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Sine, Cosine, and Tangent: 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 Three side ratios

Triangle $ABC$ is right at $C$ . Use $A$ as the acute reference angle.

The lengths are
$BC=28\text{ cm}.$
Also,
$CA=45\text{ cm}.$
The hypotenuse is
$AB=53\text{ cm}.$
Find sine, cosine, and tangent of $A$ exactly, and state whether these three answers carry length units.

The given triangle.

Text description of this figure

ABC is right at C. BC is 28 centimeters, CA is 45 centimeters, and AB is 53 centimeters. The selected reference angle is A.
1. Hint 1
  Identify the leg across from $A$ and the leg meeting $A$ .
2. Hint 2
  Each ratio compares lengths measured in the same unit, so the units cancel.
Answer

$\begin{aligned}\sin A&=\frac{28}{53}\\\cos A&=\frac{45}{53}\\\tan A&=\frac{28}{45}.\end{aligned}$
All three values are unitless.

Full solution

For angle $A$ , the opposite leg is $BC$ and the adjacent leg is $CA$ .
The hypotenuse is $AB$ .

The definitions give
$\sin A=\frac{28}{53}.$

The adjacent-to-hypotenuse ratio gives
$\cos A=\frac{45}{53}.$

The opposite-to-adjacent ratio gives
$\tan A=\frac{28}{45}.$

Each ratio divides centimeters by centimeters, leaving a number with no length unit.

Answer

$\begin{aligned}\sin A&=\frac{28}{53}\\\cos A&=\frac{45}{53}\\\tan A&=\frac{28}{45}.\end{aligned}$
All three values are unitless.

Key idea

Select the angle first, then divide the named side lengths.
Problem 2 An exact expression

Evaluate exactly:
$\frac{\sin60^\circ\cos45^\circ}{\tan30^\circ}.$
Give the result in simplest radical form.
1. Hint 1
  Use the exact values supplied by the two special right-triangle shapes.
2. Hint 2
  Divide by a fraction by multiplying by its reciprocal, then rationalize if needed.
Answer

$\frac{3\sqrt2}{4}.$

Full solution

The special triangles give
$\sin60^\circ=\frac{\sqrt3}{2}.$

They also give
$\cos45^\circ=\frac{\sqrt2}{2}.$

The remaining value is
$\tan30^\circ=\frac{\sqrt3}{3}.$

Substituting and dividing gives
$\frac{\sqrt3\sqrt2}{4}\cdot\frac3{\sqrt3}.$

Cancel the common positive radical to obtain
$\frac{3\sqrt2}{4}.$

The two special-triangle models.

Answer

$\frac{3\sqrt2}{4}.$

Key idea

Exact special-angle ratios can be combined using ordinary fraction arithmetic.
Problem 3 One side, two roles

Triangle $PQR$ is right at $Q$ .

(a) Select $P$ as the reference angle. Identify its opposite leg, adjacent leg, and hypotenuse.

(b) Select $R$ instead. Identify all three roles again.

(c) Which role of side $PR$ changes when the selected angle changes?

The two candidate reference angles.

Text description of this figure

PQR is right at Q. Both acute angles P and R are marked with distinct arc counts. No side-role labels are supplied.
1. Hint 1
  The hypotenuse is fixed by the right angle.
2. Hint 2
  The side across from one acute angle touches the other acute angle.
Answer

Part (a): opposite $QR$ , adjacent $PQ$ , hypotenuse $PR$ . Part (b): opposite $PQ$ , adjacent $QR$ , hypotenuse $PR$ . Part (c): none; $PR$ stays the hypotenuse.

Full solution

Part (a): $QR$ is across from $P$ .
Leg $PQ$ touches $P$ , while $PR$ is across from the right angle at $Q$ .

Part (b): $PQ$ is across from $R$ .
Leg $QR$ touches $R$ , and $PR$ is still opposite the right angle.

Part (c): the opposite and adjacent leg roles exchange, but $PR$ remains the hypotenuse.

Answer

Part (a): opposite $QR$ , adjacent $PQ$ , hypotenuse $PR$ . Part (b): opposite $PQ$ , adjacent $QR$ , hypotenuse $PR$ . Part (c): none; $PR$ stays the hypotenuse.

