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Right Triangles and Trigonometry: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. 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 Measurements inside a triangle

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets hypotenuse $AB$ at interior point $H$ . The pieces are $AH=17$ units and $HB=19$ units.

Find $\sin B$ , $\cos B$ , and $\tan B$ exactly. Explain how those three values would change if $A$ were selected instead, and identify the side whose role remains fixed.

Only the internal measurements are supplied.

Text description of this figure

Schematic, not for measuring. ABC is right at C. Its perpendicular CH meets AB at interior H. AH is 17 units and HB is 19 units. The whole hypotenuse and both legs have no numerical labels. B is the requested acute reference angle.
1. Hint 1
  Each hypotenuse piece is paired with the leg at its own endpoint; first identify the whole hypotenuse.
2. Hint 2
  Recover both leg lengths from their squared relationships. Relative to B, the opposite leg is CA.
3. Hint 3
  Form the three ratios after taking positive roots. Then exchange the leg roles when selecting the other acute angle.
Answer

$\begin{gathered}\sin B=\frac{\sqrt{17}}6\\\cos B=\frac{\sqrt{19}}6\\\tan B=\frac{\sqrt{323}}{19}.\end{gathered}$
At A, sine and cosine exchange, tangent reciprocates, and AB stays the hypotenuse.

Full solution

The two pieces fill the whole hypotenuse.
Adding 17 and 19 gives
$AB=36.$

Leg CA uses the piece at A.
$\begin{gathered}CA^2=36\cdot17\\CA=6\sqrt{17}.\end{gathered}$

Leg BC uses the piece at B.
$\begin{gathered}BC^2=36\cdot19\\BC=6\sqrt{19}.\end{gathered}$

For reference angle B, CA is opposite, BC is adjacent, and AB is the hypotenuse.
The sine ratio gives
$\sin B=\frac{\sqrt{17}}6.$

The adjacent-to-hypotenuse ratio gives
$\cos B=\frac{\sqrt{19}}6.$

Divide the two legs and rationalize if desired.
$\begin{gathered}\tan B=\frac{\sqrt{17}}{\sqrt{19}}\\\tan B=\frac{\sqrt{323}}{19}.\end{gathered}$

Selecting A exchanges opposite and adjacent while AB remains the hypotenuse.
Sine and cosine therefore exchange, and tangent becomes its positive reciprocal.

Answer

$\begin{gathered}\sin B=\frac{\sqrt{17}}6\\\cos B=\frac{\sqrt{19}}6\\\tan B=\frac{\sqrt{323}}{19}.\end{gathered}$
At A, sine and cosine exchange, tangent reciprocates, and AB stays the hypotenuse.

Key idea

Projection measurements can determine acute side ratios once the chosen reference angle is fixed.
Problem 2 Two outlines

Two nondegenerate right triangles have hypotenuse altitudes of the same length, $5\sqrt6$ cm. The first triangle has acute angles $45^\circ$ and $45^\circ$ . The second has acute angles $30^\circ$ and $60^\circ$ .

Find each triangle's hypotenuse exactly. Justify why the same altitude produces different hypotenuse-to-altitude multipliers in these two shapes.

Both altitudes represent the same given length h.

Text description of this figure

Two separate triangles ABC are drawn at one common scale. Each is right at C and has hypotenuse altitude CH. The first has angle A of 45 degrees; the second has angle A of 30 degrees. The two drawn altitudes have identical length h, given in the statement as five square root six centimeters. No computed hypotenuse or projection lengths are labeled.
1. Hint 1
  The altitude creates smaller right triangles whose acute angles follow from the whole triangle.
2. Hint 2
  Determine which side the altitude is in a convenient small triangle before applying a special side relationship.
3. Hint 3
  Recover an outer leg from a small triangle, then use the whole triangle to recover its hypotenuse.
Answer

First hypotenuse: $10\sqrt6$ cm. Second hypotenuse: $20\sqrt2$ cm. Their hypotenuse-to-altitude multipliers are $2$ and $4/\sqrt3$ , respectively.

Full solution

Write h for the common altitude.
In the first triangle, draw CH to AB and select the endpoint A with angle 45 degrees.
Triangle ACH has a right angle at H and another 45-degree angle at C.

