10D - The sine rule and triangle area

Can the same two side lengths and angle describe two different triangles? You will use opposite side–angle pairs to find lengths, decide how many triangles are possible, and calculate area from two sides and their included angle.

Matching sides and opposite angles

A triangle has angles $30^\circ$ and $45^\circ$ , and the side opposite $30^\circ$ is $6\,\mathrm{cm}$ . How can you find the side opposite $45^\circ$ without a right angle?

In a right triangle, a side divided by the sine of its opposite angle gives the hypotenuse. In any triangle, the corresponding ratios still agree with one another. The common value need not be one of the triangle's sides.

Label a triangle $ABC$ , with side $a=BC$ opposite angle $A$ , $b=CA$ opposite $B$ , and $c=AB$ opposite $C$ . A side's opposite angle is the angle at the vertex the side does not touch.

Sine rule

For a non-degenerate triangle, with $a,b,c>0$ and $0^\circ<A,B,C<180^\circ$ ,

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.

You may also use the reciprocal form, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

Not on the formulae sheet: learn it.

Why do these pairs match? Dropping a perpendicular height gives, for example, $h=b\sin A=a\sin B$ ; dividing gives the equal ratios. This short height argument explains the pairing. You do not need a general proof of the rule.

Use the sine rule when you have a complete opposite pair and another side or angle. If two angles are known, first use $A+B+C=180^\circ$ when necessary. Use degree mode for the examples below.

Worked example 1: an acute opposite angle

Question. In triangle $ABC$ , $A=30^\circ$ , $B=45^\circ$ and $a=6\,\mathrm{cm}$ . Find $b$ exactly.

Plan. Match $a$ with $A$ and $b$ with $B$ , then make $b$ the subject.

\begin{aligned} \frac{b}{\sin45^\circ}&=\frac{6}{\sin30^\circ} &&\text{opposite pairs}\\ b&=\frac{6\sin45^\circ}{\sin30^\circ} &&\text{multiply by }\sin45^\circ\\ &=6\sqrt2 &&\text{use exact values} \end{aligned}

Answer. $b=6\sqrt2\,\mathrm{cm}$ .

Check. $\frac{6\sqrt2}{\sqrt2/2}=12=\frac6{1/2}$ . Also, the larger angle $45^\circ$ has the longer opposite side.

Now keep the known side and angle, but change the other angle from acute to obtuse. The pairing and rearrangement stay the same.

Worked example 2: an obtuse opposite angle

Question. In triangle $ABC$ , $A=30^\circ$ , $B=105^\circ$ and $a=6\,\mathrm{cm}$ . Find $b$ to three significant figures.

Plan. Use the same opposite pairs, evaluating the sine of the obtuse angle directly.

b=\frac{6\sin105^\circ}{\sin30^\circ}=11.591109\ldots

The formula applies because $C=180^\circ-30^\circ-105^\circ=45^\circ>0^\circ$ .

Answer. $b=11.6\,\mathrm{cm}$ to three significant figures.

Check. The unrounded result gives $b/\sin105^\circ=12$ . Angle $B$ is largest, and its opposite side is longer than $a=6$ and $c=6\sqrt2$ .

Write the side–angle matches before substituting; matching a side to an angle it touches gives a different calculation.

The ambiguous case

Two sides and a non-included angle can leave the triangle free to fold into two different shapes.

Suppose $A=30^\circ$ , $b=10\,\mathrm{cm}$ and $a=7\,\mathrm{cm}$ . Fix $AC=b$ , and draw the ray from $A$ making $30^\circ$ with $AC$ . Vertex $B$ must lie on this ray and be $7\,\mathrm{cm}$ from $C$ . A circle centred at $C$ can meet the ray twice.

[DIAGRAM: asset_name: Two possible triangles from the same SSA data; asset_slug: sine-ambiguity; caption: A circle of radius 7 meets the 30 degree ray twice, giving one acute and one obtuse angle at B; description: Exact planar geometry with A at the origin, C at (10,0), and B on the ray at 30 degrees above AC. A circle of radius 7 centred at C meets the ray at B1 and B2. The two triangles share AC=10 and have CB1=CB2=7. A perpendicular from C to the ray has length 5. B1 is the nearer intersection and has an obtuse internal angle; B2 has an acute internal angle.]
A circle of radius 7 meets the 30 degree ray twice, giving one acute and one obtuse angle at B

Ambiguous case

The ambiguous case occurs when two sides and an angle opposite one of them are given, and the data allow two different triangles. It is often called SSA: side, side, angle.

The calculator value $\sin^{-1}k$ , for $0<k<1$ , is acute. But both $B=\sin^{-1}k$ and $B=180^\circ-\sin^{-1}k$ have sine $k$ . Test each candidate: all three angles must be positive and add to $180^\circ$ . An explicitly acute or obtuse angle selects the matching candidate.

Worked example 3: two valid triangles

Question. In triangle $ABC$ , $A=30^\circ$ , $a=7\,\mathrm{cm}$ and $b=10\,\mathrm{cm}$ . Find all possible values of $B$ and $C$ , to one decimal place.

Plan. Find $\sin B$ , consider both angles with that sine, then check the angle sum.

\begin{aligned} \sin B&=\frac{10\sin30^\circ}{7}=\frac57 &&\text{sine rule}\\ B_{\mathrm{acute}}&=45.584691\ldots^\circ &&\text{inverse sine}\\ B_{\mathrm{obtuse}}&=180^\circ-B_{\mathrm{acute}}=134.415308\ldots^\circ &&\text{second candidate} \end{aligned}

For $B_{\mathrm{acute}}$ , $C=180^\circ-30^\circ-B_{\mathrm{acute}}=104.415308\ldots^\circ$ . For $B_{\mathrm{obtuse}}$ , $C=15.584691\ldots^\circ$ . Both are positive.

Answer. $(B,C)=(45.6^\circ,104.4^\circ)$ or $(134.4^\circ,15.6^\circ)$ .

Check. The unrounded pairs each sum with $30^\circ$ to $180^\circ$ , and both candidate values of $B$ have sine $5/7$ . In each triangle $b>a$ , matching $B>A$ .

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10D - The sine rule and triangle area

Matching sides and opposite angles

The ambiguous case

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