Key idea

Opposite and adjacent depend on the selected acute angle; hypotenuse does not.
Problem 4 Two sizes with one angle

Triangles $ABC$ and $DEF$ are right at $C$ and $F$ . Their acute angles at $A$ and $D$ have the same measure $\theta$ .

Prove that the sine, cosine, and tangent values from these two reference angles are equal. State the ordered correspondence and explain how the length scale cancels.

Two corresponding configurations.

Text description of this figure

Two right triangles have equal selected acute angles. Their corresponding sides are labeled a,b,c and ka,kb,kc. The right angles and selected acute angles identify the ordered correspondence.
1. Hint 1
  Use the common acute angle together with the two right angles.
2. Hint 2
  Match opposite with opposite, adjacent with adjacent, and hypotenuse with hypotenuse.
3. Hint 3
  Write the second triangle's side lengths as one positive factor times the corresponding first lengths.
Answer

$\triangle ABC\sim\triangle DEF$
by AA, and
$\begin{aligned}\sin A&=\sin D\\\cos A&=\cos D\\\tan A&=\tan D.\end{aligned}$

Full solution

The angle at $A$ matches the angle at $D$ , and the right angle at $C$ matches the right angle at $F$ .
AA gives the order $ABC$ to $DEF$ , with $B$ matching $E$ .

Let the first triangle's opposite leg, adjacent leg, and hypotenuse be $a,b,c$ .
For a positive scale factor $k$ to the second triangle, the corresponding lengths are $ka,kb,kc$ .

The sine ratio is unchanged because
$\frac{ka}{kc}=\frac ac.$

The cosine ratio is unchanged because
$\frac{kb}{kc}=\frac bc.$

The tangent ratio is unchanged because
$\frac{ka}{kb}=\frac ab.$

The side lengths are positive, so these divisions are valid.
The three common values depend on the angle rather than the size.

Answer

$\triangle ABC\sim\triangle DEF$
by AA, and
$\begin{aligned}\sin A&=\sin D\\\cos A&=\cos D\\\tan A&=\tan D.\end{aligned}$

Key idea

AA establishes the correspondence before common factors are canceled.
Problem 5 A recorded set of ratios

Triangle $ABC$ is right at $C$ . The chosen reference angle is $B$ .

The lengths are
$BC=39.$
Also,
$CA=80.$
The hypotenuse is
$AB=89.$
A student writes the following ratios.
$\sin B=\frac{39}{89}.$ $\cos B=\frac{80}{89}.$ $\tan B=\frac{39}{80}.$
Explain the common side-role error and replace all three ratios with correct values.

The selected angle and given lengths.

Text description of this figure

ABC is right at C, with BC 39, CA 80, and AB 89. Angle B is the selected reference angle. The triangle is turned.
1. Hint 1
  Locate the side across from the chosen angle $B$ .
2. Hint 2
  The student's numerators would fit the other acute reference angle.
3. Hint 3
  Keep the hypotenuse fixed while exchanging the two leg roles.
Answer

The student used angle A's leg roles.
$\begin{aligned}\sin B&=\frac{80}{89}\\\cos B&=\frac{39}{89}\\\tan B&=\frac{80}{39}.\end{aligned}$

Full solution

For reference angle $B$ , the opposite leg is $CA$ , with length $80$ .
The adjacent leg is $BC$ , with length $39$ , and the hypotenuse stays $AB$ .

The student's ratios use $BC$ as opposite and $CA$ as adjacent.
Those are the roles for $A$ , not $B$ .

Using the correct opposite leg gives
$\sin B=\frac{80}{89}.$

Using the correct adjacent leg gives
$\cos B=\frac{39}{89}.$

Dividing the correct opposite leg by the adjacent leg gives
$\tan B=\frac{80}{39}.$

Answer

The student used angle A's leg roles.
$\begin{aligned}\sin B&=\frac{80}{89}\\\cos B&=\frac{39}{89}\\\tan B&=\frac{80}{39}.\end{aligned}$

Key idea

One wrong reference angle can exchange every leg ratio in a calculation.
Problem 6 Turn the picture, then change the angle

Triangle $JKL$ is right at $J$ . Initially, $K$ is selected as the reference angle.

Case A: Rotate the drawing through a half-turn while keeping $K$ selected. Name the opposite leg, adjacent leg, and hypotenuse after the turn.

Case B: Keep that turned drawing, but select $L$ instead. Name the three side roles now and explain which change came from switching the angle.

The initial named triangle.

