Law of sines — proof and applications (ambiguous case): Visual walkthrough

Parent topic: Law of Sines (parent note).

Step 0 — The one word we must define first: "sine"

WHAT. Before any proof, we need the word $sin$ (read "sine"). Take a right triangle — a triangle with one $90°$ corner. Pick one of the other two corners and call its angle $θ$ (the Greek letter "theta", just a name for the angle). Three sides get names relative to $θ$ :

the hypotenuse — the longest side, always the one facing the right angle;
the opposite side — the one facing $θ$ (it does not touch $θ$ );
the adjacent side — the remaining one, which does touch $θ$ .

WHY this tool. We want a single number that captures "how open is this angle?" independent of how big we drew the triangle. If you scale the whole triangle up by $2 \times$ , opposite and hypotenuse both double, so their ratio stays fixed. That scale-free ratio is exactly the number we want.

PICTURE. Look at the figure: the red side is the opposite, the black slanted side is the hypotenuse, and $sin θ$ is just red-length ÷ hypotenuse-length.

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

That single ratio is the only trig fact the whole proof needs. Keep it in your pocket.

Step 1 — Name the triangle we will prove things about

WHAT. Draw any triangle. Label its three corners $A$ , $B$ , $C$ . Label each side with the lowercase letter of the corner facing it:

side $a$ faces corner $A$ ,
side $b$ faces corner $B$ ,
side $c$ faces corner $C$ .

WHY. This "opposite-naming" is the heart of the Law of Sines. The law pairs each side with its facing angle — so we must lock the pairing before we start, or the algebra becomes nonsense.

PICTURE. In the figure, notice the red side $a$ sits directly across from the red corner $A$ . Each side "looks at" its own letter.

Step 2 — Drop an altitude to manufacture right triangles

WHAT. From corner $C$ , drop a straight line perpendicular (at $90°$ ) to the opposite side $A B$ . Call its length $h$ (the altitude, or height). This single line splits our triangle into two right triangles that share the height $h$ .

WHY this tool. Our only trig fact ( $sin = opposite / hypotenuse$ ) lives in right triangles. Our triangle $A B C$ need not have any right angle — so we create two right triangles by cutting with a perpendicular. That is the whole trick.

PICTURE. The red segment is $h$ . On its left is a right triangle containing angle $A$ ; on its right, one containing angle $B$ . Both are genuine right triangles because $h$ meets $A B$ at $90°$ .

Step 3 — Read $h$ two different ways

WHAT. Apply the sine definition in each right triangle.

Left triangle — angle $A$ has opposite side $h$ and hypotenuse $b$ : $ratio for A sin A = \frac{h opposite}{hypotenuse b} ⟹ h = b sin A$

Right triangle — angle $B$ has opposite side $h$ and hypotenuse $a$ : $ratio for B sin B = \frac{h opposite}{hypotenuse a} ⟹ h = a sin B$

WHY. The same physical height $h$ now has two algebraic descriptions. Two names for one thing means we can set them equal — that is where the relationship between the sides pops out.

PICTURE. The figure highlights the two hypotenuses (sides $b$ and $a$ ) in red, each with the shared height $h$ as their common "opposite" side.

Step 4 — Equate and rearrange into the ratio

WHAT. Both expressions equal $h$ , so: $b sin A = a sin B$ Divide both sides by $sin A sin B$ (both are positive for a genuine triangle angle, so this is safe): $\frac{b s i n A}{s i n A s i n B} = \frac{a s i n B}{s i n A s i n B} ⟹ \frac{b}{s i n B} = \frac{a}{s i n A}$

WHY. Dividing by $sin A sin B$ is chosen precisely to cancel one sine on each side, leaving each side paired with its own angle — the exact form the law claims. Repeating the altitude from corner $A$ onto side $B C$ gives $\frac{b}{sin B} = \frac{c}{sin C}$ by the identical argument, chaining all three together.

PICTURE. The figure shows the "swap": from $b sin A = a sin B$ we flip to the tidy equal-ratios form, each colour staying with its partner.

Recall Quick self-check on Steps 2–4

Why can we set $b sin A$ equal to $a sin B$ ? ::: Because both equal the same altitude $h$ dropped from $C$ . Why divide by $sin A sin B$ rather than something else? ::: It cancels one sine per side, leaving each side next to its own facing angle.

Step 5 — Why the ratio equals $2 R$ (enter the circle)

WHAT. Every triangle has exactly one circle passing through all three corners — its circumcircle — with radius $R$ (see Circumcircle and Circumradius R). Draw it. Now draw the diameter starting at corner $B$ ; let it hit the far side of the circle at a new point $D$ . Because it is a diameter, $B D = 2 R$ (diameter = twice the radius).

