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The ambiguous case of the sine rule

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Acute angle opposite side a
\(\displaystyle A=30^{\circ}\)
Side opposite angle A
\(\displaystyle a=BC=10\ \mathrm{cm}\)
Side opposite angle C
\(\displaystyle c=AB=14\ \mathrm{cm}\)

Two sides and a non-included angle are given. What is the first decision?

\[c\sin A=7<a=10<c=14\]
Year 11, VCE Units 1 & 2, Chapter 11, 3 difficulty levels, Calculator allowedBuild a worksheet

The ambiguous case

This example has an acute given angle and two known sides, with each side paired to its opposite angle

\[\begin{gathered}A=30^{\circ}\\a=10\ \mathrm{cm},\quad c=14\ \mathrm{cm}\end{gathered}\]
  1. Match opposite pairs Each side stays paired with the angle opposite it \(\displaystyle \frac{a}{\sin A}=\frac{c}{\sin C}\)
  2. Check the minimum for an acute given angle Here the opposite side is longer than the perpendicular height but shorter than the other given side, so two triangles fit \(\displaystyle h=14\sin30^{\circ}=7,\qquad 7<10<14\)
  3. Find the acute value Substitute the same triangle data and use inverse sine in degree mode \(\displaystyle C_{\mathrm{acute}}=\sin^{-1}\left(\frac{14\sin30^{\circ}}{10}\right)\approx44.4^{\circ}\)
  4. Take the supplement Subtract the unrounded acute value from a straight angle, then round the final answer \(\displaystyle C_{\mathrm{obtuse}}=180^{\circ}-\sin^{-1}(0.7)\approx135.6^{\circ}\)

Worked example

In a non-right-angled triangle, side a is opposite angle A and side c is opposite angle C. Angle A is not between the two given sides, and the data admits two triangles. Find both possible values of C to 1 decimal place

\[\begin{gathered}A=28^{\circ}\\a=16\ \mathrm{cm},\quad c=24\ \mathrm{cm}\\\frac{\sin C}{24}=\frac{\sin28^{\circ}}{16}\end{gathered}\]
  1. \(\displaystyle 24\sin28^{\circ}\approx11.3<16<24\)The side a is above the sine-rule minimum and below c, so two triangles are possible
  2. \(\displaystyle \sin C=\frac{24\sin28^{\circ}}{16}\approx0.7042\)Use the side-opposite-angle pairs a with A and c with C
  3. \(\displaystyle C_{\mathrm{acute}}=\sin^{-1}\left(\frac{24\sin28^{\circ}}{16}\right)\approx44.8^{\circ}\)Inverse sine gives the acute possible angle
  4. \(\displaystyle C_{\mathrm{obtuse}}=180^{\circ}-\sin^{-1}\left(\frac{24\sin28^{\circ}}{16}\right)\approx135.2^{\circ}\)The second possible angle is the supplement of the acute value

Common mistakes

Omitting the obtuse solution
✕\(\displaystyle C=\sin^{-1}(0.7)=44.4^{\circ}\)✓\(\displaystyle C_{\mathrm{acute}}\approx44.4^{\circ},\quad C_{\mathrm{obtuse}}=180^{\circ}-\sin^{-1}(0.7)\approx135.6^{\circ}\)
The non-included data can form a second triangle, so the supplementary obtuse value must also be reported
Taking the supplement of the given angle
✕\(\displaystyle C_{\mathrm{obtuse}}=180^{\circ}-28^{\circ}=152^{\circ}\)✓\(\displaystyle C_{\mathrm{obtuse}}=180^{\circ}-\sin^{-1}\left(\frac{24\sin28^{\circ}}{16}\right)\approx135.2^{\circ}\)
The supplement is taken from the acute value found by inverse sine, not from the given angle
Mismatching a side and angle
✕\(\displaystyle \sin C=\frac{16\sin28^{\circ}}{24}\)✓\(\displaystyle \sin C=\frac{24\sin28^{\circ}}{16}\)
Each side must stay paired with its opposite angle in the sine rule

Difficulty levels

standardTwo possible triangles are stated and both labelled angle values are found
challengeThe triangle count is classified across no, one and two cases before a follow-on calculation
examA multiple-choice item asks for the obtuse value and includes acute-only and supplement errors

Questions about this topic

Why can the ambiguous case have two triangles?+

For an acute given angle, the other endpoint can occupy two positions when the opposite side is longer than the perpendicular height but shorter than the other known side

Why are the two possible angles supplementary?+

Both angles have the same sine, and angles in a triangle lie between 0 and 180 degrees. The second angle is therefore 180 degrees minus the acute angle

How does the minimum classify the cases for an acute given angle?+

Below the perpendicular height there is no triangle; at exactly that height there is one right triangle; between the height and the other known side there are two triangles; at least as long as the other known side there is one triangle

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, General Mathematics, chapter 11: Trigonometry

In this chapter

Name sides and state trigonometric ratios11A Trigonometric ratios: Evaluate a trigonometric ratio, and find an angle from its valueChoose a trigonometric ratioFind an unknown side of a right-angled triangle11C Finding unknown angles: Evaluate a trigonometric ratio, and find an angle from its valueFind an angle of a right-angled triangle from two sidesSolve a practical right-angled-triangle problemAngles of elevation and depressionRead and state three-figure bearingsSolve a navigation problem with bearingsUse the sine rule in a non-right-angled triangleThe ambiguous case of the sine ruleUse the cosine rule in a non-right-angled triangle11H Cosine rule: Choose the rule a triangle problem needs11H Cosine rule: Solve a practical triangle problem end to end11I Triangle area: Choose the rule a triangle problem needsFind the area of a triangle11I Triangle area: Solve a practical triangle problem end to end