Text description of this figure

JKL is right at J, and K is initially marked as the reference angle. No numerical lengths or side-role labels are given.
1. Hint 1
  A rotation preserves the named vertices and all incidence.
2. Hint 2
  A new reference angle can change the leg roles even when the picture stays still.
Answer

Case A: opposite $JL$ , adjacent $JK$ , hypotenuse $KL$ . Case B: opposite $JK$ , adjacent $JL$ , hypotenuse $KL$ . Only switching from K to L exchanges the legs.

Full solution

Case A: the original side across from $K$ is $JL$ , and rotation does not change that incidence.
The adjacent leg remains $JK$ , and $KL$ remains the hypotenuse.

Case B: side $JK$ is across from $L$ , so it is now opposite.
Side $JL$ is the adjacent leg.
The side across from the unchanged right angle at $J$ is still $KL$ .

The turned triangle with the second reference angle.

Answer

Case A: opposite $JL$ , adjacent $JK$ , hypotenuse $KL$ . Case B: opposite $JK$ , adjacent $JL$ , hypotenuse $KL$ . Only switching from K to L exchanges the legs.

Key idea

Rotation changes placement; a reference-angle switch changes opposite and adjacent roles.
Problem 7 Compare complementary angles

Triangle $ABC$ is right at $C$ . Its acute angles are $\alpha$ at $A$ and $\beta$ at $B$ .

Prove that the sine of either acute angle equals the cosine of the other. Also prove that their tangent values multiply to $1$ .

The two selected acute angles.

Text description of this figure

ABC is right at C. The acute angles alpha at A and beta at B are marked with distinct arc counts. No unknown ratio values are supplied.
1. Hint 1
  The two acute angles total $90^\circ$ , and the hypotenuse is the same for both.
2. Hint 2
  A leg opposite one acute angle is adjacent to the other.
3. Hint 3
  Write the two tangent values using the two positive leg lengths in opposite orders.
Answer

$\begin{aligned}\sin\alpha&=\cos\beta\\\cos\alpha&=\sin\beta\\\tan\alpha\tan\beta&=1.\end{aligned}$

Full solution

Both acute angles share hypotenuse $AB$ .
The side $BC$ is opposite $\alpha$ and adjacent to $\beta$ .

Therefore the identical side ratio gives
$\sin\alpha=\cos\beta.$

Similarly, side $CA$ is adjacent to $\alpha$ and opposite $\beta$ , with the same hypotenuse.

Thus
$\cos\alpha=\sin\beta.$

The tangent product is
$\tan\alpha\tan\beta=\frac{BC}{CA}\cdot\frac{CA}{BC}.$

Both legs are positive, so both factors cancel, leaving $1$ .

Answer

$\begin{aligned}\sin\alpha&=\cos\beta\\\cos\alpha&=\sin\beta\\\tan\alpha\tan\beta&=1.\end{aligned}$

Key idea

Complementary-angle relationships come from exchanging two positive legs.
Problem 8 Two identities from three lengths

Triangle $ABC$ is right at $C$ , with acute reference angle $\theta$ at $A$ .

Use the positive length names
$a=BC.$
Also,
$b=CA.$
Finally,
$c=AB.$
Prove that tangent is sine divided by cosine and that the squares of sine and cosine add to $1$ . Explain why each division in the argument is allowed.

The length names used in the proof.

Text description of this figure

ABC is right at C, with acute theta at A. BC is labeled a, CA is b, and AB is c. The turned picture preserves the specified side names.
1. Hint 1
  Write all three ratios using $a,b,c$ .
2. Hint 2
  For the quotient, divide two fractions with the same denominator.
3. Hint 3
  For the squared-value identity, use the already established Pythagorean theorem and divide by $c^2$ .
Answer

$\begin{aligned}\tan\theta&=\frac{\sin\theta}{\cos\theta}\\\sin^2\theta+\cos^2\theta&=1.\end{aligned}$
All denominators are positive.

Full solution

The side definitions give
$\sin\theta=\frac ac.$

They give
$\cos\theta=\frac bc.$

They also give
$\tan\theta=\frac ab.$

Dividing sine by cosine cancels their common hypotenuse:
$\frac{a/c}{b/c}=\frac ab.$

The division is allowed because $b$ and $c$ are positive, making cosine positive.

