P1 3.1 - The ambiguous sine rule | Edexcel A-Level International Mathematics

Decide when two sides and a non-included angle determine no triangle, one triangle or two triangles. Find both sine-rule candidates, test them geometrically and finish each valid triangle without losing a solution.

Two possible angles

Can two different triangles have $A=30^\circ$ , $a=7\,\mathrm{cm}$ and $b=10\,\mathrm{cm}$ ? They can: a side of fixed length can swing into two positions while the given angle stays fixed.

Recall that $a$ is opposite $A$ , and $b$ is opposite $B$ . These data are two sides and an angle opposite one of them, often called SSA. They differ from two sides and their included angle, which fix a unique triangle.

The sine of a triangle angle is its height ratio. Angles $40^\circ$ and $140^\circ$ , for example, give the same positive height ratio. Thus a calculator's inverse-sine result need not be the only triangle angle.

Supplementary candidates

For $0<k<1$ , if $\sin B=k$ and $0^\circ<B<180^\circ$ , the two candidates are

B_1=\sin^{-1}k,\qquad B_2=180^\circ-B_1.

They are supplementary: they add to $180^\circ$ . If $k=1$ , both expressions give the same $90^\circ$ , so there is only one candidate. If $k>1$ , no triangle angle is possible.

These are candidates, not yet accepted solutions: each must leave room for the other two positive triangle angles. Use degree mode throughout this lesson.

Worked example 1: find both candidates

Question. A triangle is to have $A=30^\circ$ , $a=8\,\mathrm{cm}$ and $b=10\,\mathrm{cm}$ . Find the two candidate values of $B$ before testing the angle sum.

Plan. Use the sine rule to find $\sin B$ , then form the supplementary angle.

\begin{aligned} \sin B&=\frac{10\sin30^\circ}{8}=0.625\\ B_1&=\sin^{-1}(0.625)=38.682187\ldots^\circ\\ B_2&=180^\circ-B_1=141.317812\ldots^\circ \end{aligned}

Answer. The candidates are $38.7^\circ$ and $141^\circ$ to $3$ significant figures; retain the unrounded values for later work.

Check. In degree mode, both $\sin38.682187\ldots^\circ$ and $\sin141.317812\ldots^\circ$ equal $0.625$ .

Change only the opposite side $a$ from $8$ to $6$ . The sine ratio increases, so the acute candidate increases and the obtuse candidate decreases.

Worked example 2: move the candidates closer

Question. A triangle is to have $A=30^\circ$ , $a=6\,\mathrm{cm}$ and $b=10\,\mathrm{cm}$ . Find the candidate values of $B$ .

Plan. Find the new sine ratio, then use the same two-candidate procedure.

\begin{aligned} \sin B&=\frac{10\sin30^\circ}{6}=\frac56\\ B_1&=56.442690\ldots^\circ\\ B_2&=123.557309\ldots^\circ \end{aligned}

Answer. Candidates $56.4^\circ$ and $124^\circ$ , to $3$ significant figures.

Check. Each candidate gives $6\sin B/\sin30^\circ=10$ , restoring the given side $b$ .

The inverse-sine key returns an angle, not the number of triangles. That decision needs geometry.

Testing the angle sum

A candidate angle is valid only if the remaining angle $C=180^\circ-A-B$ is positive.

For example, $A=70^\circ$ and a candidate $B=120^\circ$ already total $190^\circ$ , so they cannot belong to a triangle. Equality is also invalid: $A+B=180^\circ$ leaves $C=0^\circ$ , a flattened shape with no area. If two candidates survive, keep two separate rows of working so their remaining angles and sides are not mixed.

Worked example 3: reject one candidate

Question. Solve triangle $ABC$ given $A=40^\circ$ , $a=10\,\mathrm{cm}$ and $b=8\,\mathrm{cm}$ .

Plan. Find both $B$ candidates, reject any nonpositive remaining angle, then use the sine rule for $c$ .

\begin{aligned} \sin B&=\frac{8\sin40^\circ}{10}=0.514230\ldots\\ B_1&=30.946006\ldots^\circ,\qquad B_2=149.053993\ldots^\circ \end{aligned}

Since $40^\circ+B_2>180^\circ$ , only $B_1$ is valid. Then

\begin{aligned} C&=180^\circ-40^\circ-B_1=109.053993\ldots^\circ\\ c&=\frac{10\sin C}{\sin40^\circ}=14.704878\ldots\,\mathrm{cm}. \end{aligned}

Answer. One triangle: $B=30.9^\circ$ , $C=109^\circ$ , $c=14.7\,\mathrm{cm}$ , to $3$ significant figures.

Check. With the unrounded $c$ , the independent cosine rule gives $\cos C=(10^2+8^2-c^2)/(160)$ , hence $C=109.053993\ldots^\circ$ .

Change only $a$ from $10$ to $7$ . The obtuse candidate becomes small enough to fit, and both triangles must be completed.

