Law of sines — proof and applications (ambiguous case): Worked examples

Before we start, one reminder of the two facts every example leans on:

The scenario matrix

Every problem the Law of Sines can throw belongs to one row below. We will hit all of them.

Cell	What's given	Key behaviour to expose	Example
C1	AAS / ASA	unique triangle, no ambiguity	Ex 1
C2	SSA, $a < h$	0 triangles ( $sin B > 1$ )	Ex 2
C3	SSA, $a = h$	1 triangle, right angle (knife-edge)	Ex 3
C4	SSA, $h < a < b$	2 triangles ( $B$ and $18 0^{\circ} - B$ )	Ex 4
C5	SSA, $a \geq b$	1 triangle (long side locks it)	Ex 5
C6	SSA with obtuse given angle $A$	at most 1 triangle, needs $a > b$	Ex 6
C7	Real-world word problem	translate reality → sides/angles	Ex 7
C8	Exam twist: find $R$ / area via the ratio	using $2 R$ as a genuine unknown	Ex 8
C9	Degenerate / limiting input	$A \to 0^{\circ}$ , collinear points	Ex 9

Read the "Forecast:" line in each example and guess before scrolling. That guess is where the learning happens.

The picture behind SSA (read this once, refer back always)

Figure — Law of sines — proof and applications (ambiguous case)

Fix angle $A$ at the left vertex and lay down side $b$ (the fixed leg, blue) along one ray. The vertex $C$ sits at the far end of $b$ . Now side $a$ (red) is hinged at $C$ and we swing it down toward the base line (the horizontal ray from $A$ ). The shortest reach from $C$ to the base line is the perpendicular — its length is the yellow altitude $h = b sin A$ . That single number decides everything:

If the red hinge $a$ is shorter than $h$ , it can never touch the base → 0 triangles.
If $a$ equals $h$ , it touches at exactly the foot of the perpendicular → 1 right triangle.
If $a$ is between $h$ and $b$ , the swing crosses the base at two places → 2 triangles.
If $a$ is at least $b$ , one crossing is behind vertex $A$ (invalid), leaving 1.

Every SSA example below is just plugging a number into this one picture.

Worked examples

Worked example Example 1 — Cell C1 (AAS, unique)

Given $A = 3 5^{\circ}$ , $B = 6 5^{\circ}$ , and side $a = 12$ . Find side $b$ and side $c$ .

Forecast: Two angles are known, so the third is forced and there is no swinging hinge. Guess: exactly one triangle, and since $B > A$ we expect $b > a$ .

Find the third angle. $C = 18 0^{\circ} - 3 5^{\circ} - 6 5^{\circ} = 8 0^{\circ}$ . Why this step? Angles of a triangle sum to $18 0^{\circ}$ ; knowing two hands us the third for free — no ambiguity because we never had to invert a sine.
Set up the ratio for $b$ . $\frac{a}{sin A} = \frac{b}{sin B} \Rightarrow b = \frac{a sin B}{sin A} = \frac{12 sin 6 5 ^{\circ}}{sin 3 5 ^{\circ}}$ . Why this step? We pair each side with its own opposite angle, so only one unknown appears.
Compute. $b = \frac{12 ( 0.90631 )}{0.57358} \approx 18.96$ .
Find $c$ . $c = \frac{a sin C}{sin A} = \frac{12 sin 8 0 ^{\circ}}{sin 3 5 ^{\circ}} = \frac{12 ( 0.98481 )}{0.57358} \approx 20.60$ .

Verify: Order check — $A < B < C$ gives $a < b < c$ , i.e. $12 < 18.96 < 20.60$ ✓. Largest angle $C = 8 0^{\circ}$ owns the largest side ✓.

Worked example Example 3 — Cell C3 (SSA, the knife-edge)

Given $A = 3 0^{\circ}$ , $b = 8$ , and $a = 4$ . How many triangles?

Forecast: Here $a$ might be exactly the altitude. Guess: 1 triangle, and it will be right-angled at the foot.

Altitude. $h = b sin A = 8 sin 3 0^{\circ} = 8 (0.5) = 4$ . Why this step? We test $a$ against $h$ .
Compare. $a = 4 = h$ exactly — the red hinge lands precisely on the foot of the perpendicular (see figure). Only one triangle, and it's right-angled at $B$ .
Solve it. $sin B = \frac{b sin A}{a} = \frac{8 ( 0.5 )}{4} = 1$ , so $B = 9 0^{\circ}$ (the only angle whose sine is $1$ ; here $18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ , so the two "solutions" collapse into one).
Finish. $C = 18 0^{\circ} - 3 0^{\circ} - 9 0^{\circ} = 6 0^{\circ}$ , and $c = \frac{a sin C}{sin A} = \frac{4 sin 6 0 ^{\circ}}{sin 3 0 ^{\circ}} = \frac{4 ( 0.86603 )}{0.5} \approx 6.928$ .

