a = 7, b = 10, A = 40°. Three facts, and two different triangles fit them: the third side is either 10.43 or 4.89. At Standard Level this case was excluded. At Higher Level it is the sub-topic.
On a circle of radius 1, the two coordinates of a point are the cosine and the sine of its angle.
Draw a circle of radius 1 at the origin and a radius at angle θ from the positive x-axis. The point where it lands has coordinates
(cosθ, sinθ)
That is the definition, and three things follow immediately:
The graphs are the circle, unrolled. Walk round the circle and plot the height against the angle and you draw y = sinθ. Plot the sideways position instead and you draw y = cosθ. Every feature of those two curves is a feature of the circle: the maximum of 1 is the top, the zero at 180° is the far side.
Given a = 7, b = 10 and A = 40°, the sine rule gives
sin B = 10 sin 40° / 7 = 0.9183
and here is the trouble: two angles between 0 and 180 have that sine.
Both leave a positive third angle, so both are real triangles:
The two third sides differ by more than a factor of two. This is not a rounding subtlety, it is two different triangles.
It happens when the given angle is acute, you are given two sides, and the angle is not between them, with the side opposite the given angle the shorter one. Here a = 7 is opposite the 40° and b = 10 is longer, so the short side can swing to meet the far side in two places. If the side opposite the given angle were the longer one, there would be only one triangle.
| What you are given | How many triangles |
|---|---|
| Two sides and the angle BETWEEN them | One. Use the cosine rule and nothing is ambiguous. |
| Two angles and any side | One. The third angle is forced. |
| Three sides | One, or none if they cannot close. |
| Two sides and an angle NOT between them, with the opposite side longer | One. |
| Two sides and an ACUTE angle not between them, with the opposite side shorter | Two, if sin B comes out below 1 and both third angles stay positive. |
| Two sides and an OBTUSE angle not between them, with the opposite side shorter | None. With A = 100°, a = 9.9, b = 10, sin B = 0.9948, which is under 1, and yet the third angle comes to −4.13°. Always finish the angle sum. |
What a question expects. If it says "find the two possible values", give both. If it gives you a diagram, read the diagram: an obviously obtuse angle at B settles it. If it gives neither, say that there are two and give both, with a sentence. Choosing one silently loses the mark even when the one you chose is the one they wanted.
The same two-answers problem, in equation form. Solve sinθ = 0.918 for 0 ≤ θ ≤ 180°:
Graphically, draw y = sinθ and y = 0.918 and count the crossings inside the interval. The number of crossings is the number of answers, and that is the reliable method: it cannot miss one the way an inverse key can.
An inverse trigonometric key returns one answer. A graph shows you how many there are, and that difference is the whole sub-topic.
When you may use it. Applications. A calculator is allowed in every paper. The ambiguous case is examined at Higher Level, so it is worth having a method that counts rather than one that guesses.
The mark people lose. Taking the one answer the inverse sine key gives and moving on. It returns the acute solution every time, by design, and in the ambiguous case the obtuse one is just as real. Either graph it and count, or write 180 minus your answer down beside it and then check whether the third angle stays positive. And on the Casio, G-SOLVE is SHIFT F5, not F5 alone.
1. With a = 7, b = 10, A = 40°, find sin B to 3 decimal places.
2. Give the obtuse value of B, to 1 decimal place.
3. For that obtuse triangle, find the third side c, to 2 decimal places.
4. What is cos²40° + sin²40°?
5. Why does the inverse sine key only ever give one of the two answers?
1 markThe sine rule, set up correctly.
1 markThe acute angle.
1 markThe obtuse one, and a check that the triangle still closes.
1 markBoth third sides, or the one the diagram selects, with a reason.
The third mark is the sub-topic. A question that says "find the two possible triangles" is telling you the answer has two parts.
These pages are free and stay free, but they are general and your IA is not. Send me your research question, or whatever exists so far, and I will tell you in writing whether the topic has a ceiling on it, where the marks are going, and what to change first. That costs nothing and it comes back within 24 hours.
Written by a serving IB Diploma and Career-related Programme Coordinator and Head of Mathematics, who reads internal assessments across every subject group every year. If you then want the whole draft reviewed properly against all five criteria, that is the paid one, and it is refunded if it does not name at least three specific things to fix.
Already have a full draft? Have the whole thing reviewed against all five criteria, $99.