Pythagoras gives
$a^2+b^2=c^2.$

Divide by positive $c^2$ to obtain
$\frac{a^2}{c^2}+\frac{b^2}{c^2}=1.$

The two fractions on the left are the squared sine and cosine, respectively.

Answer

$\begin{aligned}\tan\theta&=\frac{\sin\theta}{\cos\theta}\\\sin^2\theta+\cos^2\theta&=1.\end{aligned}$
All denominators are positive.

Key idea

These identities share the same side definitions and require positive denominators.
Problem 9 Four written lines

An acute angle $\theta$ is known to satisfy $\sin\theta=8/17$ . A student writes:

Line 1:
$\sin^2\theta=\frac{64}{289}.$
Line 2:
$\cos^2\theta=\frac{225}{289}.$
Line 3:
$\cos\theta=-\frac{15}{17}.$
Line 4:
$\tan\theta=-\frac8{15}.$
Identify the first invalid line. Explain why it fails, and give the corrected cosine and tangent values.
1. Hint 1
  Check both the squared-value calculation and the allowed signs.
2. Hint 2
  Sine, cosine, and tangent are all positive for acute angles.
3. Hint 3
  After choosing the allowed cosine sign, use the quotient identity for tangent.
Answer

Line 3 is first invalid because acute-angle cosine is positive.
$\begin{aligned}\cos\theta&=\frac{15}{17}\\\tan\theta&=\frac8{15}.\end{aligned}$

Full solution

Line 1 correctly squares the supplied value.
Line 2 correctly uses
$1-\frac{64}{289}=\frac{225}{289}.$

Line 3 chooses a negative cosine, but cosine is a positive adjacent leg divided by a positive hypotenuse in this acute-angle domain.

The permitted root is
$\cos\theta=\frac{15}{17}.$

The quotient identity then gives
$\tan\theta=\frac{8/17}{15/17}.$

Cancel the common denominator to get
$\tan\theta=\frac8{15}.$

A positive-length model for the corrected values.

Answer

Line 3 is first invalid because acute-angle cosine is positive.
$\begin{aligned}\cos\theta&=\frac{15}{17}\\\tan\theta&=\frac8{15}.\end{aligned}$

Key idea

A squared equality alone does not decide the sign; the angle domain does.
Problem 10 Test proposed pairs

Each case proposes a sine value first and a cosine value second for one acute angle. Decide whether such an angle can exist, and justify each decision.

Case A: $12/13$ and $5/13$ .

Case B: $3/7$ and $4/7$ .

Case C: $0$ and $1$ .

State the general positivity and squared-value tests for such a pair. Explain why passing both tests allows a right-triangle model.
1. Hint 1
  A necessary identity connects the squares of the two values.
2. Hint 2
  Strict positivity matters in the acute-angle domain, even when the squared values add to $1$ .
3. Hint 3
  For two positive numbers whose squares sum to $1$ , construct perpendicular legs of those lengths.
Answer

Case A is possible; Cases B and C are not. In general,
$\begin{aligned}s&>0\\d&>0\\s^2+d^2&=1.\end{aligned}$
These conditions give a nondegenerate right-triangle model with hypotenuse $1$ .

Full solution

Case A passes because both values are positive and
$\left(\frac{12}{13}\right)^2+\left(\frac5{13}\right)^2=1.$

A right triangle with opposite leg $12$ , adjacent leg $5$ , and hypotenuse $13$ realizes this pair.

Case B fails the identity because
$\left(\frac37\right)^2+\left(\frac47\right)^2=\frac{25}{49}.$

Case C passes the squared-value identity but fails strict positivity.
A zero opposite leg would collapse the right-triangle model, so it cannot represent an acute reference angle.

For the general test, let $s$ and $d$ be the proposed sine and cosine.
Both must be strictly positive, and their squares must sum to $1$ .

Conversely, construct perpendicular legs of positive lengths $s$ and $d$ .
Pythagoras gives hypotenuse $1$ from the assumed squared sum.
The angle opposite leg $s$ is acute, and its sine and cosine are exactly $s$ and $d$ .

A possible model for Case A.

Answer

Case A is possible; Cases B and C are not. In general,
$\begin{aligned}s&>0\\d&>0\\s^2+d^2&=1.\end{aligned}$
These conditions give a nondegenerate right-triangle model with hypotenuse $1$ .

Key idea

An identity and a domain restriction must both be checked.

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