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The Law of Sines: 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 The side across from the smaller angle

In triangle $ABC$ , angles $A$ and $C$ measure $58^\circ$ and $81^\circ$ , respectively. Side $AB$ is 20 cm long.

Find $BC$ to the nearest tenth of a centimeter. Explain why it should be shorter than $AB$ .

The known side AB faces the marked angle at C.

Text description of this figure

Triangle ABC has angles 58 degrees at A and 81 degrees at C. The base AB is 20 centimeters long. Side BC is requested.
1. Hint 1
  Identify the angle across from $AB$ .
2. Hint 2
  The requested length $BC$ is across from angle $A$ . Compare the two corresponding ratios.
Answer

$BC\approx17.2\text{ cm}.$
Side $BC$ lies opposite the smaller angle.

Full solution

Use the known pair of $AB$ and angle $C$ .

Side $BC$ faces angle $A$ , so the two ratios give
$\frac{BC}{\sin58^\circ}=\frac{20}{\sin81^\circ}.$

Solve for the length and round.
$\begin{aligned}BC&=\frac{20\sin58^\circ}{\sin81^\circ}\text{ cm}\\&\approx17.2\text{ cm}.\end{aligned}$

Angle $A$ is smaller than angle $C$ .
The side across from the smaller angle is shorter.

Answer

$BC\approx17.2\text{ cm}.$
Side $BC$ lies opposite the smaller angle.

Key idea

Label each side by the angle across from it before writing a proportion.
Problem 2 A third angle completes the known pair

In triangle $ABC$ , angles $A$ and $B$ measure $64^\circ$ and $56^\circ$ , respectively. The side joining them is 23 m long.

Find $C$ and the lengths of $BC$ and $CA$ . Round lengths to the nearest tenth of a meter.

The unknown angle is opposite the baseline.

Text description of this figure

Triangle ABC has angles 64 degrees at A and 56 degrees at B. The side between them is 23 meters long.
1. Hint 1
  The side joining $A$ and $B$ is opposite an angle that has to be found.
2. Hint 2
  Subtract the known angles from 180 degrees and use that angle with the 23 m side.
Answer

$\begin{aligned}C&=60^\circ.\\BC&\approx23.9\text{ m}.\\CA&\approx22.0\text{ m}.\end{aligned}$

Full solution

The known length is $AB$ .
The angle opposite it is
$\begin{aligned}C&=180^\circ-64^\circ-56^\circ\\&=60^\circ.\end{aligned}$

Use angle $A$ for $BC$ .
$\begin{aligned}BC&=23\frac{\sin64^\circ}{\sin60^\circ}\text{ m}\\&\approx23.9\text{ m}.\end{aligned}$

Use angle $B$ for $CA$ .
$\begin{aligned}CA&=23\frac{\sin56^\circ}{\sin60^\circ}\text{ m}\\&\approx22.0\text{ m}.\end{aligned}$

Answer

$\begin{aligned}C&=60^\circ.\\BC&\approx23.9\text{ m}.\\CA&\approx22.0\text{ m}.\end{aligned}$

Key idea

The third angle supplies the opposite partner for an included side.
Problem 3 A fixed angle and four radii

Angle $A$ measures $45^\circ$ and $AC$ is 20 units long. Point $B$ must lie on the ray from $A$ that makes this angle with $AC$ .

For each case, state how many nondegenerate triangles fit. Explain where the circle centered at $C$ meets the ray.

(a) The radius is 13 units.

(b) The radius is $10\sqrt{2}$ units.

(c) The radius is 17 units.

(d) The radius is 20 units.
1. Hint 1
  Find the perpendicular distance from $C$ to the ray.
2. Hint 2
  Compare each radius with both the perpendicular distance and the distance from $C$ to $A$ .
Answer

$\begin{aligned}\text{(a)}\quad&\text{Zero.}\\\text{(b)}\quad&\text{One, at the foot.}\\\text{(c)}\quad&\text{Two, both past }A.\\\text{(d)}\quad&\text{One past }A.\end{aligned}$

Full solution

The perpendicular distance is
$\begin{aligned}h&=20\sin45^\circ\\&=10\sqrt{2}.\end{aligned}$

(a) The radius 13 is below the height because its square is below 200.
The circle misses the line.

