4.8 - Sine rule, cosine rule and triangle area

The sine rule for lengths (Higher)

A triangle has sides $9\,\mathrm{cm}$ and $12\,\mathrm{cm}$ enclosing $48^\circ$. Can you find its area before finding the third side? Yes: by the end you will have a direct method. First, learn how to work with a triangle that has no right angle.

Use matching letters: side $a$ is opposite angle $A$, side $b$ is opposite $B$, and side $c$ is opposite $C$. For example, if $A=35^\circ$ and $B=70^\circ$, then $C=75^\circ$. The side opposite $B$ must be longer than the side opposite $A$.

[DIAGRAM: asset_slug: m072_triangle_rules; caption: Triangle ABC with sides a b and c opposite their matching angles; description: Side BC is a opposite A, CA is b opposite B, and AB is c opposite C. Angle C is enclosed by sides a and b. The figure is schematic and supplies no measurements.; file: m072_triangle_rules.png; recommended_method: matplotlib]

The sine rule compares a side with the sine of its opposite angle. A known complete pair gives you a starting point for finding another side.

Why does it work? Drop a perpendicular height from $C$ to the line $AB$. The height is both $a\sin B$ and $b\sin A$, so $a\sin B=b\sin A$. Dividing gives the sine-rule equality. For an obtuse triangle, a perpendicular may fall outside the triangle, but the supplementary-angle sine relationship keeps the equality valid.

Change only $B$ from $70^\circ$ to $80^\circ$. Keep $A$ and its opposite side fixed.

Choose only the two pairs you need, rather than substituting everything into one long equation.

Try the same length method with new data.

The sine rule and ambiguous angles (Higher)

Finding a sine-rule angle needs a second decision after the calculator step. From $a,A,b$, rearrange to $\sin B=b\sin A/a$. This rearrangement follows from the supplied Higher sine rule; it is not separately printed.

If the computed sine exceeds $1$, no triangle exists. A value strictly between $0$ and $1$ gives an acute candidate $B_1$ and its obtuse supplement $B_2=180^\circ-B_1$. These have the same sine, but each must leave a positive third angle: $A+B<180^\circ$. Any stated acute or obtuse condition also applies.

[DIAGRAM: asset_slug: m072_ambiguous_case; caption: Two different triangles fit the same two sides and non-included angle; description: Both triangles have A equal to 30 degrees, side a equal to 7 centimetres and side b equal to 10 centimetres. The angle B is acute in one and obtuse in the other. Their bases have different lengths.; file: m072_ambiguous_case.png; recommended_method: matplotlib]

This is the ambiguous case: two sides and a non-included angle may describe two triangles. A schematic's appearance cannot select between them.

Change only side $a$ from $7\,\mathrm{cm}$ to $12\,\mathrm{cm}$. Keep $A=30^\circ$ and $b=10\,\mathrm{cm}$.

The input check can settle a case even earlier. With $A=30^\circ$, $b=10\,\mathrm{cm}$ and $a=4\,\mathrm{cm}$, the sine would be $1.25$, so no triangle exists. If $a=5\,\mathrm{cm}$, the sine is $1$: there is one $90^\circ$ candidate, and $C=60^\circ$. A zero sine would give a zero or straight angle, so it cannot describe a non-degenerate triangle.

Apply both checks to new triangle data. You only need the possible values of $B$.

The cosine rule for sides and angles (Higher)

The sine rule needs a useful opposite pair. If you have two sides and their included angle, or all three sides, the cosine rule gives a route instead. An included angle is the angle between the two sides being used; in the diagram, sides $a,b$ enclose $C$.