Two circle facts do the work (both from the Inscribed Angle Theorem):

Angle in a semicircle is $90°$ . Corner $C$ sits on the circle and looks across the diameter $B D$ , so $∠ B C D = 90°$ . Triangle $B C D$ is right-angled at $C$ .
Same chord ⇒ same inscribed angle. Angles $A$ and $D$ both "stand on" the same chord $B C$ , so $∠ D = ∠ A$ .

Now use our one trig fact in right triangle $B C D$ , where angle $D$ faces $B C = a$ (opposite) and has hypotenuse $B D = 2 R$ : $sin D = \frac{a opposite}{hypotenuse 2 R} and since sin D = sin A : sin A = \frac{a}{2 R} ⟹ \frac{a}{s i n A} = 2 R$

WHY this tool. The altitude proof (Steps 2–4) only tells us the three ratios are equal to each other. It never says what number they equal. The circle upgrades "equal" into a concrete measurement: the ratio is literally the diameter of the circle the triangle is inscribed in.

PICTURE. The red diameter $B D$ , the right angle at $C$ , and the equal angles $A$ and $D$ are all marked.

Step 6 — The obtuse case: when the altitude escapes the triangle

WHAT. Suppose corner $A$ is obtuse (bigger than $90°$ ). Drop the altitude from $C$ as before — but now its foot lands outside the segment $A B$ , on the extension of the base. It looks like angle $A$ never appears in the right triangle.

WHY it still works. The right triangle you get contains the exterior angle at $A$ , which measures $180° - A$ . So the sine that appears is $sin (180° - A)$ . But sine has a beautiful symmetry (see Sine of Supplementary Angles): $sin (180° - A) = sin A$ So $h = b sin (180° - A) = b sin A$ — exactly the same equation as before. Every step of the proof survives untouched. The Law of Sines holds for all triangles: acute, right, and obtuse.

PICTURE. The foot lands left of $A$ ; the marked angle is $180° - A$ , yet its sine equals that of the interior angle $A$ .

Step 7 — The swinging-hinge picture (the ambiguous SSA case)

WHAT. Suppose you are told angle $A$ , the side $a$ facing it, and one more side $b$ — but not the angle between them (this data pattern is called SSA). Fix angle $A$ and side $b$ (from $A$ to $C$ ). The last side $a$ hangs from corner $C$ like a pendulum; you swing it until it touches the base line. Where it touches decides how many triangles exist.

The shortest possible reach from $C$ straight down to the base is the altitude $h = b sin A$ (the perpendicular is always the shortest path). Compare $a$ to that:

If $a$ is...	Picture	Triangles
shorter than $h$	can't reach the base	$0$
exactly $h$	just grazes it perpendicularly	$1$ (right)
between $h$ and $b$	reaches across, cutting the base at two spots	$2$
at least $b$	swings past, one valid cut only	$1$

WHY the two-triangle case. When $a$ reaches across, it crosses the base at two points, giving angles $B$ and $180° - B$ — both real, because $sin B = sin (180° - B)$ . Keep the second only if the angles still sum under $180°$ .

PICTURE. The red arc is side $a$ swung from $C$ ; the dashed grey circle shows all its reach. You can see it slice the base in two places when $h < a < b$ .

The one-picture summary

WHAT. This final figure compresses the whole journey: the triangle inside its circumcircle, the altitude that made two right triangles, the diameter that revealed $2 R$ , and the equal-ratio conclusion — all in one frame.

Recall Feynman retelling — the whole walkthrough in plain words

Draw any triangle and the circle that hugs all three corners. To connect a side to its opposite corner, slice the triangle with a straight drop from the top corner — a perpendicular. That slice makes two right triangles, and in a right triangle "sine" is just the far side over the slanted side. The perpendicular's height can be read off from either half, and setting those two readings equal forces side $a$ over $sin A$ to equal side $b$ over $sin B$ . So the three ratios are equal — but equal to what? Draw a straight line through one corner all the way across the circle (a diameter). A corner looking across a diameter always sees a right angle, and corners looking at the same chord see the same angle. Those two circle-facts turn the mystery ratio into an honest length: the circle's diameter, $2 R$ . If a corner is fat (obtuse) and the perpendicular slips outside, don't panic — the outside angle is the supplement, and sine treats an angle and its supplement identically, so the equation is unchanged. Finally, if someone gives you two sticks and one angle but won't say how they connect, the loose stick swings like a hinge and may hit the base twice, once, or never — that's the famous ambiguous case, and it all comes from that same "two angles share one sine" fact.

Connections

Inscribed Angle Theorem — supplies both circle-facts in Step 5.
Circumcircle and Circumradius R — defines the $R$ that becomes $2 R$ .
Sine of Supplementary Angles — the identity behind Steps 6 and 7.
Law of Cosines — the tool to reach for in SAS/SSS instead.
Area of a Triangle — $Area = \frac{ab c}{4 R}$ ties back to the same $R$ .
Solving Oblique Triangles — the master decision flowchart.