(b) The radius equals the height.
The only intersection is at the perpendicular foot, strictly ahead of $A$ .

(c) The square of 17 is above 200, so 17 exceeds the height.
Since 17 is also below 20, the two crossings are both ahead of $A$ .

(d) One crossing is at $A$ , since the radius equals $AC$ .
The farther crossing makes a triangle.

Answer

$\begin{aligned}\text{(a)}\quad&\text{Zero.}\\\text{(b)}\quad&\text{One, at the foot.}\\\text{(c)}\quad&\text{Two, both past }A.\\\text{(d)}\quad&\text{One past }A.\end{aligned}$

Key idea

For an acute given angle, compare the radius with both the perpendicular height and the distance to the ray start.
Problem 4 Survey an island station

Stations $A$ and $B$ are 118 m apart on one shore. They sight the same station $C$ on an island in the horizontal plane of $A$ and $B$ . At $A$ , the angle from the outward baseline ray pointing away from $B$ toward $C$ is $119^\circ$ . At $B$ , the interior angle is $47^\circ$ .

Find $AC$ and $BC$ to the nearest tenth of a meter. State which shore station is closer.

The dashed extension supplies the exterior reference ray at A.

Text description of this figure

Stations A and B are 118 meters apart. At A, the 119-degree exterior reading starts along the extension away from B and ends at the sight line to C. At B, the 47-degree interior reading starts along BA.
1. Hint 1
  The outward ray at $A$ and ray $AB$ form a straight angle.
2. Hint 2
  After finding the interior angle at $A$ , use the angle sum to get the angle opposite the measured baseline.
Answer

$\begin{aligned}AC&\approx90.7\text{ m}.\\BC&\approx108.5\text{ m}.\end{aligned}$
Station $A$ is closer.

Full solution

The two baseline rays from $A$ are opposite, so the interior angle is
$\begin{aligned}A&=180^\circ-119^\circ\\&=61^\circ.\end{aligned}$

The angle across from the baseline is
$\begin{aligned}C&=180^\circ-61^\circ-47^\circ\\&=72^\circ.\end{aligned}$

Segment $AC$ faces angle $B$ .
$\begin{aligned}AC&=118\frac{\sin47^\circ}{\sin72^\circ}\text{ m}\\&\approx90.7\text{ m}.\end{aligned}$

Segment $BC$ faces angle $A$ .
$\begin{aligned}BC&=118\frac{\sin61^\circ}{\sin72^\circ}\text{ m}\\&\approx108.5\text{ m}.\end{aligned}$

Station $A$ is closer because $AC$ is shorter.
This agrees with $B$ being the smaller shore angle.

Answer

$\begin{aligned}AC&\approx90.7\text{ m}.\\BC&\approx108.5\text{ m}.\end{aligned}$
Station $A$ is closer.

Key idea

Convert an exterior reading to the interior angle before finding the opposite pair for the baseline.
Problem 5 Recover both triangles

In triangle $ABC$ , angle $A$ is $22^\circ$ . The lengths of $a$ and $b$ are 15 cm and 27 cm, respectively.

Find every possible set of measurements for $B$ , $C$ , and $c$ . Give angles and lengths to the nearest tenth in degrees and centimeters.
1. Hint 1
  The sine of $B$ comes from the known opposite pair $A,a$ and the length $b$ .
2. Hint 2
  Try the inverse sine angle and its supplement. In each case, find the remaining angle before solving for $c$ .
Answer

Case 1.
$\begin{aligned}B&\approx42.4^\circ.\\C&\approx115.6^\circ.\\c&\approx36.1\text{ cm}.\end{aligned}$
Case 2.
$\begin{aligned}B&\approx137.6^\circ.\\C&\approx20.4^\circ.\\c&\approx14.0\text{ cm}.\end{aligned}$

Full solution

The opposite pairs give
$\sin B=\frac{27\sin22^\circ}{15}.$

Case 1.
Inverse sine gives about $42.4^\circ$ for $B$ .
Subtract $A$ and the unrounded value of $B$ from $180^\circ$ .
This gives a positive $C$ of about $115.6^\circ$ .