In that small equal-leg right triangle, AC is the hypotenuse and CH is a leg.
Thus
$AC=h\sqrt2.$

In the whole first triangle, AC is a leg and AB is its hypotenuse.
Therefore
$\begin{gathered}AB=(h\sqrt2)\sqrt2\\AB=2h\\AB=10\sqrt6\text{ cm}.\end{gathered}$

For the second triangle, select A as the 30-degree endpoint.
Small triangle ACH is right at H, and CH is opposite its 30-degree angle.
The small hypotenuse is therefore
$AC=2h.$

In the whole second triangle, AC is the long leg.
Its hypotenuse is twice the short leg, so
$\begin{gathered}AB=\frac{2AC}{\sqrt3}\\AB=\frac{4h}{\sqrt3}\\AB=20\sqrt2\text{ cm}.\end{gathered}$

The common altitude has a different side role in the two chosen small triangles.
That angle structure produces the different multipliers, so an altitude length alone does not fix the hypotenuse without shape information.

Answer

First hypotenuse: $10\sqrt6$ cm. Second hypotenuse: $20\sqrt2$ cm. Their hypotenuse-to-altitude multipliers are $2$ and $4/\sqrt3$ , respectively.

Key idea

A length shared by two figures must be interpreted in each figure's own angle configuration.
Problem 3 Two specifications

Specification I: Triangle $ABC$ is right at $C$ , $AB=21.6$ cm, and $\angle A=62.4^\circ$ .

Specification II: Triangle $PQR$ is right at $Q$ , $PR=21.6$ cm, and $\angle R=27.6^\circ$ . Treat the supplied measurements as exact for this calculation.

Determine whether the specifications describe congruent triangles, and justify your decision. Match the named legs between the triangles, then find each distinct leg length to the nearest tenth of a centimeter.
1. Hint 1
  The two acute angles in each right triangle complement one another; compare the complete angle information first.
2. Hint 2
  Match the hypotenuses after establishing the angle correspondence. Decide whether any size freedom remains.
3. Hint 3
  Calculate both legs from the original 21.6 cm hypotenuse and one original angle, preserving calculator precision.
Answer

$ABC\cong PRQ$ .
$\begin{gathered}BC=RQ\approx19.1\text{ cm}\\CA=QP\approx10.0\text{ cm}.\end{gathered}$

Full solution

In the first triangle, the other acute angle is
$m\angle B=27.6^\circ.$

In the second triangle, the other acute angle is
$m\angle P=62.4^\circ.$

The correspondence is A with P, B with R, and C with Q.
AA establishes similarity.
The corresponding hypotenuses AB and PR are both 21.6 cm, so the similarity factor is 1 and the triangles are congruent.

The leg opposite A matches the leg opposite P, so BC matches RQ.
The adjacent legs CA and QP also match.

Use the original first specification to calculate the opposite leg.
$\begin{gathered}BC=21.6\sin62.4^\circ\\BC\approx19.1\text{ cm}.\end{gathered}$

Calculate the adjacent leg independently from the original data.
$\begin{gathered}CA=21.6\cos62.4^\circ\\CA\approx10.0\text{ cm}.\end{gathered}$

Congruence gives the same respective values for RQ and QP.
Their rounded labels need not satisfy an exact squared equality; the original unrounded sine and cosine values do.

The correspondence is shown only after it is justified.

Answer

$ABC\cong PRQ$ .
$\begin{gathered}BC=RQ\approx19.1\text{ cm}\\CA=QP\approx10.0\text{ cm}.\end{gathered}$

Key idea

Complementary reference-angle data can describe the same shape, while an equal corresponding side fixes its size.
Problem 4 Two markers

An eye $E$ is $17.4$ m above horizontal ground. Two point markers lie on the same straight ground ray from the point directly below $E$ , in the common vertical sight plane. Their depression angles are $19^\circ$ and $46^\circ$ , measured from the true horizontal through $E$ .

