Use the sine rule in a non-right-angled triangle

Every side must match its opposite angle

The sine rule relates each side to its opposite angle. Use a known opposite pair and the desired pair. In a solid, first identify the single plane containing the three triangle vertices; an angle between unrelated edges is not automatically an interior angle of that triangle

A triangle has A=30°, B=90° and a=5 cm. Find b in cm

A triangle has A=40° and B=65°. Find C in degrees

Inverse sine can give two candidates

If sin B=k with 0<k<1, candidates in (0°,180°) are arcsin k and 180°−arcsin k. Check each against A+B<180° and all given information. SSA can produce zero, one or two triangles. A calculator’s principal inverse is not the complete geometric argument

For A=30°, a=6 and b=9, find sin B

Original triangle

\(\displaystyle A=30^\circ,\quad a=6,\quad b=9\)

How many valid triangles satisfy those data?

Use a face in three dimensions

Suppose A,B,S lie on one triangular face of a solid, with ∠ASB=30°, ∠SAB=45° and AB=6 cm. The opposite-pair rule gives SB=6 sin45°/sin30°=6√2 cm. The face may be tilted; its internal triangle angles and distances still obey planar trigonometry

On a triangular face, ∠ASB=30°, ∠SAB=45° and AB=6 cm. Find the coefficient of √2 in SB

Resolve SSA and justify the section, ungraded

Construct both triangles for A=30°, a=6, b=9, label opposite pairs and explain the two possible angles. Separately explain why a sine-rule calculation on a named triangular face is valid but substituting a line–plane angle into it may not be

Model: preserve the full geometry

For A=30°, a=6 and b=9, let β=arcsin(3/4). The candidates B=β and B=180°−β both satisfy A+B<180°, giving C=150°−β or C=β−30°. A face is a plane containing all three vertices; its interior angles belong to that triangle. A line–plane angle uses an orthogonal projection, so it needs a different stated section unless the same angle is proved to occur

Prove the sine rule, ungraded

Let ABC be a nondegenerate triangle with positive side lengths a, b, c opposite angles A, B, C and area K>0. Draw an altitude to the line through a side, extending that side if needed. Use base and perpendicular height to derive 2K=bc sin A and the corresponding expressions at B and C. Divide by abc and justify taking reciprocals to prove the sine rule. Explain why the argument remains valid for an obtuse triangle

Model: one area, three side–angle pairs

In a nondegenerate triangle ABC, let a, b, c>0 be opposite A, B, C and let its area be K>0. Use base AB=c and drop the perpendicular from C to the line AB. The height is h=b sin A, even if the foot is outside AB. For obtuse A, h=b sin(180°−A)=b sin A; for A=90°, h=b. Thus 2K=ch=bc sin A. Repeating with the other two bases gives the same area. Divide these three identities by the positive product abc to obtain the common ratio below. All interior angles lie strictly between 0° and 180°, so every sine and this common ratio are positive. Taking reciprocals gives the sine rule shown below. The proof does not assume an acute triangle