3.1.19 · D5Advanced Trigonometry
Question bank — Law of sines — proof and applications (ambiguous case)
Before the traps, four quick pictures re-build every symbol used below on this page, so nothing has to be recalled from a link.
Figure 1: the circumcircle (through corners , , , radius ) versus the incircle (inside, tangent to the sides). The dashed segment is a diameter drawn through corner to the opposite point on the circle, so — this is the construction the "why " proof uses below. When a trap says "", it means the outer circle's full width.
Figure 2: the altitude dropped from corner onto side (the side , opposite ). In the right triangle on the left, , so . Keep this picture in mind for every "critical height" trap.
Figure 3: the swinging-hinge SSA (Side-Side-Angle) picture. Side hangs from like a hinge and sweeps an arc onto the base line. Compare its length to : too short misses (0), just reaches (1 right triangle), medium cuts twice (2), long as or more locks one (1).
Figure 4: why is symmetric about . On the unit circle, the height (the sine) at angle and at is the same — both points sit at the same height, mirrored left–right. So sine climbs from to , peaks at , then falls back on as a mirror image. This is the engine behind both "largest side ↔ largest angle" and the two-answer SSA trap.
Recall Mini-proof recap: why the ratio equals
(so the "why" answers below are self-contained) Look at figure 1. Draw the diameter from corner across to the point on the circle. Two facts finish it:
- Angle in a semicircle is : any angle standing on a diameter is a right angle, so — triangle is right-angled at .
- Inscribed Angle Theorem: angles (at corner ) and (at corner ) both "look at" the same chord , so they are equal, . In the right triangle , . Since , we get , i.e. . That is the whole "" story — no need to leave the page. See Inscribed Angle Theorem and Circumcircle and Circumradius R for more.
True or false — justify
The ratio can be different for the three sides of the same triangle.
False — all three ratios are equal, and each equals , the diameter of the one circumcircle (the outer circle through corners , figure 1) the triangle lives inside.
In a triangle, the largest side is always opposite the largest angle.
True — since , and (figure 4) rises on then falls back symmetrically on because the unit-circle height at equals the height at , a bigger angle still means a bigger within a triangle; the Law of Sines makes this exact.
The Law of Sines fails for obtuse triangles because the altitude falls outside.
False — the foot of (figure 2) landing outside the base only changes the picture, not the algebra: (the mirror-about- symmetry of figure 4), so still holds.
If in an SSA (Side-Side-Angle) problem, then must be .
False — by the mirror symmetry of figure 4, could also be , since . Both must be tested against the angle sum.
The value in the Law of Sines depends on which side you start from.
False — is a single fixed number for the whole triangle (its circumdiameter, figure 1), which is exactly why the three ratios are equal.
An SSA setup with gives exactly one triangle, provided is acute.
True — for acute (figure 3), when the swinging side is at least as long as , the arc meets the base line in only one valid point; the obtuse alternative would break the angle sum. (If is right or obtuse the count is handled separately — see the edge cases.)
is possible if the sides are large enough.
False — the unit-circle height (figure 4) never exceeds , so signals no triangle exists for those SSA data.
An SSA problem can force , giving a genuine triangle with .
False — means , a degenerate "flat" triangle where the arc in figure 3 only grazes the base line at itself; the three corners become collinear, so it is a limiting case, not a real triangle.
You can always use the Law of Sines to solve an SAS (Side-Angle-Side) triangle directly.
False — SAS has no side paired with its known opposite angle, so every Sines ratio has two unknowns; start with the Law of Cosines instead.
The two triangles in the ambiguous case have the same side and side .
True — both share the given , , and angle ; only the third side and angles differ, because and (figure 4) are both admissible.
If , the ambiguous case can still produce two triangles.
False — with obtuse or right, the swinging side (figure 3) must be the longest, so at most one triangle forms; the two-triangle situation needs acute.
Spot the error
"Given , , , then , so ."
Error: is impossible (figure 4 caps the height at ), so you must stop and conclude no triangle — you can't invert a sine value above .