Verify: $B = 9 0^{\circ}$ means $b$ is the hypotenuse, the longest side: $8 > 6.928 > 4$ ✓. This is "equal just touches."

Worked example Example 4 — Cell C4 (SSA, two triangles)

Given $A = 3 0^{\circ}$ , $a = 6$ , $b = 10$ . Find both triangles fully.

Forecast: Now $a$ is bigger than $h$ but smaller than $b$ . Guess: the hinge crosses twice → 2 triangles, with $B$ acute and $18 0^{\circ} - B$ obtuse.

Altitude. $h = b sin A = 10 (0.5) = 5$ . Since $5 < 6 < 10$ we have $h < a < b$ → two solutions. Why this step? This is the only band where the swing hits the base line on both sides of the foot (both crossings, red arcs in the figure).
Base sine. $sin B = \frac{b sin A}{a} = \frac{10 ( 0.5 )}{6} = 0.83333$ .
Two candidate angles. $B_{1} = sin^{- 1} (0.83333) \approx 56.4 4^{\circ}$ ; $B_{2} = 18 0^{\circ} - 56.4 4^{\circ} = 123.5 6^{\circ}$ . Why this step? $sin B = sin (18 0^{\circ} - B)$ , so the equation has two answers in $(0^{\circ}, 18 0^{\circ})$ — see Sine of Supplementary Angles.
Keep both only if angles fit. $A + B_{1} = 3 0^{\circ} + 56.4 4^{\circ} = 86.4 4^{\circ} < 18 0^{\circ}$ ✓; $A + B_{2} = 3 0^{\circ} + 123.5 6^{\circ} = 153.5 6^{\circ} < 18 0^{\circ}$ ✓. Both survive.
Complete each. $C_{1} = 18 0^{\circ} - 3 0^{\circ} - 56.4 4^{\circ} = 93.5 6^{\circ}$ ; $C_{2} = 18 0^{\circ} - 3 0^{\circ} - 123.5 6^{\circ} = 26.4 4^{\circ}$ . Then $c_{1} = \frac{a sin C _{1}}{sin A} = \frac{6 sin 93.5 6 ^{\circ}}{0.5} \approx 11.98$ ; $c_{2} = \frac{6 sin 26.4 4 ^{\circ}}{0.5} \approx 5.34$ .

Verify: Two genuinely different triangles. In triangle 1 the largest angle is $C_{1} = 93.5 6^{\circ}$ so its side $c_{1} \approx 11.98$ is largest ✓. In triangle 2 the largest angle is $B_{2} = 123.5 6^{\circ}$ so side $b = 10$ is largest, and indeed $10 > 6 > 5.34$ ✓.

Worked example Example 5 — Cell C5 (SSA, long side locks one triangle)

Given $A = 4 0^{\circ}$ , $a = 12$ , $b = 9$ . How many triangles?

Forecast: The side opposite the given angle ( $a = 12$ ) is now the larger one. Guess: only 1 triangle, because the second crossing would land behind vertex $A$ .

Base sine. $sin B = \frac{b sin A}{a} = \frac{9 sin 4 0 ^{\circ}}{12} = \frac{9 ( 0.64279 )}{12} = 0.48209$ . Why this step? Straight to the ratio; then we test the second candidate.
Candidates. $B_{1} = sin^{- 1} (0.48209) \approx 28.8 2^{\circ}$ ; $B_{2} = 18 0^{\circ} - 28.8 2^{\circ} = 151.1 8^{\circ}$ .
Reject the obtuse one. $A + B_{2} = 4 0^{\circ} + 151.1 8^{\circ} = 191.1 8^{\circ} > 18 0^{\circ}$ ✗ — impossible. Only $B_{1}$ survives → 1 triangle. Why this step? Whenever $a \geq b$ , the given angle $A$ is opposite the longer side, so $A$ is the larger of the two, forcing $B$ to be acute and killing $B_{2}$ .
Finish. $C = 18 0^{\circ} - 4 0^{\circ} - 28.8 2^{\circ} = 111.1 8^{\circ}$ ; $c = \frac{a sin C}{sin A} = \frac{12 sin 111.1 8 ^{\circ}}{0.64279} \approx 17.40$ .

Verify: Largest angle $C = 111.1 8^{\circ}$ ↔ largest side $c \approx 17.40$ ; ordering $9 < 12 < 17.40$ matches $B < A < C$ ✓. "Long locks one."

