The ambiguous case

Two possible triangles

In 2 of 20 recent exams
Paper 2, calculator
3 to 5 marks

How it is set in the exam

Paper 2 (calculator) only, 3–5 marks a part, as a Section A question or the start of a Section B one. Rare (2 of 20 exam versions, 3 parts), but it is the trap waiting inside every sine rule for an angle.

Spot it in the exam

Exam phrases to spot (4)

“find the two possible values of angle $ACB$ ”
“there are two possible triangles”
“find the least possible perimeter”
“not drawn to scale”

In disguise (2)

one angle is asked for, with no hint that there are two: the data (the known angle acute and facing the shorter side) is the only signal
an equation for the third side from the cosine rule turns out to have two positive roots

Variants (4)

both angles, from the sine rule
both areas, one for each triangle
the least (or greatest) possible perimeter: choose the triangle with the shorter (or longer) third side
the cosine rule as a quadratic in the third side: two positive roots are the two triangles

Looks like it, but isn't (2)

the known angle is obtuse, or faces the longer of the two known sides, so there is only one triangle: Find a side or an angle: choose the rule
an area is given and the angle between two known sides is asked (two answers, but not from SSA): Area of a triangle

Worked example

Both angles, then both areas

Paper 2 (calculator)

Question

In triangle $ABC$ , $AB = 13$ cm, $BC = 9$ cm and angle $BAC = 32°$ .
(a) Find the two possible values of angle $ACB$ . [4]
(b) Find the two possible areas of triangle $ABC$ . [3]

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Which method?

Asked for	Route
the two angles	the sine rule: $B_1 = \sin^{-1}(\ldots)$ and $B_2 = 180° - B_1$ , each checked with the angle sum
the third side, or a perimeter	the cosine rule with the unknown side in it: a quadratic (solve a polynomial equation)
the two areas	either route, then $\frac{1}{2}ab\sin C$ for each triangle, with its own included angle

Both routes score full marks; the cosine-rule route is quicker when a side is wanted, and it shows the number of triangles at once (the number of positive roots).

Steps

Spot the case:
two sides and an angle that is not between them, the angle acute and facing the shorter side. Check the height, with the test's labels (The ambiguous case): $b\sin A < a < b$ means two triangles.
Sine rule
for the angle that faces the other known side (the longer one; $B$ in the test's labels, angle $ACB$ in Example 3.1): $\sin B = \frac{b\sin A}{a}$ , then $B_1 = \sin^{-1}(\ldots)$ (Angle from a ratio) and $B_2 = 180° - B_1$ .
Check each
with the angle sum: $A + B_2 < 180°$ .
Finish both triangles:
the third angle, then the third side, the area or the perimeter, each from its own triangle.
If only one is wanted
("least perimeter", "the obtuse angle"), choose it and say why.

Common mistakes

Stopping at $\sin^{-1}$
the GDC gives only the acute angle.
Giving $180° - B$ without checking
that it fits with the known angle.
The wrong angle in the area
each triangle's area uses the angle between its own two known sides, and that angle is different in the two triangles.

More mistakes (3)

Rounding $B_1$ before working out $180° - B_1$ ,
which carries the error into both triangles.
The wrong triangle for "least perimeter"
compare the two third sides. The obtuse value of the angle found by the sine rule goes with the smaller triangle.
The wrong mode
in radian mode Example 3.1 gives $\sin A\hat{C}B = 0.797$ and an "angle" of $0.921$ , so both triangles are lost. The angles show °, so use degree mode (check the settings).

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