SOLVETUTORMATH SOLVER

Instrument MI-01-641 · Mathematics

Triangle Side Calculator

Two known angles plus the length trapped between them are enough to close a triangle completely; this sheet returns the missing third angle and both remaining sides from that trio.

Instrument MI-01-641
Sheet 1 OF 1
Rev A
Verified
Type 05 — Trigonometry SER. 2026-01641

Side a (opposite A)

8.15207469

C = 180° − A − B

70.00000000 Angle C (deg)
9.21604985 Side b (opposite B)
The working Every figure verified twice
  1. angleC = π − 0.872665 − 1.047198 = 1.22173048
  2. sideA = 10·sin(0.872665) ⁄ sin(π − 0.872665 − 1.047198) = 8.15207469
  3. sideB = 10·sin(1.047198) ⁄ sin(π − 0.872665 − 1.047198) = 9.21604985
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

Two angles and the length of the side pinned directly between them make for a different kind of triangle recipe than the ones already live on this site. The side-side-side sheet starts from three lengths and works backward to every corner, and the side-angle-side sheet starts from two lengths squeezed around one angle; here, neither known measurement is a second side at all. Both are angles, and the single length supplied is the one stretched directly between them, never one of the triangle's other two sides.

Once two angles are fixed, the third one is free: a flat triangle's three corners always add to 180°, so subtracting the given pair from that total hands over the last angle before any length is even touched. What remains is purely a matter of scale, and the sandwiched side supplies exactly that. Feed it into the Law of Sines ratio, sinA ⁄ a = sinB ⁄ b = sinC ⁄ c, and the two still-missing lengths fall out of one division apiece — no second formula, no guessing which of two candidate shapes fits.

Push the two known angles close enough together and the third one shrinks toward nothing — a limit this sheet checks for directly, since the pair has to sum to under a full 180° or no triangle survives at all. Near that ceiling the two recovered lengths swing toward extreme, elongated values, a reminder that this configuration answers to the same triangle limits as any other, even though its two starting knowns happen to be angles rather than sides.

C=180ABC = 180^{\circ} - A - Ba=csinAsinCa = \dfrac{c \sin A}{\sin C}b=csinBsinCb = \dfrac{c \sin B}{\sin C}
A, B — the two known angles entered above; C — the third angle, found by subtraction; c — the given side lying between A and B; a, b — the two sides this sheet solves for, opposite A and B in turn.
  • Enter your first known angle into Angle A, choosing degrees, radians, or turns from its unit menu.
  • Type the length of the side trapped directly between your two angles into Included side c (between A and B).
  • Enter the second known angle into Angle B — together, A and B must stay under 180°.
  • Read Angle C for the recovered third angle, computed first from the other two by subtraction.
  • Read Side a (opposite A) and Side b (opposite B) for the two lengths the Law of Sines returns.

Worked example — 50°, a 10-unit included side, 60°

Set Angle A to 50°, Included side c (between A and B) to 10 units, and Angle B to 60°. Angle C falls out first, by subtraction alone: 180° minus 50° minus 60° leaves 70°, the only third angle any triangle carrying this A and B could ever have, settled before either remaining side is touched.

Side a follows from a = c·sinA ⁄ sinC = 10 × sin(50°) ⁄ sin(70°) = 10 × 0.76604444 ⁄ 0.93969262 ≈ 8.15207469 units, matching what Side a (opposite A) shows for these exact inputs. Side b uses the equivalent ratio for B: b = 10 × sin(60°) ⁄ sin(70°) = 10 × 0.86602540 ⁄ 0.93969262 ≈ 9.21604985 units, the figure Side b (opposite B) displays once Angle C and Side a are already settled.

Questions

What makes this ASA case different from the SAS and SSS pages on this site?

This one starts from two angles and the side sandwiched between them, never a second side — the SAS page starts from two sides plus the angle between them, and the SSS page starts from three sides with no angle given at all. All three are classic triangle-congruence conditions, but each hands the solver a different pair of known quantities, so the arithmetic path to the missing pieces differs even though every one lands on a single, uniquely determined triangle.

Why must the known side lie between the two known angles?

Because that position is what pins the triangle to one exact shape and size. Two angles alone fix a shape but not a scale — any similarly proportioned copy, larger or smaller, shares those same two angles — and the side sandwiched between them sets that missing scale. A length measured outside the two angles belongs to a related but distinct case, and needs the third angle found first before the Law of Sines can be applied to it.

How is the third angle, C, actually found?

By plain subtraction. Every flat triangle's three angles sum to exactly 180°, so C works out to 180° minus A minus B, with no trigonometry needed for that step at all. Only after C is known does this sheet turn to the Law of Sines to recover the two remaining side lengths from the included side you supplied.

What happens if Angle A and Angle B already add to 180° or more?

No triangle exists, and the sheet flags the input rather than returning a number. A flat or negative third angle means the two entered angles have already used up the entire straight angle available to any plane triangle, so their sum has to land strictly under 180° before the two remaining sides mean anything at all.

Does the unit used for the included side matter?

Only in that it must stay consistent. The Law of Sines is a pure ratio, so an included side entered in centimetres returns both remaining sides in centimetres, and the same measurement entered in feet returns feet — mixing units partway through the calculation is the one reliable way to turn a correct formula into a wrong answer.

References