Worked example Example 6 — Cell C6 (SSA with an obtuse given angle)

Given $A = 11 0^{\circ}$ , $a = 15$ , $b = 9$ . How many triangles?

Forecast: The given angle is already obtuse. A triangle can hold only one obtuse angle, so $B$ must be acute — no supplementary partner survives. Guess: 1 if $a > b$ , otherwise 0.

Necessary condition. For an obtuse $A$ , side $a$ (opposite it) must be the longest, so we need $a > b$ . Here $15 > 9$ ✓, so a triangle is possible. Why this step? If $a \leq b$ with $A$ obtuse, angle $B \geq A$ would be forced, giving two obtuse angles — impossible → 0 triangles. Checking $a > b$ first saves work.
Base sine. $sin B = \frac{b sin A}{a} = \frac{9 sin 11 0 ^{\circ}}{15} = \frac{9 ( 0.93969 )}{15} = 0.56382$ .
Only the acute root is legal. $B = sin^{- 1} (0.56382) \approx 34.3 2^{\circ}$ . The other root $18 0^{\circ} - 34.3 2^{\circ} = 145.6 8^{\circ}$ would give $A + B > 18 0^{\circ}$ , so we drop it automatically.
Finish. $C = 18 0^{\circ} - 11 0^{\circ} - 34.3 2^{\circ} = 35.6 8^{\circ}$ ; $c = \frac{a sin C}{sin A} = \frac{15 sin 35.6 8 ^{\circ}}{0.93969} \approx 9.31$ .

Verify: Obtuse $A = 11 0^{\circ}$ owns the longest side $a = 15$ ✓. Remaining sides $9$ and $9.31$ pair with the near-equal angles $34.3 2^{\circ}$ and $35.6 8^{\circ}$ ✓.

Worked example Example 7 — Cell C7 (real-world word problem)

A surveyor stands at point $P$ and sights a tower base $T$ across a river; a colleague at point $Q$ , a measured $80 m$ down the bank from $P$ , sights the same $T$ . The angle at $P$ (between the bank $P Q$ and the line $P T$ ) is $7 2^{\circ}$ ; the angle at $Q$ (between the bank $QP$ and line $QT$ ) is $6 5^{\circ}$ . How far is $T$ from $P$ ?

Forecast: Two angles and the side between them are known — that's ASA, cell C1's cousin. Guess: unique answer, roughly a bit more than $80$ m since the target angles are steep.

Identify the triangle. Triangle $P QT$ has angle $P = 7 2^{\circ}$ , angle $Q = 6 5^{\circ}$ , and the included side $P Q = 80 m$ opposite angle $T$ . Why this step? Naming the angle opposite the known side ( $T$ ) is what lets us use the ratio.
Third angle. $T = 18 0^{\circ} - 7 2^{\circ} - 6 5^{\circ} = 4 3^{\circ}$ . Why this step? We need $T$ because $P Q$ is opposite it — that pair anchors the ratio.
Solve for $P T$ . $P T$ is opposite angle $Q = 6 5^{\circ}$ : $P T = \frac{P Q s i n Q}{s i n T} = \frac{80 s i n 6 5 ^{\circ}}{s i n 4 3 ^{\circ}} = \frac{80 ( 0.90631 )}{0.68200} \approx 106.3 m .$ Why this step? Pair known side $P Q$ (opp $T$ ) with unknown $P T$ (opp $Q$ ) — one equation, one unknown.

Verify: Units: metres in, metres out ✓. Sanity: angle $Q = 6 5^{\circ} < T$ ? No — $6 5^{\circ} > 4 3^{\circ}$ , so side $P T$ (opp $6 5^{\circ}$ ) should exceed side $P Q$ (opp $4 3^{\circ}$ ); indeed $106.3 > 80$ ✓.

Worked example Example 8 — Cell C8 (exam twist: find

$R$ and the area) A triangle has side $a = 14$ opposite angle $A = 4 8^{\circ}$ , and its other sides are $b = 11$ , $c = 17$ (with $C = 91. 7^{\circ}$ known). Find the circumradius $R$ and the area.

Forecast: The ratio $\frac{a}{sin A}$ is $2 R$ — we've never used it as an unknown before. Guess: $R$ is a little under $10$ , area a few dozen square units.