Find the distance between the markers to the nearest tenth of a meter. Identify which marker is nearer and check that both horizontal runs are positive.
1. Hint 1
  Both sight triangles have the same vertical drop to ground, but their horizontal runs differ.
2. Hint 2
  Relate the common drop to each horizontal run using its own depression angle and the parallel horizontal references.
3. Hint 3
  The markers lie on the same ground ray, so their separation is the larger run minus the smaller run. Retain unrounded runs for the subtraction.
Answer

The markers are approximately $33.7$ m apart. The $46^\circ$ marker is nearer. Its run is approximately $16.8$ m; the $19^\circ$ marker's run is approximately $50.5$ m, both positive.

Full solution

The true horizontal through E is parallel to the ground.
Each depression angle equals the corresponding elevation angle at its marker.
The vertical drop in both right triangles is 17.4 m.

For the marker at 19 degrees, write its horizontal run as d.
$\begin{gathered}\tan19^\circ=\frac{17.4}{d}\\d=\frac{17.4}{\tan19^\circ}.\end{gathered}$

For the other marker, write its run as n.
$n=\frac{17.4}{\tan46^\circ}.$

Both tangents are positive, so both runs are positive.
Their unrounded values are approximately 50.533269 m and 16.802985 m.
The larger depression belongs to the nearer marker.

Subtract the unrounded runs because the two markers lie on the same ray.
If s is their separation,
$\begin{gathered}s=d-n\\s\approx33.7\text{ m}.\end{gathered}$

The marker at 46 degrees is therefore nearer.
Adding the runs would describe markers on opposite sides, contrary to the stated ray.

The model gives two positive runs and their separation.

Answer

The markers are approximately $33.7$ m apart. The $46^\circ$ marker is nearer. Its run is approximately $16.8$ m; the $19^\circ$ marker's run is approximately $50.5$ m, both positive.

Key idea

A common drop can determine two separate ground runs whose relationship depends on the stated marker order.
Problem 5 An objection to an argument

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets the interior of $AB$ at $H$ . Use length names $a=BC$ , $b=CA$ , $c=AB$ , $p=AH$ , and $q=HB$ .

Leah claims: "Any proof of the Pythagorean theorem that starts with the leg-square relationships must be circular, because those relationships always assume the theorem."

Decide whether Leah's claim is correct. Give an argument obtaining both leg-square relationships independently, then complete the sum-of-squares proof without assuming its conclusion. Explain where the interior position of $H$ is used.

The configuration supplies the angle and incidence facts.

Text description of this figure

ABC is right at C and CH is perpendicular to AB at interior H. The labels are a for BC, b for CA, c for AB, p for AH, and q for HB. No squared-side conclusion is supplied by the drawing.
1. Hint 1
  Matching right angles and a shared acute angle can establish similarity without any squared-side formula.
2. Hint 2
  Compare the whole triangle with each small triangle whose hypotenuse is one of the whole triangle's legs. Form corresponding ratios.
3. Hint 3
  The two resulting products share the same whole hypotenuse. The interior pieces fill that whole segment.
Answer

Leah's claim is false. AA gives the leg-square relationships independently. Their sum yields $a^2+b^2=c^2$ , using $p+q=c$ because $H$ is interior.

Full solution

The right angles at C and H and the shared endpoint angles establish
$\begin{gathered}\triangle ABC\sim\triangle ACH\\\triangle ABC\sim\triangle CBH.\end{gathered}$

The first comparison matches AB with AC and AC with AH.
Thus
$\begin{gathered}\frac cb=\frac bp\\b^2=cp.\end{gathered}$

The other comparison matches AB with CB and BC with BH.
Thus
$\begin{gathered}\frac ca=\frac aq\\a^2=cq.\end{gathered}$

Both equalities follow from similarity and positive-length cross multiplication, without invoking the Pythagorean theorem.
Add them.
$a^2+b^2=cq+cp.$

Factor the common whole hypotenuse.
$a^2+b^2=c(q+p).$

The interior point H partitions AB into AH and HB.
Segment addition supplies
$q+p=c.$

Substitute this whole length.
$a^2+b^2=c^2.$

This provides a noncircular proof and refutes Leah's universal claim.
A different argument that obtained the leg relationships from Pythagoras would indeed need its circular dependency repaired.

Answer

Leah's claim is false. AA gives the leg-square relationships independently. Their sum yields $a^2+b^2=c^2$ , using $p+q=c$ because $H$ is interior.