Case 1 length.
$\begin{aligned}c&=\frac{15\sin C}{\sin22^\circ}\text{ cm}\\&\approx36.1\text{ cm}.\end{aligned}$

Case 2.
The supplement gives about $137.6^\circ$ for $B$ .
It leaves a positive $C$ of about $20.4^\circ$ .

Case 2 length.
$\begin{aligned}c&=\frac{15\sin C}{\sin22^\circ}\text{ cm}\\&\approx14.0\text{ cm}.\end{aligned}$

There are exactly two triangle angles with the calculated sine, so these cases exhaust the possibilities.

Answer

Case 1.
$\begin{aligned}B&\approx42.4^\circ.\\C&\approx115.6^\circ.\\c&\approx36.1\text{ cm}.\end{aligned}$
Case 2.
$\begin{aligned}B&\approx137.6^\circ.\\C&\approx20.4^\circ.\\c&\approx14.0\text{ cm}.\end{aligned}$

Key idea

For each candidate that leaves a positive third angle, calculate the remaining length from that candidate alone.
Problem 6 Two bases and one height

Angle $A$ is $30^\circ$ , side $AC$ is 40 cm, and side $BC$ is 29 cm. The two possible triangles share the same $AC$ , with both choices of $B$ on the same ray from $A$ .

(a) Find the two exact lengths of $AB$ .

(b) Find the exact difference between the areas of the two triangles.

The same circle and ray determine both configurations.

Text description of this figure

The ray starts at A and makes a 30-degree angle with AC. AC is 40 centimeters long. The circle centered at C has radius 29 centimeters.
1. Hint 1
  Drop the perpendicular from $C$ to the ray and name its foot $H$ .
2. Hint 2
  Use the right triangle with hypotenuse $BC$ to find the distance from $H$ to either crossing.
3. Hint 3
  Both areas use the same height. Find the difference between the two bases.
Answer

(a)
$\begin{aligned}c_{\text{near}}&=(20\sqrt{3}-21)\text{ cm}.\\c_{\text{far}}&=(20\sqrt{3}+21)\text{ cm}.\end{aligned}$
(b)
$420\text{ cm}^2.$

Full solution

(a) The perpendicular height is
$\begin{aligned}CH&=40\sin30^\circ\\&=20\text{ cm}.\end{aligned}$

The signed projection is positive and gives
$\begin{aligned}AH&=40\cos30^\circ\\&=20\sqrt{3}\text{ cm}.\end{aligned}$

Both crossings are the same distance from $H$ .
With $BC$ as hypotenuse,
$\begin{aligned}HB&=\sqrt{29^2-20^2}\text{ cm}\\&=21\text{ cm}.\end{aligned}$

Since $AH$ exceeds 21 cm, both crossings are past $A$ .
Subtract 21 cm to get the nearer base and add it for the farther base.

(b) The farther base is 42 cm longer.
Multiply half that difference by the common height.
$\begin{aligned}K_{\text{far}}-K_{\text{near}}&=\tfrac12(42)(20)\text{ cm}^2\\&=420\text{ cm}^2.\end{aligned}$

Answer

(a)
$\begin{aligned}c_{\text{near}}&=(20\sqrt{3}-21)\text{ cm}.\\c_{\text{far}}&=(20\sqrt{3}+21)\text{ cm}.\end{aligned}$
(b)
$420\text{ cm}^2.$

Key idea

Two crossing points equidistant from the foot give bases that differ by twice that distance.
Problem 7 A sine value that passes only the first check

The given angle $A$ is $135^\circ$ . The lengths of sides $a$ and $b$ are 10 and 12 units, respectively.

(a) Find both angle candidates from the sine equation for $B$ . Report them to the nearest tenth of a degree.

(b) Test whether either candidate gives a nondegenerate triangle.

