The Law of Sines Advanced. This lesson goes beyond core Geometry. You can skip it.

Watch the video explainer

Learning goals

Derive the law of sines by writing one triangle's area in three ways
Solve ASA and AAS triangles using a known opposite pair
Model an inaccessible distance from a measured baseline and two angles
Classify SSA data by counting the valid intersections of a circle and a ray
Test both angle candidates and find all missing measurements in SSA triangles

One area connects the opposite pairs

The sine of an angle and the length across from it form the same ratio at every corner of a triangle. To see where the ratio comes from, start with a triangle whose legs are 5 and 12 and whose hypotenuse is 13. Use capital letters for angles and lowercase letters for their opposite side lengths. The lengths of $a$ , $b$ , and $c$ are 5, 12, and 13, respectively.

The side labels stay fixed when we choose a different included angle.

The legs are perpendicular, so half their product gives an area $K$ of 30 square units. Multiply each included-angle area expression by 2, then divide by the product of the three side lengths. Each calculation gives the same quotient.

\begin{aligned} \frac{2K}{5\cdot12\cdot13}&=\frac{1}{13}.\\[2pt] \frac{\sin A}{5}&=\frac{1}{13}.\\[2pt] \frac{\sin B}{12}&=\frac{1}{13}.\\[2pt] \frac{\sin C}{13}&=\frac{1}{13}. \end{aligned}

For example, using sides 12 and 13 gives twice the area as $12\cdot13\sin A$ . Dividing by all three lengths cancels 12 and 13, so the 5 across from $A$ stays below $\sin A$ . The same cancellation works at the other two corners. The argument uses the included-angle formula from Triangle Area with Sine Advanced. This lesson goes beyond core Geometry. You can skip it., which works at acute, right, and obtuse angles.

Why opposite side and sine ratios agree#

Take any nondegenerate triangle, with positive sides $a$ , $b$ , and $c$ . Apply the area formula once at each corner.

\begin{aligned} 2K&=bc\sin A.\\ 2K&=ca\sin B.\\ 2K&=ab\sin C. \end{aligned}

Divide each equation by $abc$ . The same quotient on the left appears three times.

\begin{aligned} \frac{2K}{abc}&=\frac{\sin A}{a}.\\[2pt] \frac{2K}{abc}&=\frac{\sin B}{b}.\\[2pt] \frac{2K}{abc}&=\frac{\sin C}{c}. \end{aligned}

This common value is positive. Taking reciprocals gives the law of sines in side-over-sine form.

\begin{aligned} \frac{a}{\sin A}&=\frac{b}{\sin B}.\\[2pt] \frac{b}{\sin B}&=\frac{c}{\sin C}. \end{aligned}

Check your understanding

Which proportion compares two opposite pairs in triangle $ABC$ ?

a/\sin B=b/\sin A

b/\sin B=c/\sin C

a/\sin A=c/\sin B

a/\sin C=b/\sin B

Two angles and a side fix the triangle

In ASA, the known side joins the two known angle vertices. In AAS, it joins one known angle vertex to the unknown one. Either way, subtract the known angles from $180^\circ$ . A positive third angle gives the shape, and the known side fixes its scale. Use the known side and its opposite angle to find the missing sides.

Worked example 1 Use the side between two angles

Problem. In triangle $ABC$ , the angles at $A$ and $B$ measure $48^\circ$ and $67^\circ$ , respectively. The side $AB$ is 19 cm long. Find $C$ , $a$ , and $b$ , rounding lengths to the nearest tenth of a centimeter.

The known angles sit at the endpoints of the known side.

Solution. The known side lies across from $C$ . Find its partner angle.

\begin{aligned} C&=180^\circ-48^\circ-67^\circ\\ &=65^\circ. \end{aligned}

Use the known pair $c=19$ cm and $C=65^\circ$ in each proportion.

\begin{aligned} a&=\frac{19\sin48^\circ}{\sin65^\circ}\text{ cm}\\ &\approx15.6\text{ cm}. \end{aligned}

\begin{aligned} b&=\frac{19\sin67^\circ}{\sin65^\circ}\text{ cm}\\ &\approx19.3\text{ cm}. \end{aligned}

The length order matches the angle order, with $A$ smallest and $B$ largest.

Worked example 2 Start with an opposite pair already given

Problem. In triangle $ABC$ , angle $A$ measures $37^\circ$ , angle $C$ measures $108^\circ$ , and $BC$ is 11 m long. Find $B$ , $b$ , and $c$ , rounding lengths to the nearest tenth of a meter.

The known side BC pairs with the angle at A.