"Since , every SSA problem always has two answers."
Error: the second angle is kept only if . Often it breaks the angle sum and must be rejected, leaving one triangle.
"I used with , , and got ."
Error: the ratio is , so , not — you swapped and in the cross-multiplication.
"The critical altitude is , so compare against it."
Error: the critical height (figure 2, for given ) is — built from the known angle and the fixed side ; we compare the swinging side against it.
" means equals the inradius of the triangle."
Error: is the circumradius (the outer circle through the vertices , figure 1), not the inradius (inner circle tangent to the sides).
" doesn't matter in an AAS problem, so I can't find side ."
Error: is found instantly as ; then gives . AAS is fully determined and unique.
Why questions
Why does the common ratio equal the diameter and not the radius ?
Because the proof (recapped above, figure 1) builds a right triangle on a full diameter of the circumcircle; then , and the diameter — not the radius — is the hypotenuse doing the work. It rests on the Inscribed Angle Theorem.
Why do we drop an altitude to prove the ratio equality?
The altitude (figure 2) splits the triangle into two right triangles, where the single fact applies; both halves share the same , letting us equate two expressions.
Why is SSA (Side-Side-Angle) the only "ambiguous" case among AAS, ASA, SAS, SSS?
Because in SSA the given angle isn't wedged between the two given sides, so one side is free to "swing" like a hinge (figure 3) and can meet the base at two, one, or zero points — the other cases lock the shape uniquely.
Why must we sometimes reject the obtuse solution even though its sine matches?
Because a valid triangle needs with all angles positive; if , no room is left for , so that solution is impossible despite the matching sine (figure 4).
Why does the "short side can't reach" case () give zero triangles?
The swinging side (figure 3) is shorter than the perpendicular distance from to the base line, so its tip never touches the base — no closing side can form.
Why is the Law of Cosines preferred for SSS but the Law of Sines convenient afterward?
SSS has no known angle, so Cosines is needed to extract the first angle; once one angle is known, Sines quickly finds the rest because now a side is paired with its opposite angle. See Solving Oblique Triangles.
Why does the area formula contain the same as the Law of Sines?
Because both describe the same circumcircle (figure 1): substituting (etc.) into area relations reveals as the shared geometric constant. See Area of a Triangle.
Edge cases
What happens to the ambiguous count exactly when ?
The swinging side (figure 3) just reaches the base perpendicularly — exactly one (right) triangle forms, the boundary between "zero" and "two".
What if the given angle is obtuse and ?
No triangle — an obtuse angle must sit opposite the longest side, so if isn't the longest, the data are inconsistent.
What if comes out exactly in an SSA problem?
Then uniquely (its supplement is the same angle, the peak of figure 4), giving exactly one right triangle — no second solution to weigh.
What if comes out exactly ?
Then : the arc in figure 3 grazes the base line right at , so the three corners fall on one line — a degenerate limit, not a genuine triangle.
In a degenerate "triangle" where two vertices coincide, what does the Law of Sines say?
It breaks down: a side is and the opposite angle is (so ), giving — degenerate figures lie outside the law's scope.
As angle with fixed, what does do?
Since and , the circumradius — the circle (figure 1) flattens toward a straight line, matching a triangle collapsing to a segment.
For an equilateral triangle, what is the common ratio in terms of one side ?
, so — every side gives the identical value, confirming full symmetry.
If two different triangles arise in the ambiguous case, are their circumradii equal?
Yes — they share and , so is identical, meaning both sit on the same-size circumcircle, an easy surprise to miss.
Connections
- Law of Cosines — the correct first tool for SAS/SSS traps above.
- Inscribed Angle Theorem — justifies why (not ) appears.
- Circumcircle and Circumradius R — clears the circum- vs in-radius confusion.
- Sine of Supplementary Angles — the root of every SSA two-answer trap.
- Area of a Triangle — the shared- edge case.
- Solving Oblique Triangles — the decision flowchart these traps live inside.