Worked example 4: keep both triangles

Question. Solve triangle $ABC$ given $A=40^\circ$ , $a=7\,\mathrm{cm}$ and $b=8\,\mathrm{cm}$ .

Plan. Test each sine-rule angle separately, then pair each accepted $B$ with its own $C$ and $c$ .

\sin B=\frac{8\sin40^\circ}{7}=0.734614\ldots

Both rows below have $A+B<180^\circ$ :

	First triangle	Second triangle
$B$	$47.274644\ldots^\circ$	$132.725355\ldots^\circ$
$C=180^\circ-A-B$	$92.725355\ldots^\circ$	$7.274644\ldots^\circ$
$c=7\sin C/\sin40^\circ$	$10.877749\ldots\,\mathrm{cm}$	$1.378961\ldots\,\mathrm{cm}$

Answer. Two triangles: $(B,C,c)=(47.3^\circ,92.7^\circ,10.9\,\mathrm{cm})$ , or $(133^\circ,7.27^\circ,1.38\,\mathrm{cm})$ , to $3$ significant figures.

Check. The side inequalities give $1<c<15$ ; both values satisfy this. Independently, the cosine rule using $a=7$ , $b=8$ and each unrounded $c$ returns its own listed $C$ .

Rounding the obtuse angle to a whole degree before subtraction can badly affect a small remaining angle. Keep the calculator value until all sides are found.

Zero one or two triangles

The number of solutions can often be predicted from a perpendicular height before any inverse-sine calculation.

Fix an acute angle $A$ , draw side $AC=b$ , and let $B$ lie on the ray forming angle $A$ with $AC$ . The perpendicular distance from $C$ to that ray is $h=b\sin A$ . The unknown point $B$ must be at distance $a$ from $C$ , so it lies where a circle of radius $a$ , centred on $C$ , meets the ray.

[DIAGRAM: asset_name: Two positions for the third vertex; asset_slug: ambiguous-triangles; caption: With A equal to thirty degrees, b equal to ten and a equal to seven, the circle reaches the ray twice.; description: Two triangles share A and C. A ray from A contains two possible vertices B1 and B2, each seven units from C. AC is ten units and makes thirty degrees with the ray. A dashed perpendicular from C to the ray has height five.]
With A equal to thirty degrees, b equal to ten and a equal to seven, the circle reaches the ray twice.

For acute $A$ , the geometry gives the following complete classification. All side lengths are positive.

Comparison	Number of triangles	Reason
$a<h$	None	The circle cannot reach the ray
$a=h$	One	The circle touches the ray; $B=90^\circ$
$h<a<b$	Two	Two intersections lie ahead of $A$
$a\ge b$	One	The second intersection is at $A$ or behind it

When $A\ge90^\circ$ , its opposite side must be the longest: there is one triangle if $a>b$ , and none if $a\le b$ . Do not apply the acute-angle table without first checking $A$ .

Worked example 5: the side cannot reach

Question. Can a triangle have $A=30^\circ$ , $b=10\,\mathrm{cm}$ and $a=4\,\mathrm{cm}$ ?

Plan. Compare the opposite side with the perpendicular distance to the ray.

h=10\sin30^\circ=5\,\mathrm{cm},\qquad a=4<h

Answer. No triangle exists.

Check. The sine rule would require $\sin B=10\sin30^\circ/4=1.25$ , which is impossible for a real angle.

Change only $a$ from $4$ to $5$ . The circle now just touches the ray: two candidates meet at a single right angle.

Worked example 6: the boundary gives one triangle

Question. Can a triangle have $A=30^\circ$ , $b=10\,\mathrm{cm}$ and $a=5\,\mathrm{cm}$ ? If so, find its remaining angles and side.

Plan. Recognise $a=h$ , then complete the right triangle.

\begin{aligned} \sin B&=\frac{10\sin30^\circ}{5}=1\\ B&=90^\circ,\qquad C=60^\circ\\ c&=\frac{5\sin60^\circ}{\sin30^\circ}=5\sqrt3\,\mathrm{cm} \end{aligned}

Answer. Exactly one triangle: $B=90^\circ$ , $C=60^\circ$ , $c=5\sqrt3\,\mathrm{cm}$ .

Check. Pythagoras gives $5^2+(5\sqrt3)^2=100=10^2$ , with $b$ correctly the hypotenuse.

For our opening data, $h=5<7<10$ , so two triangles exist. Their $B$ angles are $45.584691\ldots^\circ$ and $134.415308\ldots^\circ$ ; both leave a positive third angle.

Quick Recap

In an SSA problem, find both supplementary sine candidates.
Retain only candidates leaving a positive third angle; keep each triangle's values together.
A sine ratio greater than $1$ gives no triangle; a ratio of $1$ gives one candidate, not two.
For an acute given angle, compare $a$ with $b\sin A$ and $b$ ; for an obtuse given angle, its opposite side must be longest.

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P1 3.1 - The ambiguous sine rule

Two possible angles

Testing the angle sum

Zero one or two triangles

Quick Recap

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