each fraction equation before entering numbers into the calculator.
Answer

$\begin{aligned}PR&\approx8.0\text{ m},\\PQ&\approx4.2\text{ m}.\end{aligned}$

Full solution

The hypotenuse is $PR$ , opposite the right angle at $Q$ .
Sine gives
$\begin{aligned}\sin58^\circ&=\frac{6.8}{PR},\\PR&=\frac{6.8}{\sin58^\circ},\\PR&\approx8.0\text{ m}.\end{aligned}$

Relative to $P$ , the adjacent leg is $PQ$ .
Use tangent.
$\begin{aligned}\tan58^\circ&=\frac{6.8}{PQ},\\PQ&=\frac{6.8}{\tan58^\circ},\\PQ&\approx4.2\text{ m}.\end{aligned}$

The opposite leg is longer than the adjacent leg, consistent with the given angle exceeding $45^\circ$ .
Both legs are shorter than the hypotenuse.

Answer

$\begin{aligned}PR&\approx8.0\text{ m},\\PQ&\approx4.2\text{ m}.\end{aligned}$

Key idea

When an unknown lies below the fraction bar, multiply through by that unknown before dividing by the function value.
Problem 3 Three starting points

For each case, give a direct equation or expression that uses the original given data to find the requested quantity. Explain the side roles or theorem that justify it. Do not evaluate the expressions.

Case A: Triangle $ABC$ is right at $C$ , with hypotenuse $AB$ of $27$ units and leg $BC$ of $11$ units. Find angle $A$ .

Case B: Triangle $DEF$ is right at $E$ , angle $D$ measures $34^\circ$ , and $DE$ is $15$ units. Find $EF$ .

Case C: Triangle $JKL$ is right at $K$ , with legs $JK$ of $9$ units and $KL$ of $14$ units. Find $JL$ .
1. Hint 1
  For each case, decide first whether the requested result is an angle or a length.
2. Hint 2
  Use the known side pair directly for an inverse operation; use a known angle with the relevant ratio for a side.
3. Hint 3
  Two known legs and a missing third side give a direct Pythagorean calculation.
Answer

Case A:
$m\angle A=\sin^{-1}(11/27)$
Case B:
$EF=15\tan34^\circ$
Case C:
$JL=\sqrt{9^2+14^2}$
Equivalent correct equations are acceptable.

Full solution

Case A: $BC$ is opposite $A$ and $AB$ is the hypotenuse.
Their ratio is between zero and one.
$m\angle A=\sin^{-1}(11/27)$

Case B: $DE$ is adjacent to $D$ , and $EF$ is opposite it.
The tangent equation is
$\tan34^\circ=EF/15$
Thus
$EF=15\tan34^\circ$

Case C: $JL$ is the hypotenuse.
Pythagoras uses the two given legs.
$JL=\sqrt{9^2+14^2}$

Other valid routes are possible, but these equations avoid finding an extra side or angle first.

Answer

Case A:
$m\angle A=\sin^{-1}(11/27)$
Case B:
$EF=15\tan34^\circ$
Case C:
$JL=\sqrt{9^2+14^2}$
Equivalent correct equations are acceptable.

Key idea

Choose a relation containing the requested quantity and the original known measurements.
Problem 4 Triangle UVW

Triangle $UVW$ is right at $V$ . Its hypotenuse $UW$ is $31$ cm and leg $VW$ is $12$ cm.

Find the remaining side to the nearest tenth of a centimeter and both acute angles to the nearest tenth of a degree.

The given leg faces the requested angle U.

Text description of this figure

Triangle UWV is right at V. Side VW is labeled 12 cm, side UV is labeled UV, and hypotenuse UW is labeled 31 cm. The marked angle at U is labeled ?.
1. Hint 1
  The remaining side is a leg; subtract squares for its exact length.
2. Hint 2
  The original $12$ cm leg lies opposite angle $U$ . Use its ratio with the original hypotenuse to find that angle.
3. Hint 3
  Keep the inverse result unrounded when subtracting it from $90^\circ$ .
Answer

$\begin{aligned}UV&\approx28.6\text{ cm},\\m\angle U&\approx22.8^\circ,\\m\angle W&\approx67.2^\circ.\end{aligned}$

Full solution

Find the other leg by Pythagoras.
$\begin{aligned}UV^2&=31^2-12^2,\\UV&=\sqrt{817},\\UV&\approx28.6\text{ cm}.\end{aligned}$

The known opposite-to-hypotenuse ratio gives angle $U$ directly.
$m\angle U=\sin^{-1}(12/31)$

In degree mode, the angle is approximately $22.773972^\circ$ .
Keep the stored inverse value to obtain its complement.
$m\angle W=90^\circ-m\angle U$

The requested angle values are $22.8^\circ$ and $67.2^\circ$ .
The longer leg $UV$ faces the larger angle $W$ , consistent with side-angle ordering.