Key idea

A proof's dependencies matter: relations obtained from earlier similarity can establish a later squared-side theorem.
Problem 6 A ratio record

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets the interior of hypotenuse $AB$ at $H$ , and $AH:HB=49:72$ .

Find $\cos A$ and $\tan B$ exactly. Find angle $A$ to the nearest tenth of a degree, and explain why the record needs no absolute side length for these tasks.

The labels represent relative parts, not measured units.

Text description of this figure

Schematic, not for measuring. ABC is right at C, and CH is perpendicular to hypotenuse AB at interior H. The two pieces are labeled 49 parts at AH and 72 parts at HB. These are ratio parts rather than fixed length units. No leg ratio, angle value, or whole numerical hypotenuse is supplied.
1. Hint 1
  The projection relationships connect ratios of squared legs with the given hypotenuse-piece ratio.
2. Hint 2
  Write the two pieces as 49k and 72k for a positive common k. Compare a leg square with the whole hypotenuse square, or compare the two leg squares.
3. Hint 3
  Take positive roots for the acute ratios, then select the matching inverse operation in degree mode. The common size factor should cancel.
Answer

$\begin{gathered}\cos A=\frac7{11}\\\tan B=\frac{7\sqrt2}{12}\\m\angle A\approx50.5^\circ.\end{gathered}$
The common positive scale cancels, so absolute side lengths are unnecessary.

Full solution

Write p for AH, q for HB, and c for AB.
With a positive common part size k, the pieces are 49k and 72k, so
$c=121k.$

Write b for CA and a for BC.
The leg relationship at A gives b squared equal to cp.
Therefore the squared adjacent-to-hypotenuse ratio is
$\cos^2 A=\frac pc.$

Substitute the part expressions and cancel k.
$\cos^2 A=\frac{49}{121}.$

Cosine is positive for the acute angle A, so
$\cos A=\frac7{11}.$

For B, CA is opposite and BC is adjacent.
Dividing the leg-square relationships gives
$\tan^2 B=\frac pq.$

Thus the positive tangent can be simplified as
$\begin{gathered}\tan B=\sqrt{\frac{49}{72}}\\\tan B=\frac{7\sqrt2}{12}.\end{gathered}$

Recover A with its original exact cosine ratio in degree mode.
$\begin{gathered}m\angle A=\cos^{-1}(7/11)\\m\angle A\approx50.5^\circ.\end{gathered}$

The ratio inputs depend only on the relative pieces; k cancels in each calculation.
Positive dilations change all side lengths while preserving the requested ratios and angle.

One representative size gives the same ratios.

Answer

$\begin{gathered}\cos A=\frac7{11}\\\tan B=\frac{7\sqrt2}{12}\\m\angle A\approx50.5^\circ.\end{gathered}$
The common positive scale cancels, so absolute side lengths are unnecessary.

Key idea

Dimensionless projection data can determine shape without determining an absolute size.
Problem 7 A changing frame

For fixed $s>0$ and variable $k>0$ , three proposed side lengths are $s$ , $s$ , and $ks$ .

Give the complete range of $k$ for which a nondegenerate triangle exists. Within that range, give all acute, right, and obtuse cases, with each boundary included or excluded correctly. Explain how your comparison handles the change of longest side when $k$ passes through $1$ .
1. Hint 1
  A common positive factor can be removed from both a length inequality and a comparison of squares.
2. Hint 2
  The longest side is a leg of length s when k is below 1, but it is the side ks when k is above 1.
3. Hint 3
  Establish the strict existence range first. Then make the squared-side comparison in each longest-side case and check the equality boundaries.
Answer

Existence:
$0<k<2.$
Acute:
$0<k<\sqrt2.$
Right:
$k=\sqrt2.$
Obtuse:
$\sqrt2<k<2.$
For $k\ge2$ , no nondegenerate triangle exists.

Full solution

Since k and s are positive, the only restrictive triangle inequality is that the two equal sides reach beyond the third.
$\begin{gathered}2s>ks\\k<2.\end{gathered}$

Together with the given positivity, the existence range is
$0<k<2.$

For k at most 1, a longest side has length s.
Its square is smaller than the other two squares added, because the additional square is positive.
$s^2<s^2+(ks)^2.$

Therefore every value with k positive and at most 1 is acute, including the equilateral case k equal to 1.