Solution. The side $BC$ lies across from $A$ , so $a$ and $A$ form the known pair. The endpoints of the known side are $B$ and $C$ , so this is AAS.

\begin{aligned} B&=180^\circ-37^\circ-108^\circ\\ &=35^\circ. \end{aligned}

\begin{aligned} b&=\frac{11\sin35^\circ}{\sin37^\circ}\text{ m}\\ &\approx10.5\text{ m}. \end{aligned}

\begin{aligned} c&=\frac{11\sin108^\circ}{\sin37^\circ}\text{ m}\\ &\approx17.4\text{ m}. \end{aligned}

Check your understanding

In triangle $ABC$ , $B$ is $44^\circ$ , $C$ is $71^\circ$ , and $AC$ is 9 cm long. Which expression gives $AB$ in centimeters?

9\sin44^\circ/\sin71^\circ

9\sin65^\circ/\sin44^\circ

9\sin71^\circ/\sin65^\circ

9\sin71^\circ/\sin44^\circ

Measure from a baseline

The triangle needs a straight measured baseline and two directions to the same target. Use the interior angle from the baseline ray pointing toward the other observer at each station. A compass reading by itself gives a direction from north, so it would first need to be converted to an interior angle.

Worked example 3 Find the distance to a lookout

Problem. Stations $A$ and $B$ lie 76 m apart on a straight trail segment. Both stations and lookout $C$ lie in one horizontal plane. The interior angle from $AB$ to $AC$ at $A$ is $38^\circ$ . The interior angle from $BA$ to $BC$ at $B$ is $54^\circ$ . How far is the lookout from station $A$ , to the nearest tenth of a meter?

Both readings are measured from the baseline toward the same lookout.

Solution. The baseline $AB$ is opposite the lookout angle.

\begin{aligned} C&=180^\circ-38^\circ-54^\circ\\ &=88^\circ. \end{aligned}

The requested distance is $AC$ , across from $B$ .

\begin{aligned} AC&=\frac{76\sin54^\circ}{\sin88^\circ}\text{ m}\\ &\approx61.5\text{ m}. \end{aligned}

The angle at $B$ is smaller than the angle at $C$ , so $AC$ should be shorter than the baseline. The result passes that check.

Check your understanding

Two stations aim at one target. At station $B$ , a reading of $104^\circ$ is measured from the extension of $AB$ beyond $B$ toward the target. What interior triangle angle should the surveyor use at $B$ ?

14^\circ

104^\circ

76^\circ

86^\circ

Two sides and a nonincluded angle can leave a choice

With SSA, you know two sides and an angle opposite one of them. Suppose the data are an angle $A$ strictly between $0^\circ$ and $180^\circ$ and positive lengths $a$ and $b$ . Fix $AC$ with length $b$ and draw the ray from $A$ on which $B$ must lie. The point $B$ must be exactly $a$ units from $C$ , so it must also lie on the circle of radius $a$ centered at $C$ . Each crossing on the ray past $A$ gives a triangle. A reflected drawing gives the same measurements and adds no new triangle up to congruence.

Compare the circles for a fixed acute angle

In the next five panels, $A$ stays at $30^\circ$ and $AC$ stays 6 units long. The perpendicular distance $h$ from $C$ to the line of the ray is 3 units. The perpendicular foot $H$ lies ahead of $A$ .

\begin{aligned} h&=6\sin30^\circ\\ &=3. \end{aligned}

Predict how many times the circle meets the ray past $A$ for one of these radii. Select that radius to check its panel.

2 units · 3 units · 5 units · 6 units · 7 units

Radius 2

Radius 2. The circle cannot reach the line.

There is no possible angle $B$ .

Radius 3

Radius 3. The circle touches the ray once at H.

At the point of contact, $B=90^\circ$ .

Radius 5

Radius 5. Both crossings lie past A.

At the nearer crossing $N$ , the angle is about $143.1^\circ$ . At the farther crossing $F$ , it is about $36.9^\circ$ .

Radius 6

Radius 6. Only B gives a triangle, since a crossing at A would give AB length zero.

The surviving angle at $B$ is $30^\circ$ .

Radius 7

Radius 7. Only the crossing ahead of A can be B.

The surviving angle at $B$ is about $25.4^\circ$ .

Explain the thresholds

For any fixed $0^\circ<A<180^\circ$ , the height is $h=b\sin A$ , and the signed position of $H$ along the line from $A$ is $b\cos A$ . A positive projection puts $H$ ahead of $A$ . Zero puts $H$ at $A$ , and a negative projection puts $H$ behind $A$ .

Acute $A$ . No point on the line is closer to $C$ than $H$ . When the radius reaches $h$ , only $H$ is available. For a larger radius, two line intersections sit on opposite sides of $H$ , equally far from it. As the radius grows, the farther intersection stays on the ray, while the nearer one moves toward $A$ . It reaches $A$ exactly when the radius equals $AC$ , which is $b$ . At that endpoint, $B=A$ would give a zero-length side, so count only the farther intersection.

Right or obtuse $A$ . Here $H$ is at $A$ or behind it. The distance from $C$ to a point on the ray is shortest at $A$ and grows strictly as the point moves forward. To see this, the perpendicular height stays fixed while the horizontal distance from $H$ increases. The Pythagorean theorem then makes the distance from $C$ increase as well. Every distance greater than $b$ occurs exactly once, since the distance increases continuously from $b$ without bound.