Answer

$\begin{aligned}UV&\approx28.6\text{ cm},\\m\angle U&\approx22.8^\circ,\\m\angle W&\approx67.2^\circ.\end{aligned}$

Key idea

Original side measurements give a more reliable inverse input than a newly rounded side does.
Problem 5 Two proposed routes

A right triangle has an acute angle of $38^\circ$ and an adjacent leg of $18$ units. Its hypotenuse is $h$ .

Plan A uses $h=18/\cos38^\circ$ .

Plan B first finds the opposite leg as $18\tan38^\circ$ , then uses $h=\sqrt{18^2+(18\tan38^\circ)^2}$ .

Decide whether each plan is mathematically valid without intermediate rounding. Choose a more direct plan and explain one precision risk in the other plan. You do not need a decimal result.

Both plans refer to this same side assignment.

Text description of this figure

Triangle ABC is right at C. Side CB is labeled opposite, side AC is labeled 18, and hypotenuse AB is labeled h. The marked angle at A is labeled 38°.
1. Hint 1
  Check each equation against the roles of the sides, rather than assuming one plan must be wrong.
2. Hint 2
  One plan uses the adjacent leg and hypotenuse directly. The other builds a third side first.
3. Hint 3
  A rounded opposite leg changes what is squared in the second plan.
Answer

Both plans are valid without intermediate rounding. Plan A is more direct. Rounding the opposite leg before applying Pythagoras in Plan B can alter the final result.

Full solution

Plan A correctly uses cosine because $18$ is the adjacent leg and $h$ is the hypotenuse.
$\cos38^\circ=18/h$
Solving gives the displayed Plan A expression.

Plan B correctly uses tangent to find the opposite leg from the adjacent leg.
Applying Pythagoras to those two legs then gives its displayed expression.

Both plans describe the same triangle.
Their exact expressions agree.
Plan A finds the requested side in one ratio equation and does not require an extra side.

Plan B remains valid if its full expression or unrounded stored opposite leg is retained.
Replacing that leg by a rounded decimal before squaring introduces avoidable error.

Answer

Both plans are valid without intermediate rounding. Plan A is more direct. Rounding the opposite leg before applying Pythagoras in Plan B can alter the final result.

Key idea

A more direct method can reduce intermediate calculations without making other exact methods invalid.
Problem 6 A ratio and a length

A right triangle has hypotenuse $34$ m. Relative to an acute angle $\theta$ , its opposite-to-adjacent leg ratio is $7:9$ .

Find both acute angles to the nearest tenth of a degree and both legs to the nearest tenth of a meter.
1. Hint 1
  The leg ratio fixes the shape; the hypotenuse fixes its size.
2. Hint 2
  Use the inverse operation associated with opposite divided by adjacent.
3. Hint 3
  Keep that unrounded angle when using the given hypotenuse to recover each leg.
Answer

$\begin{aligned}\theta&\approx37.9^\circ,\\\phi&\approx52.1^\circ,\\O&\approx20.9\text{ m},\\A&\approx26.8\text{ m}.\end{aligned}$
Here $\phi$ is the other acute angle, and $O,A$ are opposite and adjacent to $\theta$ .

Full solution

The opposite-to-adjacent ratio gives
$\theta=\tan^{-1}(7/9)$
This positive input produces an acute angle.

Keep its stored value.
The complementary angle is
$\phi=90^\circ-\theta$
The angles round to $37.9^\circ$ and $52.1^\circ$ .

Use the original hypotenuse and unrounded angle to find the opposite leg.
$O=34\sin\theta$
This rounds to $20.9$ m.

For the adjacent leg,
$A=34\cos\theta$
This rounds to $26.8$ m.

An exact check can use a common positive part size $s$ .
The legs are $7s$ and $9s$ , so
$130s^2=34^2$
Their positive lengths therefore have ratio $7:9$ and the stated hypotenuse.

The side values are rounded; their original ratio is 7 to 9.