For k greater than 1, the longest side is ks.
Compare its square with the sum of the equal-side squares.
Dividing by positive s squared reduces that comparison to k squared versus 2.

The strict smaller-square case gives
$1<k<\sqrt2.$

At equality, the converse establishes a right triangle.
$k=\sqrt2.$

The greater-square case must still satisfy the strict existence bound.
$\sqrt2<k<2.$

Combining the two acute subcases gives all positive k below square root two.
At k equal to 2 the arrangement is flat, and larger k cannot close.

Examples illustrate the separate longest-side cases after the classification.

Answer

Existence:
$0<k<2.$
Acute:
$0<k<\sqrt2.$
Right:
$k=\sqrt2.$
Obtuse:
$\sqrt2<k<2.$
For $k\ge2$ , no nondegenerate triangle exists.

Key idea

A parameter classification must track both triangle existence and which side is actually longest.
Problem 8 An angle claim

Acute angles $\theta$ and $\phi$ satisfy $\tan\phi=3\tan\theta$ . A student claims that every such pair must satisfy $\phi=3\theta$ .

Decide whether the claim is true. If it is false, give a concrete pair of acute angles and verify both the required tangent relation and the failure of the student's angle claim.
1. Hint 1
  Multiplying an angle measure and multiplying a side ratio are different operations; test whether the claimed implication respects that distinction.
2. Hint 2
  Look for two acute special angles whose exact tangent values have the required ratio, then compare their degree measures.
Answer

The claim is false. One counterexample is $\theta=30^\circ$ , $\phi=60^\circ$ : $\tan\phi=3\tan\theta$ , but $\phi\ne3\theta$ .

Full solution

Choose theta as 30 degrees and phi as 60 degrees.
Both angles lie strictly between zero and 90 degrees.

The exact special-angle values give
$\tan30^\circ=\frac{\sqrt3}{3}.$

Multiplying this value by 3 gives
$3\tan30^\circ=\sqrt3.$

The other chosen angle has
$\tan60^\circ=\sqrt3.$

Thus the required tangent relation holds exactly.
But three times theta is 90 degrees, while phi is 60 degrees, so the student's asserted angle equality fails.

One pair satisfying the hypothesis and failing the conclusion disproves the universal claim.
No angle-addition formula is required.

Answer

The claim is false. One counterexample is $\theta=30^\circ$ , $\phi=60^\circ$ : $\tan\phi=3\tan\theta$ , but $\phi\ne3\theta$ .

Key idea

Scaling a tangent value need not scale its angle measure by the same factor.
Problem 9 An internal condition

Triangle $ABC$ is nondegenerate. Point $H$ lies inside $AB$ , and $CH\perp AB$ . The lengths satisfy $CH^2=AH\cdot HB$ . No angle at $C$ is given.

Do these facts force $\angle ACB$ to be right? Prove your conclusion without assuming that the whole triangle is right at the start.

Only the right angle at the perpendicular foot is given.

Text description of this figure

Schematic, not for measuring. In nondegenerate ABC, H lies inside AB and CH is perpendicular to AB. AH is p, HB is q, and CH is h. The only right-angle mark is at H; no right angle is assumed or marked at C.
1. Hint 1
  The perpendicular gives two small triangles that are known right, even though the whole triangle has not yet been classified.
2. Hint 2
  Write a squared-side equality in each small triangle. Add them and substitute the supplied relationship involving the altitude.
3. Hint 3
  Compare the resulting sum with the square of the whole side AB before deciding whether a converse applies.
Answer

Yes. The stated interior-foot, perpendicularity, and length conditions force $\angle ACB=90^\circ$ .

Full solution

Let p be AH, q be HB, and h be CH.
These are positive lengths.
The two small triangles are right at H, so the established theorem applies to them separately.

