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The Law of Cosines: 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: Advanced (beyond the core course) Advanced. This problem set goes beyond core Geometry. You can skip it.

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Problem 1 of 10

Problem 1 A corner of a frame

A rigid triangular frame $XYZ$ has sides $XZ$ , $YZ$ , and $XY$ of lengths $10$ , $16$ , and $21$ units, respectively.

Write the expression for the cosine of the angle at $Z$ using the given lengths.

The side lengths are given. The opening at Z is unknown.

Text description of this figure

Frame XYZ has side lengths XZ = 10 units, YZ = 16 units, and XY = 21 units. The angle at Z is the target and is not labeled.
1. Hint 1
  Locate the side opposite the angle at $Z$ .
2. Hint 2
  Use the squares of the other two sides as the positive terms in the numerator.
Answer

$\cos Z=\frac{10^2+16^2-21^2}{2(10)(16)}.$

Full solution

The two shorter lengths sum to $26$ , so the three lengths can form a triangle.

Side $XY$ faces $Z$ .
The sides meeting at $Z$ are $XZ$ and $YZ$ .

Subtract the square of $XY$ from the sum of the squares of the other sides, then divide by twice their product.
$\cos Z=\frac{10^2+16^2-21^2}{2(10)(16)}.$

Answer

$\cos Z=\frac{10^2+16^2-21^2}{2(10)(16)}.$

Key idea

The sides that meet at the target angle give the denominator.
Problem 2 Opening a pair of arms

Two arms of lengths $9$ cm and $14$ cm meet at an angle of $135^\circ$ . Find the exact distance between the free ends.
1. Hint 1
  The unknown distance is across from the opening.
2. Hint 2
  The cosine of $135^\circ$ is the negative of the cosine of $45^\circ$ .
Answer

The free ends are $\sqrt{277+126\sqrt2}$ cm apart.

Full solution

The angle between the arms is the included angle.
Its cosine is $-\tfrac{\sqrt2}{2}$ .

Substitute the cosine with its sign.
$\begin{aligned}d^2&=9^2+14^2\\&\quad-2(9)(14)(-\tfrac{\sqrt2}{2})\\&=277+126\sqrt2.\end{aligned}$

Taking the positive square root gives $\sqrt{277+126\sqrt2}$ cm.

Answer

The free ends are $\sqrt{277+126\sqrt2}$ cm apart.

Key idea

Keeping an exact cosine value gives an exact length.
Problem 3 Classify a corner from side squares

A triangular sign has side lengths $18$ , $32$ , and $38$ units.

Classify the angle opposite the longest side as acute, right, or obtuse. Justify your decision without evaluating an inverse trigonometric function.
1. Hint 1
  Use the sign of the cosine to classify the angle.
2. Hint 2
  The denominator of the cosine quotient is positive. Which squares appear in its numerator?
Answer

The angle opposite the side of length $38$ is obtuse.

Full solution

The two shorter lengths add to $50$ , so the sides close to make a nondegenerate triangle.

The cosine numerator for the angle opposite $38$ is negative.
$18^2+32^2-38^2=-96.$

The denominator is positive.
The cosine is negative, which puts this angle between a right angle and a straight angle.

Answer

The angle opposite the side of length $38$ is obtuse.

Key idea

The sign of the cosine classifies the angle without giving its measure.
Problem 4 Recover a triangle from three lengths

Triangle $ABC$ has lengths $20$ , $31$ , and $42$ units for sides $a$ , $b$ , and $c$ , respectively.

Find all three angles to the nearest tenth of a degree and check their order against the side lengths.
1. Hint 1
  Start with the angle opposite the longest side.
2. Hint 2
  Find a second angle using its own adjacent sides. Use unrounded values when subtracting to find the last angle.
Answer

$\begin{aligned}A&\approx26.8^\circ,\\B&\approx44.3^\circ,\\C&\approx109.0^\circ.\end{aligned}$
The order from smallest to largest is $A$ , $B$ , $C$ , matching the order of the opposite lengths.

Full solution

The two shorter sides sum to $51$ , which exceeds $42$ .

Angle $C$ is across from the longest side.
$\begin{aligned}\cos C&=\frac{20^2+31^2-42^2}{2(20)(31)},\\C&\approx109.0^\circ.\end{aligned}$

Use the two sides that meet at $A$ to find that angle.
$\begin{aligned}\cos A&=\frac{31^2+42^2-20^2}{2(31)(42)},\\A&\approx26.8^\circ.\end{aligned}$

Subtract the unrounded measures of $A$ and $C$ from the angle sum.
$\begin{aligned}B&=180^\circ-A-C\\&\approx44.3^\circ.\end{aligned}$

The order of the angles agrees with their opposite sides.
Rounding each angle separately gives a displayed sum of $180.1^\circ$ .
The unrounded measures sum to $180^\circ$ .