Answer

$\begin{aligned}\theta&\approx37.9^\circ,\\\phi&\approx52.1^\circ,\\O&\approx20.9\text{ m},\\A&\approx26.8\text{ m}.\end{aligned}$
Here $\phi$ is the other acute angle, and $O,A$ are opposite and adjacent to $\theta$ .

Key idea

A dimensionless leg ratio determines an angle; one actual side length then determines the scale.
Problem 7 A calculator interpretation

A right triangle has hypotenuse $20$ cm and a $12$ cm leg opposite acute angle $\theta$ .

Noor writes $\sin\theta=12/20$ .

She then writes $\theta=(12/20)^{-1}$ , interpreting the inverse-sine key as a reciprocal, and reports about $1.7^\circ$ .

Explain why the second line does not undo sine, find the correct angle to the nearest tenth of a degree, and describe a forward check using the original ratio.

The unknown angle is opposite the labeled 12 cm leg.

Text description of this figure

Triangle ABC is right at C. Side CB is labeled 12 cm, side AC is labeled adjacent, and hypotenuse AB is labeled 20 cm. The marked angle at A is labeled θ.
1. Hint 1
  An inverse function reverses the angle-to-ratio operation; a reciprocal reverses a division.
2. Hint 2
  Use the inverse associated with the ratio in Noor's first line, in degree mode.
3. Hint 3
  Apply sine to the unrounded angle and compare with the original ratio.
Answer

The reciprocal of $12/20$ is a number, not its inverse-sine angle. The correct angle is $\theta\approx36.9^\circ$ . Its unrounded value has sine $0.6$ , the original ratio.

Full solution

The first line is valid because $12$ is opposite and $20$ is the hypotenuse.
The ratio is $0.6$ , strictly between zero and one.

Noor's second line takes the reciprocal of that number.
$(12/20)^{-1}=5/3$
It does not find the angle whose sine is $12/20$ .

Use the inverse-sine operation in degree mode.
$\theta=\sin^{-1}(12/20)$

The unrounded result is approximately $36.869898^\circ$ , which rounds to $36.9^\circ$ .

A forward check applies sine to the stored inverse result.
$\sin(\sin^{-1}(12/20))=0.6$
Using the rounded angle instead gives a value close to $0.6$ , rather than exactly equal to it.

Answer

The reciprocal of $12/20$ is a number, not its inverse-sine angle. The correct angle is $\theta\approx36.9^\circ$ . Its unrounded value has sine $0.6$ , the original ratio.

Key idea

A reciprocal of a side ratio cannot replace the inverse trigonometric operation that recovers an angle.
Problem 8 An incomplete record

Triangle $ABC$ is right at $C$ , and side $BC$ is $8$ units. No other side or angle measurement is given.

Is this enough information to determine the other two sides and acute angles? Justify your decision by giving two noncongruent right triangles that satisfy the record. For each example, state the other leg and the hypotenuse exactly; angle calculations are not required.
1. Hint 1
  The other leg has not been fixed by the record.
2. Hint 2
  Choose two different positive values for that leg and use Pythagoras to complete each triangle.
3. Hint 3
  Each example must retain the given $8$ -unit leg, while a different side length makes the two triangles noncongruent.
Answer

The data are insufficient. One example has other leg $6$ units and hypotenuse $10$ units. A second has other leg $15$ units and hypotenuse $17$ units. Other valid pairs of distinct completions are acceptable.

Full solution

The right angle and one leg leave the other positive leg free.
Choose $CA$ as $6$ units in the first completion.
Pythagoras gives
$AB=\sqrt{8^2+6^2}$
Its positive length is $10$ units.

Choose $CA$ as $15$ units in the second completion.
Then
$AB=\sqrt{8^2+15^2}$
The hypotenuse is $17$ units.

Each construction has perpendicular positive legs and the required $8$ -unit side.
Their different hypotenuses show that they are not congruent.

Their opposite-to-adjacent ratios at $A$ are also different, so the acute angles are not fixed either.
Thus the record does not determine all missing quantities.

Both completions retain an 8-unit leg; their different side lengths establish noncongruence.

Answer

The data are insufficient. One example has other leg $6$ units and hypotenuse $10$ units. A second has other leg $15$ units and hypotenuse $17$ units. Other valid pairs of distinct completions are acceptable.

Key idea

Two valid noncongruent completions establish that the given measurements do not determine one triangle.
Problem 9 Two saved values

A right triangle has legs $11$ and $28$ cm. Angle $\theta$ is opposite the $11$ cm leg.