(c) Check the conclusion using the given angle type and the two side lengths.
1. Hint 1
  First check whether the right side of the sine equation is at most 1.
2. Hint 2
  Angle $A$ already uses three quarters of the triangle angle sum.
3. Hint 3
  For an obtuse $A$ , the distance from $C$ to points of the ray starts at $AC$ and increases.
Answer

$\begin{aligned}\text{(a)}\quad&58.1^\circ,\ 121.9^\circ.\\\text{(b)}\quad&\text{Neither angle fits.}\\\text{(c)}\quad&\text{No ray intersection.}\end{aligned}$

Full solution

(a) Match $A$ with $a$ to get
$\begin{aligned}\sin B&=\frac{12\sin135^\circ}{10}\\&=\frac{3\sqrt{2}}{5}.\end{aligned}$

Inverse sine gives about $58.1^\circ$ .
Subtract the unrounded angle from $180^\circ$ to get about $121.9^\circ$ .

(b) Angle $A$ leaves only $45^\circ$ for $B$ and $C$ together.
Both candidates for $B$ exceed that amount, so each leaves a negative angle at $C$ .

(c) The perpendicular foot lies behind the start.
All points past $A$ are farther from $C$ than $A$ is, so a radius of 10 cannot reach the ray.

Answer

$\begin{aligned}\text{(a)}\quad&58.1^\circ,\ 121.9^\circ.\\\text{(b)}\quad&\text{Neither angle fits.}\\\text{(c)}\quad&\text{No ray intersection.}\end{aligned}$

Key idea

A sine value within range supplies candidates, and the angle sum decides whether they fit.
Problem 8 An area identity with a missing step

Triangle $ABC$ is nondegenerate. Twice its area can be written in one way at $A$ and another way at $C$ .
$2K=bc\sin A.$ $2K=ab\sin C.$
(a) Use these equal expressions to prove that $\sin A/a$ equals $\sin C/c$ . Explain why the division is allowed.

(b) Apply the relationship to a triangle with $a$ of length 15, $c$ of length 19, and sine of $C$ equal to $4/5$ . Give the sine of $A$ exactly.
1. Hint 1
  Which product contains all three side lengths?
2. Hint 2
  For part (b), solve the equation between the sine-over-side ratios for the missing sine.
Answer

(a) Division by the positive product $abc$ gives
$\frac{\sin A}{a}=\frac{\sin C}{c}.$
(b)
$\sin A=\frac{12}{19}.$

Full solution

(a) The two expressions equal the same $2K$ .
All three side lengths are positive, so dividing each expression by their product preserves equality.

The factors $b$ and $c$ cancel from the first expression, leaving $\sin A/a$ .
The factors $a$ and $b$ cancel from the second, leaving $\sin C/c$ .
The two quotients are equal.

(b) Substitute the lengths and the sine of $C$ , then multiply by 15.
$\begin{aligned}\sin A&=\frac{15}{19}\cdot\frac45\\&=\frac{12}{19}.\end{aligned}$

Answer

(a) Division by the positive product $abc$ gives
$\frac{\sin A}{a}=\frac{\sin C}{c}.$
(b)
$\sin A=\frac{12}{19}.$

Key idea

Dividing equivalent area expressions by the product of the sides leaves the opposite-pair ratios.
Problem 9 Line tangency behind the start

Angle $A$ is $135^\circ$ and $AC$ is 14 units long. Point $B$ must lie on the ray from $A$ making that angle with $AC$ . The diagram draws the ray solid and its backward extension dashed.

For each radius of the circle centered at $C$ , say whether any point of the ray past $A$ can be $B$ . Explain any contact at the foot of the perpendicular or at the start.

(a) The radius is $7\sqrt{2}$ units.

(b) The radius is 14 units.

(c) The radius is 15 units.

The dashed extension is part of the line behind A.