Here the radius exceeds AC. The left crossing lies on the extension, so only B counts.

Given angle	Condition	Valid triangles
Acute	$a<h$	Zero
Acute	$a=h$	One, with $B=90^\circ$
Acute	$h<a<b$	Two
Acute	$a=b$	One, with the start of the ray excluded
Acute	$a>b$	One
Right or obtuse	$a\le b$	Zero, including the excluded start when $a=b$
Right or obtuse	$a>b$	One

Check your understanding

The known angle $A$ is right, and the two known side lengths satisfy $a=b$ . How many valid triangles have these data?

A Zero B One acute triangle C One right triangle D Two right triangles

Find and check the SSA angles

The circle explains how many solutions to expect. The law of sines gives the angle values.

\sin B=\frac{b\sin A}{a}.

Call the value on the right $q$ . For valid givens it is positive. If $q>1$ , no angle works. If $0<q\le1$ , use degree mode to compute $B_1=\arcsin q$ . The inverse sine key gives the angle up to $90^\circ$ . On the unit semicircle, a horizontal line at height $q$ can meet the arc at most twice. The two angles are supplementary, so also try $180^\circ-B_1$ when it differs from $B_1$ . At $q=1$ the line touches the arc at its top. Test $90^\circ$ once.

For each distinct candidate, find the remaining angle and keep the candidate only if that remaining angle is positive.

C=180^\circ-A-B.

A candidate that passes this check really does give a triangle. Construct it from $A$ , $B$ , and the given side $a$ using two angles and a side. The sine law then gives $b=a\sin B/\sin A$ , exactly the supplied length because the candidate was chosen to satisfy the sine equation. Find $c$ using that triangle’s value of $C$ . Keep full precision through the angle tests and side calculation, especially near a boundary.

Worked example 4 Keep both angles when both fit

Problem. Triangle $ABC$ has these measurements.

\begin{gathered} A=34^\circ\\ a=10\text{ cm}\\ b=14\text{ cm} \end{gathered}

Find every possible set of values for $B$ , $C$ , and $c$ . Round angles and lengths to the nearest tenth in degrees and centimeters, respectively.

Solution. The sine equation gives two candidates for $B$ .

\begin{aligned} \sin B&=\frac{14\sin34^\circ}{10}\\ &\approx0.7829. \end{aligned}

Case 1. Use the inverse sine angle. It leaves a positive angle at $C$ .

\begin{aligned} B_1&\approx51.5^\circ.\\ C_1&\approx94.5^\circ.\\ c_1&=\frac{10\sin C_1}{\sin34^\circ}\text{ cm}\\ &\approx17.8\text{ cm}. \end{aligned}

Case 2. Use the supplement of the first angle. It also leaves a positive angle at $C$ .

\begin{aligned} B_2&\approx128.5^\circ.\\ C_2&\approx17.5^\circ.\\ c_2&=\frac{10\sin C_2}{\sin34^\circ}\text{ cm}\\ &\approx5.4\text{ cm}. \end{aligned}

The shorter base belongs to the triangle with the larger angle at $B$ . As a length check, each base is strictly between the difference of the two known sides (4 cm) and their sum (24 cm).

Worked example 5 Reject the supplementary angle

Problem. Triangle $ABC$ has these measurements.

\begin{gathered} A=117^\circ\\ a=13\text{ m}\\ b=9\text{ m} \end{gathered}

Find every possible set of values for $B$ , $C$ , and $c$ . Round angles and lengths to the nearest tenth in degrees and meters, respectively.

Solution. The known angle is obtuse, and its opposite side is longer than $b$ . The circle meets the ray once.

\sin B=\frac{9\sin117^\circ}{13}.

Case 1. The inverse sine angle leaves room for $C$ .

\begin{aligned} B_1&\approx38.1^\circ.\\ C_1&\approx24.9^\circ.\\ c_1&=\frac{13\sin C_1}{\sin117^\circ}\text{ m}\\ &\approx6.1\text{ m}. \end{aligned}

Case 2. The supplement is about $141.9^\circ$ . It gives a negative value for $C$ , so discard it.

Check your understanding

The sine equation for an SSA triangle with $A=57^\circ$ gives two candidates for $B$ , measuring $31^\circ$ and $149^\circ$ . Which can be used?

A Only

149^\circ

B Only

31^\circ

C Both candidates D Neither candidate

The Law of Sines Advanced. This lesson goes beyond core Geometry. You can skip it.

Learning goals

One area connects the opposite pairs

Two angles and a side fix the triangle

Measure from a baseline

Two sides and a nonincluded angle can leave a choice

Compare the circles for a fixed acute angle

Radius 2

Radius 3

Radius 5

Radius 6

Radius 7

Explain the thresholds

Find and check the SSA angles

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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