Non-right-angled triangles

Labelling a triangle Extended

SOH CAH TOA only works in a right-angled triangle. For any other triangle there are three formulas, and they all use the same labelling:

• the vertices (corners) are capital letters $A$, $B$, $C$, and the angle at each vertex is called by the same letter;
• each side is the small letter of the angle opposite it: $a$ is the side opposite angle $A$, which is $BC$; $b$ is opposite $B$; $c$ is opposite $C$.

The formulas are given in the List of formulas on the Extended papers, but they are written for a triangle labelled $ABC$. In a triangle called $PQR$ you must match the letters up yourself: the side opposite angle $P$ is $QR$. Getting a side and its opposite angle paired correctly is the first step of every question.

The sine rule Extended

In any triangle $ABC$,

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$

and, turned upside down for finding angles,

$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$$

Why it is true. Drop a perpendicular of height $h$ from $C$ to the side $AB$. It splits the triangle into two right-angled triangles. In the left one, $h=b\sin A$; in the right one, $h=a\sin B$. Both are the same height, so $b\sin A=a\sin B$, and dividing by $\sin A\sin B$ gives $\frac{a}{\sin A}=\frac{b}{\sin B}$. The same argument from another vertex brings in $c$.

When to use it. The sine rule links two sides and the two angles opposite them. Use it when you know a side and the angle opposite it, plus one more side or angle. If you know two angles, the third is ${180}^{°}$ minus the other two, so you can always find the angle opposite a known side.

To find a side, use the version with sides on top: there is then only one step of rearranging. To find an angle, use the version with sines on top, then ${\sin }^{-1}$.

Obtuse angles and the sine rule Extended

From the graph of $y=\sin x$, an angle and its supplement have the same sine: $\sin ({180}^{°}-x)=\sin x$. For example $\sin {30}^{°}=\sin {150}^{°}=0.5$. So when you find an angle from its sine, there are always two candidates between $0^{°}$ and ${180}^{°}$: the acute one your calculator gives, $\theta$, and the obtuse one, ${180}^{°}-\theta$.

Usually only one of them can be in the triangle. Two quick tests:

• the largest angle of a triangle is opposite the longest side, so an angle opposite a side that is not the longest cannot be obtuse;
• the three angles add up to ${180}^{°}$, so if ${180}^{°}-\theta$ plus the angle you already know is ${180}^{°}$ or more, the obtuse value is impossible.

The cosine rule does not have this problem: every angle from $0^{°}$ to ${180}^{°}$ has a different cosine, and an obtuse angle has a negative cosine. That is why the cosine rule is the safer tool for finding the largest angle of a triangle.

The ambiguous case Extended

Suppose you are told two sides and an angle that is not between them. Sometimes two different triangles fit.

In the diagram, $AB=8$ cm and angle $A={45}^{°}$, and $BC$ must be 6 cm. Swing a 6 cm arm about $B$: it meets the base line from $A$ at two points, $C_1$ and $C_2$. Both triangles $AB{C}_1$ and $AB{C}_2$ have $AB=8$, $BC=6$ and angle $A={45}^{°}$.

The two possible angles at $C$ add up to ${180}^{°}$, because triangle $B{C}_1{C}_2$ is isosceles. That is exactly the pair $\theta$ and ${180}^{°}-\theta$ that the sine rule gives.

When does it happen? When the side opposite the given angle (here $BC=6$) is shorter than the other given side ($AB=8$), but long enough to reach the line. If the side opposite the given angle is the longer one, the arm only reaches the line once on the correct side, and there is one triangle.

In an exam, "find the two possible values" or a diagram showing both positions tells you it is ambiguous. Otherwise, check the obtuse value with the angle sum before you discard it.

The cosine rule Extended

In any triangle $ABC$,

$$a^2=b^2+c^2-2bc\cos A$$

and, rearranged for finding an angle,

$$\cos A=\frac{b^2+c^2-a^2}{2bc}$$

Why it is true. It is Pythagoras' theorem with a correction. If angle $A$ is ${90}^{°}$, $\cos A=0$ and the formula is $a^2=b^2+c^2$. If $A$ is acute, the side $a$ opposite it is shorter than Pythagoras would give, and $-2bc\cos A$ is negative, taking some off. If $A$ is obtuse, $\cos A$ is negative, so $-2bc\cos A$ is positive and $a$ is longer. (The proof drops a perpendicular, as for the sine rule, and uses Pythagoras in both right-angled triangles.)

When to use it.

• To find a side when you know the other two sides and the angle between them (two sides and the included angle).
• To find an angle when you know all three sides.

The side on the left, $a$, is always the one opposite the angle in the cosine. In the angle form, the side opposite the angle you want is the one subtracted.

Calculator order. Work out $b^2+c^2-2bc\cos A$ as one expression, or work out $2bc\cos A$ first and subtract it. A common error is to work out $b^2+c^2-2bc$ and then multiply by $\cos A$. Then remember to square root: the formula gives $a^2$.

Area of a triangle Extended

For any triangle,

$$\text{Area}=\frac{1}{2}ab\sin C$$

where $a$ and $b$ are two sides and $C$ is the angle between them.

Why it is true. Take $a$ as the base. The perpendicular height from the opposite vertex is $b\sin C$ (it is the opposite side in a right-angled triangle with hypotenuse $b$). Then $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}a\times b\sin C$.

Read it as "half the product of two sides times the sine of the angle between them". The letters do not matter: in triangle $PQR$, the area is $\frac{1}{2}\times PQ\times PR\times \sin P$.

The formula also works backwards. If you know the area and two sides, $\sin C=\frac{2\times \text{Area}}{ab}$. Remember that this gives two possible angles, $C$ and ${180}^{°}-C$, as both have the same sine: a triangle with an obtuse angle can have the same area.

On the non-calculator paper, the angle will have an exact sine, such as $\sin {30}^{°}=\sin {150}^{°}=\frac{1}{2}$ or $\sin {60}^{°}=\frac{\sqrt{3}}{2}$.

Choosing the right rule Extended

Before you write anything, list what you know and what you want, and look for a side with its opposite angle.

• A side and its opposite angle are both known: sine rule.
• Two sides and the angle between them, want the third side: cosine rule.
• Three sides, want an angle: cosine rule (angle form). Find the largest angle this way first, as it is the only one that might be obtuse.
• Two sides and the angle between them, want the area: $\frac{1}{2}ab\sin C$.
• The triangle has a right angle: SOH CAH TOA and Pythagoras are quicker.

Longer problems chain these together: the cosine rule gives a side, which then pairs with an angle for the sine rule. Keep unrounded values in your calculator between steps and round only the final answer, or your last figure may be wrong.