Text description of this figure

AC has length 14. Angle A measures 135 degrees between AC and the forward ray. The perpendicular from C to the ray line meets the dashed extension at H.
1. Hint 1
  Where is the perpendicular foot from $C$ to the full line through the ray?
2. Hint 2
  Compare the distance from $C$ to $A$ with distances from $C$ to points farther along the forward ray.
Answer

$\begin{aligned}\text{(a)}\quad&\text{No. Tangent behind }A.\\\text{(b)}\quad&\text{No. Contact only at }A.\\\text{(c)}\quad&\text{Yes. One crossing past }A.\end{aligned}$

Full solution

The perpendicular foot $H$ is behind $A$ because the signed projection of $AC$ is negative.

(a) The perpendicular height is
$\begin{aligned}CH&=14\sin135^\circ\\&=7\sqrt{2}.\end{aligned}$

The circle touches the full line only at $H$ , so it misses the ray.

(b) The radius equals $CA$ , so the circle contains $A$ .
For points strictly past $A$ , the distance from $H$ increases while $CH$ stays fixed.
The Pythagorean theorem makes the distance to $C$ strictly greater than 14 there.

(c) The distance from $C$ increases continuously from 14 and becomes arbitrarily large along the ray.
Exactly one point has distance 15.

Answer

$\begin{aligned}\text{(a)}\quad&\text{No. Tangent behind }A.\\\text{(b)}\quad&\text{No. Contact only at }A.\\\text{(c)}\quad&\text{Yes. One crossing past }A.\end{aligned}$

Key idea

For an obtuse given angle, the ray begins beyond the perpendicular foot, so the start sets the threshold.
Problem 10 Choose the next field check

A marker $B$ is somewhere along a straight maintenance route starting at $A$ . Station $C$ is 36 m from $A$ , and the interior angle between the forward route and the sight line from $A$ to $C$ is $28^\circ$ . A range reading puts the marker 21 m from $C$ . All three points lie in the same horizontal plane.

(a) Find all possible distances from $A$ to the marker, to the nearest tenth of a meter.

(b) The crew can verify one of two claims. The marker is farther than 25 m from $A$ , or it is less than 50 m from $A$ . Which claim, if confirmed, would identify the position uniquely? Give that position and explain why the other claim would leave the position unresolved.

The route starts at A and extends along the arrow.

Text description of this figure

A is at the beginning of the route. The sight line to station C makes a 28-degree angle with the route. Station C is 36 meters from A. The marker lies somewhere on the route.
1. Hint 1
  The distance from $C$ gives a circle, and the marker must be on the forward route.
2. Hint 2
  Find both sine candidates for the angle at the marker, or use the equal offsets from the perpendicular foot.
3. Hint 3
  Compare both distances with each proposed field check.
Answer

(a)
$AB\approx19.3\text{ m or }44.3\text{ m}.$
(b) Farther than 25 m identifies the position about 44.3 m from $A$ . Both positions are less than 50 m away.

Full solution

(a) The angle at the marker satisfies
$\sin B=\frac{36\sin28^\circ}{21}.$

The acute candidate for $B$ is about $53.6^\circ$ .
It leaves $C$ about $98.4^\circ$ .
Use the unrounded value of $C$ to find the farther distance.
$\begin{aligned}c_{\text{far}}&=\frac{21\sin C}{\sin28^\circ}\text{ m}\\&\approx44.3\text{ m}.\end{aligned}$

The supplementary candidate for $B$ is about $126.4^\circ$ .
It leaves $C$ about $25.6^\circ$ .
Use that unrounded value of $C$ for the nearer distance.
$\begin{aligned}c_{\text{near}}&=\frac{21\sin C}{\sin28^\circ}\text{ m}\\&\approx19.3\text{ m}.\end{aligned}$

(b) A verified distance beyond 25 m rules out the nearer point and keeps the farther one.

Both distances are below 50 m, so that second check would keep both points.

The crossing at B₁ is the only one farther than 25 m from A.

Answer

(a)
$AB\approx19.3\text{ m or }44.3\text{ m}.$
(b) Farther than 25 m identifies the position about 44.3 m from $A$ . Both positions are less than 50 m away.

Key idea

When two positions fit the original readings, a follow-up observation resolves the ambiguity only if the positions predict different outcomes.

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