Answer

$\begin{aligned}A&\approx26.8^\circ,\\B&\approx44.3^\circ,\\C&\approx109.0^\circ.\end{aligned}$
The order from smallest to largest is $A$ , $B$ , $C$ , matching the order of the opposite lengths.

Key idea

Keep full precision for angle subtraction even when each displayed angle is rounded.
Problem 5 Complete a layout

In triangle $ABC$ , sides $a$ and $b$ are $52$ cm and $61$ cm, respectively. Angle $C$ is $74^\circ$ .

Find $c$ to the nearest tenth of a centimeter and find $A$ and $B$ to the nearest tenth of a degree.
1. Hint 1
  The known angle is between the two given sides.
2. Hint 2
  After finding $c$ , keep it unrounded and use the three sides to find $B$ . The angle sum then gives $A$ .
Answer

$\begin{aligned}c&\approx68.4\text{ cm},\\A&\approx47.0^\circ,\\B&\approx59.0^\circ.\end{aligned}$

Full solution

The missing length is across from the included angle.
$\begin{aligned}c^2&=52^2+61^2\\&\quad-2(52)(61)\cos74^\circ,\\c&\approx68.4\text{ cm}.\end{aligned}$

Keep the unrounded positive root in the expression for $B$ .
$\begin{aligned}\cos B&=\frac{52^2+c^2-61^2}{2(52)c},\\B&\approx59.0^\circ.\end{aligned}$

Subtract the known angle and the unrounded value of $B$ to get $A$ .
$\begin{aligned}A&=180^\circ-74^\circ-B\\&\approx47.0^\circ.\end{aligned}$

The longest side is across from $C$ , the largest angle.
The shortest side is across from $A$ , the smallest angle.

Answer

$\begin{aligned}c&\approx68.4\text{ cm},\\A&\approx47.0^\circ,\\B&\approx59.0^\circ.\end{aligned}$

Key idea

The new side turns SAS into three known lengths, enough to use the cosine equation again.
Problem 6 Two directions from a survey station

From survey station $S$ , marker $E$ is $63$ m away in a direction $52^\circ$ east of north. Marker $W$ is $87$ m away in a direction $67^\circ$ west of north. All three points are on level ground.

Find the straight distance between $E$ and $W$ to the nearest tenth of a meter.

The dashed north ray splits the opening at the survey station.

Text description of this figure

Two lines leave station S, with a dashed north ray between them. The sight line to E measures 63 m at 52 degrees east of north. The line to W measures 87 m at 67 degrees west of north. Segment WE is the unknown separation.
1. Hint 1
  The two sight lines lie on opposite sides of the north ray.
2. Hint 2
  Add the two given angles to obtain the included angle at $S$ .
Answer

The markers are approximately $129.8$ m apart.

Full solution

The sight lines turn away from north on different sides.
$\begin{aligned}\angle ESW&=52^\circ+67^\circ\\&=119^\circ.\end{aligned}$

The sought separation is opposite that angle.
$\begin{aligned}EW^2&=63^2+87^2\\&\quad-2(63)(87)\cos119^\circ,\\EW&\approx129.8\text{ m}.\end{aligned}$

The separation is less than the sum of the two sight-line lengths, $150$ m.

Answer

The markers are approximately $129.8$ m apart.

Key idea

When two directions are measured from opposite sides of the same ray, add their angles to find the opening.
Problem 7 Two field records

Each field record describes a different triangle $ABC$ . For each record, name the first triangle law that can find $a$ , and explain why.

Record A. Side $b$ is $46$ m, angle $A$ is $57^\circ$ , and angle $B$ is $82^\circ$ . Also find $a$ to the nearest tenth of a meter.

Record B. Side $b$ is $33$ m, side $c$ is $47$ m, and angle $A$ is $73^\circ$ .
1. Hint 1
  Mark every known side and look directly across from it.
2. Hint 2
  Record A gives a complete opposite pair. In Record B, notice the common endpoint of the two given sides.
Answer

Record A uses the Law of Sines because $b$ and $B$ are a known opposite pair. The length $a$ is approximately $39.0$ m. Record B uses the Law of Cosines because $A$ is the angle included between the known sides.

Full solution

Record A.
The known side $b$ faces the known angle $B$ , so the sine law pairs it with side $a$ and angle $A$ .
$\begin{aligned}\frac{a}{\sin57^\circ}&=\frac{46}{\sin82^\circ},\\a&\approx39.0\text{ m}.\end{aligned}$

Record B.
Sides $b$ and $c$ meet at $A$ .
Those two lengths and their included angle supply all three inputs to the cosine equation for $a^2$ .

Answer

Record A uses the Law of Sines because $b$ and $B$ are a known opposite pair. The length $a$ is approximately $39.0$ m. Record B uses the Law of Cosines because $A$ is the angle included between the known sides.