A solver rounds $\tan^{-1}(11/28)$ to $21^\circ$ , then uses $h=28/\cos21^\circ$ to report a hypotenuse of $30.0$ cm to the nearest tenth.

Find the hypotenuse to the nearest tenth while preserving the original data. Use an independent side calculation to check it, and explain why the two rounded answers differ.
1. Hint 1
  The original two sides already determine an exact side calculation.
2. Hint 2
  Compare the result from Pythagoras with the result that keeps the inverse expression inside cosine.
3. Hint 3
  Rounding the angle first changes the triangle used in the hypotenuse equation.
Answer

The hypotenuse is $30.1$ cm to the nearest tenth, exactly $\sqrt{905}$ cm. The reported $30.0$ cm comes from changing the angle to $21^\circ$ before calculating the side.

Full solution

Use the original legs for an independent calculation.
$\begin{aligned}h&=\sqrt{11^2+28^2},\\h&=\sqrt{905},\\h&\approx30.083218\text{ cm}.\end{aligned}$
Thus the requested length is $30.1$ cm.

The unrounded inverse angle is approximately $21.447736^\circ$ .
Keeping it stored, or nesting the inverse expression, gives
$h=\frac{28}{\cos(\tan^{-1}(11/28))}$
This equals the same positive hypotenuse as the Pythagorean calculation.

The altered angle produces
$28/\cos21^\circ\approx29.992060$
That value rounds to $30.0$ cm, but it came from a changed angle.

The original ratio is not the tangent of exactly $21^\circ$ .
Preserving the inverse value until the final rounding avoids this discrepancy.

The original two leg lengths permit an independent side check.

Answer

The hypotenuse is $30.1$ cm to the nearest tenth, exactly $\sqrt{905}$ cm. The reported $30.0$ cm comes from changing the angle to $21^\circ$ before calculating the side.

Key idea

A rounded intermediate angle can change the final rounded side, even when the calculator arithmetic is correct.
Problem 10 Two summaries

Triangle $ABC$ is right at $C$ . The hypotenuse $AB$ is exactly $16$ cm, and angle $A$ is exactly $27^\circ$ .

Report I lists $BC$ as $7.3$ cm, $CA$ as $14.3$ cm, and angle $B$ as $63^\circ$ .

Report II lists $BC$ as $14.3$ cm, $CA$ as $7.3$ cm, and angle $B$ as $63^\circ$ .

Each listed leg is intended to be rounded to the nearest tenth. Choose the report consistent with the original data. Explain why a Pythagorean check alone cannot distinguish these two reports, then use the selected reference angle to justify your choice.

The reports must respect the marked reference angle.

Text description of this figure

Triangle ABC is right at C. Side CB is labeled BC, side AC is labeled CA, and hypotenuse AB is labeled 16 cm. The marked angle at A is labeled 27°.
1. Hint 1
  The two reports have the same pair of leg values, assigned to different sides.
2. Hint 2
  Relate the smaller given acute angle to the smaller opposite side.
3. Hint 3
  Use the original hypotenuse with the sine and cosine of the original angle to check the side assignments.
Answer

Report I is consistent. Pythagoras is unchanged when the two leg values are exchanged. Relative to $A$ , $BC$ is the opposite leg and rounds to $7.3$ cm; $CA$ is adjacent and rounds to $14.3$ cm.

Full solution

Both reports give the correct complementary angle because
$90^\circ-27^\circ=63^\circ$

Pythagoras adds the two leg squares, so swapping their names leaves that sum unchanged.
Moreover, the reported sides are rounded; an exact squared equality is not expected from them.

Angle $A$ is smaller than angle $B$ , so its opposite side $BC$ must be the smaller leg.
This already supports Report I.

Check using the original angle and hypotenuse.
$\begin{aligned}BC&=16\sin27^\circ,\\BC&\approx7.263848\text{ cm},\\CA&=16\cos27^\circ,\\CA&\approx14.256104\text{ cm}.\end{aligned}$

These values round to the side assignments in Report I.
Report II exchanges opposite and adjacent without changing the reference angle.

Answer

Report I is consistent. Pythagoras is unchanged when the two leg values are exchanged. Relative to $A$ , $BC$ is the opposite leg and rounds to $7.3$ cm; $CA$ is adjacent and rounds to $14.3$ cm.

Key idea

A side-square check verifies lengths together, while the reference angle checks which leg has which name.

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