The Sine and Cosine Rule Worksheets, Questions and Revision

With this triangle in mind, the sine rule is

\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}

The cosine rule is

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

In this topic, we’ll go through examples of how to use both of these rules.

Example: Use the cosine rule to find the side-length marked x to 1dp

To answer this question, we have to match up the information in the question to the letters in the formula.

You want to make a equal to the thing you’re looking for, and then it doesn’t matter which way round you assign the other two sides, let’s say b=5 and c=7.

The only remaining bit of information is the angle, and looking at the cosine rule formula, we see that this must be the opposite angle to whichever side a is.

Since a=x, we know this is the case, so we have that A=44\degree (if this isn’t the case, then you’re not in a position to use the cosine rule).

So, we’re ready to substitute the values into the formula. Doing so, we get

x^2=5^2+7^2-(2\times5\times7\times\cos(44))=25+49-70\cos(44)

Taking the square root of both sides, and putting it into the calculator, we get

x=\sqrt{25+49-70\cos(44)}=4.9\text{ (1dp)}.

We don’t need to evaluate the cosine until the final step, it’d only bog us down with long decimals.

Example: Use the sine rule to find the side-length marked x to 3sf.

Like with the cosine rule, we need to match up the letters in the formula with the information in the question. Again, let a=x, the thing that we’re looking for. Then, the angle opposite it must be A, so A=21\degree.

Then, the remaining information should be a pair: a side and an angle opposite to it. Respectively, they will be b and B, so we have b=23 and B=35\degree (if they aren’t an opposite pair, then you’re not in a position to use the sine rule).

Now, we’re ready to substitute the values into the formula. Doing so, we get

\dfrac{x}{\sin(21)}=\dfrac{23}{\sin(35)}

Multiply both side by \sin(21) to get

x=\dfrac{23}{\sin(35)}\times\sin(21)

Putting this into a calculator, we get x=14.4 to 3sf. Like before, no need to evaluate the sine functions until the final step. Note: we only ever use 2 parts of the sine rule at a time.

In both of these examples, we found side-lengths. If you want to use the cosine rule to find a missing angle, you can rearrange it after subbing the numbers in, or you can use the rearranged formula

\cos A=\dfrac{b^2+c^2-a^2}{2bc},

and then take \cos^{-1} of the whole thing once it’s in your calculator. To use the sine rule to find a missing angle, just flip the whole thing over. However, when finding an angle, we must beware the ambiguous case of the sine rule. We’ll see here what that means, and how to deal with it.

Example: Use the sine rule to find the obtuse angle marked x to 2sf.

So, the flipped-over sine rule looks like

\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}

You don’t have to do this, but it makes life easier. Then, as usual, we label the triangle. A should be the thing we want this time, so A=x, and so the side opposite to it is 43m and so we get a=43.

Then, the remaining information is a pair, and we will let b=25 and its opposite angle B=33\degree. Subbing these values into the formula, we get

\dfrac{\sin x}{43}=\dfrac{\sin(33)}{25}

Multiply both sides by 43 to get

\sin x=\dfrac{43\sin(33)}{25}

Then, taking \sin^{-1} of both sides, and putting it into the calculator, we get x=70\degree to 2sf. However, the question asked for an obtuse angle, but we got an acute answer – why? It’s because we can draw two different (but both correct) triangles using the information we were given at the start.

This triangle on the left also has 2 sides of 43m and 25m and an angle of 33\degree, all in the same positions as the original triangle. However, as you can see, it has a different angle, y, and this angle is acute. This is the ambiguous case of the sine rule, and it occurs when you have 2 sides and an angle that doesn’t lie between them. To find the obtuse angle, simply subtract the acute angle from 180, so the answer here is 180-70=110\degree.

Note: if the sum obtuse answer and original angle (here, ) is above 180, then it is not ambiguous. Angles in a triangle cannot go above 180, so the acute answer must be the only correct one.