For triangle ACH,
$AC^2=p^2+h^2.$

For triangle BCH,
$BC^2=q^2+h^2.$

Add the two equalities.
$AC^2+BC^2=p^2+q^2+2h^2.$

The given condition is h squared equal to pq.
Substitute it and recognize the square of a sum.
$AC^2+BC^2=(p+q)^2.$

Because H is inside AB, segment addition gives
$p+q=AB.$

Consequently, the original triangle satisfies
$AC^2+BC^2=AB^2.$

The converse of the Pythagorean theorem now establishes a right angle opposite AB, at C.
The theorem was used initially only on the two triangles whose right angles were given by perpendicularity, so no right angle in the whole triangle was assumed.

Answer

Yes. The stated interior-foot, perpendicularity, and length conditions force $\angle ACB=90^\circ$ .

Key idea

A converse can establish a missing right angle after earlier right triangles supply the needed side equality.
Problem 10 Two instrument elevations

Far station $F$ , near station $N$ , and the base $B$ of a vertical mast lie in that order on one horizontal line. The station separation is $FN=23.0$ m.

The far instrument eye is $1.25$ m above the common ground level; the near instrument eye is $2.35$ m above it. They sight the same mast top in one vertical plane, at elevations $29.0^\circ$ from $F$ and $51.0^\circ$ from $N$ , measured from their own true horizontals.

Find $NB$ and the whole mast height, each to the nearest tenth of a meter. Check both observation equations with unrounded values, and explain why a model using one shared rise above the two eyes is invalid.

Each horizontal reference begins at its own instrument eye.

Text description of this figure

Far station F, near station N, and vertical mast base B lie in that order on one horizontal ground line, drawn at one common scale from the unrounded observation model. The F instrument eye is 1.25 meters above ground; the N eye is 2.35 meters above that same level. Their elevations to the same top T are 29 and 51 degrees, shown from each eye's own true horizontal. FN is labeled 23 meters and NB is labeled x. No computed run or mast height is labeled, and no common-rise equality is marked.
1. Hint 1
  The two eye elevations differ, so identify the height quantity that the observations actually share.
2. Hint 2
  If x is NB, the far horizontal run is x plus 23. Write a whole-height equation for each station using its own eye elevation.
3. Hint 3
  Equate the whole heights and collect the terms containing x, retaining the difference between the eye elevations.
4. Hint 4
  Use the unrounded run in both original ratio equations. Check that the rises above the individual eyes are positive and produce the same whole height.
Answer

$\begin{gathered}NB\approx17.1\text{ m}\\H\approx23.5\text{ m}.\end{gathered}$
The two eye-level rises differ by $1.10$ m, so they are not one shared rise.

Full solution

Let x be NB and H be the whole mast height above the common ground.
The far run is x plus 23.
Its rise starts at the 1.25 m eye, so
$\tan29^\circ=\frac{H-1.25}{x+23}.$

The near rise starts at the different 2.35 m eye.
$\tan51^\circ=\frac{H-2.35}{x}.$

Each equation determines the same whole height, but the rises differ.
Rearranging the equations and equating their whole-height values retains the 1.10 m difference between eye elevations.
Solving for x gives
$x=\frac{23\tan29^\circ-1.10}{\tan51^\circ-\tan29^\circ}.$

Keep the original expression while evaluating in degree mode.
The unrounded near run is approximately 17.116238 m, so
$NB\approx17.1\text{ m}.$

Use that unrounded run with the near eye elevation.
$\begin{gathered}H=2.35+x\tan51^\circ\\H\approx23.5\text{ m}.\end{gathered}$

For a check, the unrounded whole height is approximately 23.486794 m.
Substituting the unrounded x and H into both original ratios returns their original tangents.
$\begin{gathered}\frac{H-1.25}{x+23}=\tan29^\circ\\\frac{H-2.35}{x}=\tan51^\circ.\end{gathered}$

Both runs and both rises are positive.
The far and near rises are approximately 22.236794 m and 21.136794 m, respectively.
They differ by 1.10 m because the eyes have different elevations; treating them as equal would discard a stated measurement.

The complete model uses unequal eye elevations and one whole height.

Answer

$\begin{gathered}NB\approx17.1\text{ m}\\H\approx23.5\text{ m}.\end{gathered}$
The two eye-level rises differ by $1.10$ m, so they are not one shared rise.

Key idea

Different eye elevations require different triangle rises even when both observations determine one whole height.

Multiple choice

Core practice