Higher Level only. Standard Level 3.2 excludes the ambiguous case, and the page there teaches choosing a rule that cannot go wrong. This is the completion of it.
Step to the second view, then the third, and say nothing in between. Side b and the 40° angle do not move. The dashed arc does not move. Only which crossing is solid changes, and the third side goes from 10.43 to 4.89.
The dashed arc is the whole explanation and it is worth naming out loud once: side a is 7 long and it is pinned at one end, so its free end sweeps a circle, and that circle cuts the base line twice. Two crossings, two triangles. No rule to remember.
Then ask the question that does the work: which one is wrong? Let them argue. Neither is. That is the sub-topic.
| Question | Answer |
|---|---|
| 1. sin B, with a = 7, b = 10, A = 40° | 10 sin 40° / 7 = 0.918. |
| 2. The obtuse B | 180 − 66.7 = 113.3°. |
| 3. The third side of that triangle | C = 26.7°, so c = 7 sin 26.7° / sin 40° = 4.89. |
| 4. cos²40° + sin²40° | 1, and 1 at every angle. |
| 5. Why the inverse sine key gives only one | B. An inverse has to return a single value, so it is defined on −90° to 90°. |
The acute triangle, for completeness: B = 66.7°, C = 73.3°, c = 10.43. The two third sides differ by more than a factor of two, which is what makes this worth examining rather than a curiosity.
1 markThe sine rule, set up with each side against its opposite angle.
1 markThe acute angle.
1 markThe obtuse one, with the third angle checked as positive.
1 markBoth third sides, or the one the diagram selects, with a reason given.
The third mark is the whole point. A student who writes 66.7 and stops has not made an arithmetic slip, they have answered a different question, and it is worth telling them so in those words.
| They wrote | What happened |
|---|---|
| 0.45 | On question 1, 7 sin 40° / 10. The sides are the wrong way up. The side opposite B belongs on top. |
| 0.643 | sin 40° itself, unscaled. They have written the rule and not rearranged it. |
| 66.7 for question 2 | The acute value. The acute answer, taken as the only one. They have done the mathematics and taken the calculator at its word. |
| 140° | 180 − 40. Supplement of the wrong angle. Worth catching, because 140 + 40 is already 180, leaving nothing at all for C. |
| 23.3° | 90 − 66.7. They are thinking of complementary angles, which belong to right-angled triangles. |
| 10.43 on question 3 | The acute triangle's third side. They have solved the other case. Right working, wrong triangle. |
| 3.42 | Divided by sin B instead of sin A. Using sin 113.3° in the numerator would have given 10.00, so this one is specifically the wrong denominator. |
| 1.41 on question 4 | cos 40° + sin 40°, unsquared. Also tells you they reached for the calculator on a question that did not need one. |
"So which answer do I write?" Both, unless something chooses. A diagram chooses. A sentence like "angle B is obtuse" chooses. A context sometimes chooses: if the triangle is a plot of land with a stated perimeter, only one fits. If nothing chooses, write both and say that both are valid. That is the full answer, not a hedge.
"How do I know when to look for two?" When you are given two sides and an angle that is not between them. If the side opposite the given angle is the shorter of the two, check for a second solution. If it is the longer, there is only one. The arc picture is the reason: a short arm reaches the line twice, a long one reaches it once.
"Why does the cosine rule not do this?" Because the cosine rule is given the angle BETWEEN the two known sides, which pins the triangle completely. There is nothing left to swing. If a question can be done with the cosine rule, it has no ambiguity in it at all, which is the practical reason to prefer it.
"Can sin B come out above 1?" Yes, and it means no triangle exists. Worth demonstrating once with a = 4, b = 10, A = 40°: sin B = 1.607, and the arc of radius 4 never reaches the base line. Zero solutions, one solution, two solutions, all three happen.
| Stage | What to do |
|---|---|
| Demonstrate | Compute sin⁻¹(Ans) straight off the stored sine and get 66.7. Typing the rounded 0.918 back in gives 66.6, which is worth doing second, on purpose, as the rounding lesson. Then compute sin(113.3) and get 0.918 back. Do those two in that order, on the board. The machine has just confirmed that the answer it refused to give is correct, which is a more convincing argument than any explanation of restricted ranges. |
| Where they stick | Graphing y = sin x against y = 0.918 and finding both crossings. The window is the obstacle: the default x range is in the wrong units or far too wide, so neither crossing is findable. Set x from 0 to 180 explicitly. On the Casio, G-SOLVE is SHIFT F5, and the right arrow steps between intersections rather than needing a new search. |
| The check | After any obtuse solution, add the three angles. 40 + 113.3 + 26.7 = 180. It catches the 140° error instantly, and it is the only check that works on every ambiguous-case question regardless of what is asked. |
Degrees mode throughout this sub-topic. Radians belong to 3.7 and to the sectors; a triangle question written with a degree symbol is a degrees question.
| Step | What |
|---|---|
| 1 | Unit circle view. Define cos and sin as the two coordinates, and get the identity out of Pythagoras in one line. |
| 2 | Set a = 7, b = 10, A = 40° and ask for the third side. Collect answers. Two numbers will come back. |
| 3 | Step the figure through both triangles. Ask which is wrong. |
| 4 | The dashed arc as the explanation: a short arm crosses twice. |
| 5 | sin⁻¹ then sin, on the machine, in that order. |
| 6 | The zero-solution case, a = 4, so they see the third possibility. |
| 7 | The angle-sum check, then the questions. |
Do not say "take the obtuse one when the answer looks wrong". There is no looks-wrong here. Both triangles close, both have positive angles, and 10.43 and 4.89 are both plausible lengths. The decision comes from the question, never from the appearance of the number.
Do not say "the calculator is wrong". It is not. It returns the value its function is defined to return, and the restriction is what makes the inverse a function at all. Telling a class the machine is wrong teaches them to distrust it generally, when what they need is to know exactly which one thing it will not do for them.