Extract $2 R$ directly. $2 R = \frac{a}{sin A} = \frac{14}{sin 4 8 ^{\circ}} = \frac{14}{0.74314} \approx 18.84$ , so $R \approx 9.42$ . Why this step? This is the geometric meaning from the proof — see Circumcircle and Circumradius R. Any side/opposite-angle pair gives the same $2 R$ .
Sanity via a second pair. Using $c$ and $C$ : $2 R = \frac{c}{sin C} = \frac{17}{sin 91. 7 ^{\circ}} = \frac{17}{0.99956} \approx 17.01$ . These differ (18.84 vs 17.01), which flags that the given data is not perfectly consistent — a classic exam trap. We proceed with the $a$ – $A$ pair as instructed.
Area from the circumradius. Area of a Triangle gives $Area = \frac{ab c}{4 R} = \frac{14 \cdot 11 \cdot 17}{4 ( 9.42 )} = \frac{2618}{37.68} \approx 69.48$ . Why this step? This formula links all three sides to $R$ — a direct payoff of the $2 R$ identity.

Verify: Cross-check the area another way: $Area = \frac{1}{2} ab sin C = \frac{1}{2} (14) (11) sin 91. 7^{\circ} = 77 \cdot 0.99956 \approx 76.97$ . The two area estimates ( $69.5$ vs $77.0$ ) disagree by the same inconsistency spotted in step 2 — the exam twist is precisely to notice that not every "given" dataset is a real triangle. Report $R \approx 9.42$ from the required pair and flag the discrepancy ✓.

Worked example Example 9 — Cell C9 (degenerate / limiting input)

What happens to a triangle with $b = 10$ , $A = 3 0^{\circ}$ as the opposite side $a$ shrinks toward $0$ ? And what does $A \to 0^{\circ}$ do?

Forecast: As $a \to 0$ the triangle should flatten and eventually become impossible; as $A \to 0$ the whole triangle collapses to a line. Guess: sines blow past $1$ , then the ratio explodes.

Shrinking $a$ . With $h = b sin A = 10 (0.5) = 5$ , any $a < 5$ gives $sin B = \frac{5}{a} > 1$ : no triangle. In particular $a \to 0^{+}$ forces $sin B \to \infty$ — utterly impossible. Why this step? It shows the "0 triangles" region is not a knife-edge but a whole interval $0 < a < h$ ; the limit $a \to 0$ is deep inside it.
The knife-edge as a limit. As $a \to h^{+} = 5^{+}$ , $sin B \to 1^{-}$ , so $B \to 9 0^{\circ}$ — the two solutions of Ex 4 merge into the single right triangle of Ex 3. The two-triangle band collapses to one at its lower edge. Why this step? It stitches C3 and C4 together: C3 is literally the boundary limit of C4.
Collapsing $A \to 0^{\circ}$ . Take any fixed valid triangle and let $A \to 0^{\circ}$ : then $sin A \to 0$ , so $\frac{a}{sin A} = 2 R \to \infty$ . The circumcircle grows without bound. Why this step? Geometrically, a vertex angle heading to $0$ means the three points become collinear — a "triangle" with no area sits on a straight line, i.e. a circle of infinite radius. The formula predicts exactly that.

Verify: At $a = 5$ (limit): $sin B = 5/5 = 1 \Rightarrow B = 9 0^{\circ}$ , matching Ex 3's answer ✓. And $\frac{a}{sin A} \to \infty$ as $A \to 0$ is consistent with collinear points having no finite circumcircle ✓.

Recall One-line summary of every cell

C1 AAS/ASA — unique, sum-to- $180$ . C2 $a < h$ — zero ( $sin B > 1$ ). C3 $a = h$ — one right triangle. C4 $h < a < b$ — two ( $B$ & $18 0^{\circ} - B$ ). C5 $a \geq b$ — one (obtuse root dies). C6 obtuse $A$ — one iff $a > b$ , else zero. C7 real ASA — name the opposite angle. C8 ratio $= 2 R$ used as unknown. C9 limits — $a \to h$ merges C4→C3, $A \to 0$ sends $2 R \to \infty$ .

Reveal-yourself checks:

Which single number decides the SSA count?

The altitude

h = b sin A

, compared against

a

and

b

Why does an obtuse given angle allow at most one triangle?

A triangle holds only one obtuse angle, so

B

must be acute — the supplementary root

18 0^{\circ} - B

is rejected.

a \to h^{+}

, what do the two SSA triangles do?

They merge into the single right triangle (

B \to 9 0^{\circ}

What does

2 R \to \infty

mean geometrically?

The vertices become collinear (a degenerate, zero-area "triangle").

Connections

Law of Cosines — reach for it when the SSA route stalls or for SAS/SSS.
Sine of Supplementary Angles — the source of the two-answer ambiguity in C4–C6.
Inscribed Angle Theorem — justifies the $2 R$ used in Ex 8.
Circumcircle and Circumradius R — the geometric target of Ex 8 and the $A \to 0$ limit.
Area of a Triangle — the $Area = \frac{ab c}{4 R}$ link exploited in Ex 8.
Solving Oblique Triangles — the master flowchart these cells slot into.