Establish and apply the sine rule $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ (including the ambiguous case when finding an angle), the cosine rule $a^2 = b^2 + c^2 - 2bc\cos A$ to find a side and its rearrangement to find an angle, and the area rule $\text{Area} = \dfrac12 ab\sin C$, to solve problems involving non-right-angled triangles

What this dot point is asking

Right-angled triangle trigonometry only works when there is a right angle to anchor the ratios. Most triangles in surveying, navigation and design have no right angle, and this dot point gives you the three tools that handle them: the sine rule, the cosine rule and the area rule. Together they let you find any missing side or angle, and the area, of a triangle from the right starting information, all in degrees. The exam-worthy skill is not memorising three formulas; it is reading the triangle to decide which rule to reach for, and knowing the one place the sine rule can trip you, the ambiguous case, where two different triangles fit the same numbers.

First, the labelling that every one of these rules assumes. Name the three vertices (corners) with capital letters $A$, $B$, $C$, and name each side with the lower-case letter of the angle it sits opposite. So side $a$ is the side across the triangle from angle $A$, side $b$ is opposite angle $B$, and side $c$ is opposite angle $C$. Getting this opposite-side pairing right is the whole game: every rule below pairs a side with the angle facing it.

The decision the whole topic turns on is which rule to use, and it is settled by what you are given:

• Two angles and any side (so you can find the third angle), or two sides and an angle opposite one of them: use the sine rule.
• Two sides and the included angle (the angle between them), and you want the third side: use the cosine rule.
• All three sides, and you want an angle: use the cosine rule rearranged.
• Two sides and the included angle, and you want the area: use the area rule.

The answer

The sine rule and why it is true

The sine rule says that in any triangle the ratio of a side to the sine of its opposite angle is the same for all three sides:

Here is why it holds. Drop a perpendicular of height $h$ from vertex $C$ straight down to the base $AB$. That splits the triangle into two right-angled triangles that share the height $h$. In the left one, $\sin A = \dfrac{h}{b}$, so $h = b\sin A$. In the right one, $\sin B = \dfrac{h}{a}$, so $h = a\sin B$. The same height equals both, so $b\sin A = a\sin B$, which rearranges to $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$. Dropping a perpendicular from a different vertex brings in the third ratio. So the sine rule is just right-angled trigonometry applied twice to a shared height.

You use the rule as a pair of equal fractions, picking the two that involve your knowns and the one unknown. When the unknown is a side, write that side over its angle at the top left and solve. When the unknown is an angle, it is tidier to flip every fraction and use the reciprocal form $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$, so the unknown $\sin$ sits on top.

The cosine rule and why it is true

The cosine rule generalises Pythagoras to triangles with no right angle. For the side $a$ opposite angle $A$,

Compare it with Pythagoras, $a^2 = b^2 + c^2$. The extra term $-2bc\cos A$ is a correction for the angle not being $90\degree$. When $A = 90\degree$, $\cos A = 0$ and the correction vanishes, leaving Pythagoras exactly. When $A$ is acute ($\cos A > 0$) the correction is subtracted, so $a^2 < b^2 + c^2$; when $A$ is obtuse ($\cos A < 0$) the correction adds on, so $a^2 > b^2 + c^2$. That sign sense is a quick reality check on your answer.

To find an angle from the three sides, rearrange the same rule to make the cosine the subject:

There is a hidden bonus here. The cosine of an obtuse angle is negative, so if this formula gives a negative value the angle is obtuse, and if it gives a positive value the angle is acute. Either way the inverse cosine returns exactly one angle between $0\degree$ and $180\degree$, so the cosine rule never has the two-answer problem the sine rule does. That is why, when you have a free choice, the cosine rule is the safer way to find an angle.

The area rule

The familiar area $= \tfrac12 \times \text{base} \times \text{height}$ becomes, once you write the height as $a\sin C$ (from the same dropped perpendicular as before),

In words: the area of a triangle is half the product of two sides times the sine of the angle between them. The angle must be the included angle, the one sitting between the two sides you use; any of the three pairings works as long as you keep that pairing. This is exactly the SAS information, two sides and the angle between them, so whenever a question hands you SAS and asks for area, this is the one line you need.

The ambiguous case: when the sine rule gives two triangles

This is the edge case textbooks flag and exams love. It happens only when you use the sine rule to find an angle, and never when you find a side or use the cosine rule. The reason is that two different angles share the same sine: $\sin\theta = \sin(180\degree - \theta)$. So $\sin B = 0.5$ is satisfied by both $B = 30\degree$ and $B = 150\degree$, and the sine rule cannot tell which one your triangle has.

The geometry behind it is the SSA (side, side, angle) configuration: you know two sides and an angle that is not between them. Picture fixing the known angle at a vertex, drawing one known side along a ray, and then swinging the other known side like a compass to meet the base. If that swung side is short enough, the arc can cross the base in two places, giving two genuinely different triangles, one with an acute second angle and one with an obtuse one. If it is just long enough it touches once (a right angle); if it is too long it crosses once on the far side only.

The practical test is simple. When the sine rule gives you $\sin B = k$ with $0 < k < 1$, there are two candidate angles, the acute $B_1 = \sin^{-1} k$ and the obtuse $B_2 = 180\degree - B_1$. Both are geometrically real only if they still leave room for the third angle, that is, only if the known angle plus the candidate is less than $180\degree$. So you always compute both and then test each against the angle sum: if $A + B_2 \ge 180\degree$ the obtuse option is impossible and you keep just the acute one; if both pass, the question genuinely has two answers and you must give both triangles. A common shortcut: if the side opposite the known angle is the longer of the two given sides, the obtuse option always fails, so there is only one triangle.

How exam questions ask about the sine and cosine rules

The wording tells you which rule the markers expect:

• "Find the length of ..." with a side-and-opposite-angle pair given points to the sine rule for a side. Put the unknown side over its angle at the top left.
• "Find the size of the angle ..." with all three sides given is the cosine rule for an angle (make $\cos$ the subject); with two sides and a non-included angle given it is the sine rule for an angle, so watch for the ambiguous case.
• "Find the third side" with two sides and the angle between them is the cosine rule for a side, the SAS situation.
• "Find the area" with two sides and the included angle is the area rule $\frac12 ab\sin C$. If only some of that is given, find the missing piece first.
• "Show that there are two possible triangles" or "find both values" is an explicit ambiguous-case prompt: give the acute and obtuse angles and carry each through to its own side or area.
• "Correct to the nearest minute" wants an angle in degrees and minutes; "correct to ... decimal places" wants a length. Keep full precision in the calculator and round only the final line.
• A bearings or angle of elevation problem usually hides a non-right triangle; sketch it, mark the known side-angle pairs, then choose the rule as above.