Key idea

An opposite pair starts the sine law, while two sides and the angle between them start the cosine law.
Problem 8 Check a reported distance

Two sides of lengths $5$ m and $12$ m meet at $130^\circ$ . Riley reports that the opposite side is $9.6$ m after subtracting the absolute value of the cosine correction.

Explain how to rule out the reported value without a calculator, and find the opposite side to the nearest tenth of a meter.
1. Hint 1
  A right opening with the same sides gives a comparison length.
2. Hint 2
  What happens when you subtract the negative cosine correction with its sign intact?
Answer

The obtuse opening requires a length greater than $13$ m. The opposite side is approximately $15.7$ m.

Full solution

At a right opening, the opposite length would be $13$ m.
$5^2+12^2=169.$

The obtuse angle has negative cosine.
Subtracting the signed correction makes the opposite side square greater than $169$ , so the length must exceed $13$ m.

Use the signed cosine and take the positive square root.
$\begin{aligned}d^2&=169-120\cos130^\circ,\\d&\approx15.7\text{ m}.\end{aligned}$

Answer

The obtuse opening requires a length greater than $13$ m. The opposite side is approximately $15.7$ m.

Key idea

A right opening gives a benchmark for judging a result at an obtuse opening.
Problem 9 Recover the foot from three sides

Triangle $ABC$ has sides $BC$ , $CA$ , and $AB$ of lengths $29$ , $12$ , and $25$ units, respectively.

Put $C$ at $(0,0)$ and $A$ on the positive horizontal axis. Place $B$ above that axis.

Find the signed horizontal coordinate of $B$ . Is its perpendicular foot between $C$ and $A$ , beyond $A$ , or to the left of $C$ ?
1. Hint 1
  The signed coordinate $x$ is $a$ times the cosine of $C$ .
2. Hint 2
  You can also isolate $x$ directly in the squared-distance formula before replacing it by the cosine expression.
Answer

The signed horizontal coordinate is $15$ . The perpendicular foot lies beyond $A$ .

Full solution

For the horizontal coordinate $x$ of $B$ , the cosine derivation gives
$c^2=a^2+b^2-2bx.$

Substitute the three given lengths and isolate $x$ .
$\begin{aligned}25^2&=29^2+12^2-24x,\\24x&=360,\\x&=15.\end{aligned}$

Point $A$ has horizontal coordinate $12$ .
The coordinate of the perpendicular foot is $15$ , which puts it beyond $A$ on the positive axis.

Answer

The signed horizontal coordinate is $15$ . The perpendicular foot lies beyond $A$ .

Key idea

A positive projection can extend past the endpoint of the base.
Problem 10 Which measurement fixes the opening?

Two rigid arms $CA$ and $CB$ have known positive lengths $b$ and $a$ . Their opening at $C$ lies strictly between zero and a straight angle.

Plan A measures the straight distance between the free ends $A$ and $B$ .

Plan B drops a perpendicular from $B$ to the line through $CA$ . It records the distance $p$ from $C$ to the foot, where $p$ is positive and smaller than $a$ . It leaves out which side of $C$ contains the foot.

Decide which plan always determines the opening uniquely. Prove your decision and explain how two openings can fit the other plan.
1. Hint 1
  Compare the inputs for the cosine quotient with the information in each plan.
2. Hint 2
  Place $CA$ along the positive horizontal axis. The foot's unsigned distance from $C$ allows two signed coordinates.
3. Hint 3
  Check whether each of those coordinates can belong to a point $B$ above the axis at distance $a$ from $C$ .
Answer

Plan A determines one opening. Plan B allows two openings, one acute and one obtuse.

Full solution

Plan A.
The measured distance is $c$ , opposite the opening.
The three sides give a single cosine.
$\cos C=\frac{a^2+b^2-c^2}{2ab}.$

Cosine takes each value strictly between $-1$ and $1$ exactly once for an interior triangle angle, so the actual measured triangle has one possible opening.

Plan B.
With $C$ at the origin and $CA$ pointing to the right, the signed coordinate of the foot can be $p$ or $-p$ .

Either choice can put $B$ above the axis at distance $a$ from $C$ .
Use the same positive height in both cases.
$y=\sqrt{a^2-p^2}.$

The two possible cosine values come from dividing each signed projection by $a$ .
$\begin{aligned}\cos C_1&=\frac{p}{a},\\\cos C_2&=-\frac{p}{a}.\end{aligned}$

Both ratios lie strictly between $-1$ and $1$ .
One is positive and the other negative, so they produce distinct valid angles, one acute and one obtuse.

The same unsigned foot distance fits two different openings.

Answer

Plan A determines one opening. Plan B allows two openings, one acute and one obtuse.

Key idea

A horizontal projection determines the cosine only when its direction is known.

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