Non-Right-Angled Triangles

Two rules, one triangle, and the labels that make them work

Everything in this topic happens in a triangle with no right angle in it. Pythagoras is no use, SOHCAHTOA is no use, and the two rules that replace them - the sine rule and the cosine rule - both depend on the triangle being labelled the Cambridge way: side $a$ is opposite angle $A$, side $b$ is opposite angle $B$, and side $c$ is opposite angle $C$. Label the triangle before you write a formula, because nearly every mistake in this topic starts with a side paired with the wrong angle.

The sine rule links a side to the angle opposite it. The cosine rule links all three sides to one angle. Between them they solve any triangle, as long as you know three of its six measurements and at least one of the three is a side.

Both rules, and the area formula, are printed in the List of formulas at the front of the question paper. You are not expected to recall them from memory, but you are expected to copy them accurately - a mis-copied $2bc\cos A$ loses every mark in the question.

The sine rule: finding a side

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

Use it when you have a matching pair: one side together with the angle opposite it, plus one more side or angle. That is the whole test for choosing it. In the triangle above, $BC=9.4$ cm is the side opposite angle A, so it is $a$, and the side you are asked for, $AC$, is opposite angle B, so it is $b$.

In triangle ABC, angle A = 42 degrees, angle B = 65 degrees and BC = 9.4 cm. Then $\frac{b}{\sin {65}^{\circ }}=\frac{9.4}{\sin {42}^{\circ }}$, so $b=\frac{9.4\times \sin {65}^{\circ }}{\sin {42}^{\circ }}=12.7$ cm.

Look at the shape of that calculation, because it is the same every time: multiply by the sine of the angle you are heading for, then divide by the sine of the angle you already have. The side you know goes on top.

• Find the third angle first: $C={180}^{\circ }-{42}^{\circ }-{65}^{\circ }={73}^{\circ }$
• Then the third side: $c=\frac{9.4\times \sin {73}^{\circ }}{\sin {42}^{\circ }}=13.4$ cm

The sine rule: finding an angle

Turn the rule upside down and it finds angles instead: $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$. The sides go on the bottom. Writing it this way saves a rearrangement and removes the commonest error in the topic.

In triangle ABC, BC = 8.6 cm, AB = 12.4 cm and angle A = 35 degrees. Side $a$ is BC = 8.6 cm, opposite A, and side $c$ is AB = 12.4 cm, opposite C. So $\frac{\sin C}{12.4}=\frac{\sin {35}^{\circ }}{8.6}$, giving $\sin C=\frac{12.4\sin {35}^{\circ }}{8.6}=0.8270$, and $C={55.8}^{\circ }$.

That last step is the inverse sine, ${\sin }^{-1}$ on the calculator, usually shift then sin. Angles are written correct to one decimal place unless the question says otherwise.

The cosine rule: finding a side

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

Use it when the three things you know do not include a matching pair: two sides and the angle between them. That angle is called the included angle, and without it the sine rule has nothing to start from.

In triangle ABC, AB = 10 cm, AC = 7 cm and angle A = 62 degrees. The side BC is opposite angle A, so it is $a$: $a^2=7^2+{10}^2-2\times 7\times 10\times \cos {62}^{\circ }=149-65.73=83.27$, so $a=\sqrt{83.27}=9.13$ cm.

The minus sign is doing real work. If the included angle is obtuse then $\cos A$ is negative, so the last term adds on instead of subtracting, and the third side comes out longer than either of the sides you started with.

• Substitute into the printed formula rather than a rearranged version of it - the rearrangement is where sign errors live
• Square root at the end: $a^2=83.27$ is not the answer, $a=9.13$ cm is

The cosine rule: finding an angle

Rearranged, the same rule gives an angle: $\cos A=\frac{b^2+c^2-a^2}{2bc}$, where $a$ is the side opposite the angle you want.

Use this form when all three sides are known and no angle is. Then $A={\cos }^{-1}$ of the fraction your calculator produces.

A triangle has sides 5 cm, 7 cm and 9 cm. Its largest angle is opposite its longest side, so $a=9$, $b=5$ and $c=7$: $\cos A=\frac{5^2+7^2-9^2}{2\times 5\times 7}=\frac{-7}{70}=-0.1$, so $A={95.7}^{\circ }$.

A negative cosine is the signal that the angle is obtuse, which is exactly what you should expect opposite the longest side of a triangle whose sides are as different as 5, 7 and 9. This is also the only reliable way to find an obtuse angle: the inverse sine would have given you 84.3 degrees and hidden the real answer.

• Put the side opposite the angle you want first in the numerator, on its own, with a minus sign in front of it
• Denominator: twice the product of the other two sides
• If $\cos A$ comes out positive, the angle is acute, so check you have not put the wrong side in the numerator

The area of a triangle: $\frac{1}{2}ab\sin C$

The area of a triangle is half the product of two sides multiplied by the sine of the angle between them: $A=\frac{1}{2}ab\sin C$. The C in the formula is a warning label - the angle must be the one included between the two sides you use.

A triangle has sides 6.5 cm and 9.2 cm with an included angle of 48 degrees. The area is $\frac{1}{2}\times 6.5\times 9.2\times \sin {48}^{\circ }=29.9\times 0.7431=22.2$ cm2.

It works with an obtuse included angle, because the sine of an obtuse angle is still positive, and it is often the quickest route to a perpendicular height when a later part of the question needs one.

• $\sin {90}^{\circ }=1$, so for a right-angled triangle the formula collapses to $\frac{1}{2}ab$ - a useful sanity check on the method
• If the angle you have is not between the two sides you have, find the third side with the cosine rule first, or find the missing angle

Choosing the rule, and the ambiguous case

Two decisions, in this order. First: is there a right angle? If there is, Pythagoras or SOHCAHTOA is quicker and you should use it. If there is not, look for a matching side-and-opposite-angle pair. With one, the sine rule will start; without one, the cosine rule is the only way in.

The ambiguous case. When you are given two sides and an angle that is not between them, the sine rule may return two different triangles that both fit. $\sin B=0.7143$ has two solutions between 0 and 180 degrees, $B={45.6}^{\circ }$ and $B={180}^{\circ }-{45.6}^{\circ }={134.4}^{\circ }$, and both of them can be the angle of a triangle built from the information given.

In triangle ABC, BC = 7 cm, AC = 10 cm and angle A = 30 degrees. Then $\frac{\sin B}{10}=\frac{\sin {30}^{\circ }}{7}$, so $\sin B=0.7143$, and $B={45.6}^{\circ }$ or ${134.4}^{\circ }$. Both triangles exist: with B = 45.6 degrees the third angle is 104.4 degrees, and with B = 134.4 degrees it is 15.6 degrees.

The question decides whether you need both. 'Find the two possible values' or 'find all possible values' means both. 'The obtuse angle' or 'one possible value' means one. Always check the third angle: if 180 degrees minus your two known angles comes out zero or negative, that triangle does not exist and that solution has to be discarded.