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More trigonometry rules

Trigonometry - sine and cosine rule

We will now move away from using trigonometry only with right-angled triangles and instead look at how we can find unknown values for triangles which can be of any shape and size. So the triangle does not strictly have to contain an angle of 90^\circ. To do this we must label different sides and angles of the triangle.

The following formulae are also given in your exam for you to use:

Triangle trigonometry rules

As you may have noticed, the lower case letters ab and c refer to the different side lengths of the triangle and the upper case A, B and C are the angles with each pair of the same letter being opposite. So the angle B is directly opposite from the side b. All of these letters could be moved around as long as the same letters are always opposite one another. Therefore, this can be used for any triangle.

The sine rule

From the previous page we can see that the sine rule tells us:

\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}

So the relation between each side and its opposite angle is equal to all of the others. This can be a very useful technique as it will allow us to find unknown side lengths and angles. The sine rule can really be split into three different equations:

The Sine rule

Using any one of these three separate rules we can plug in values that we know and then find any unknowns. For example, if we knew the values for a, A and b we could use the first equation and put these values in, rearranging to find the value of the unknown angle B.

Example

Find the unknown side length a.

Sine rule example

Since we have the correct values for one set of opposites (the side and angle) we can use the sine rule. Putting the correct values into \frac{a}{sinA}=\frac{b}{sinB} we find \frac{a}{sin(47)}=\frac{16}{sin(39)}  Since we have only one unknown, this equation can then be rearranged and the trigonometric parts can be found on a calculator and leave us with the answer of a=18.59^\circ

Using the sine rule all that we need to really do is to label on the correct sides and angles and then carefully put these into the equation and rearrange for the unknown. As long as you have the values for one pair of opposites (one angle and its corresponding side length at the other side of the triangle) we can then find other values for a triangle. One of the only ways in which people make mistakes on this type of question is by labelling the triangle incorrectly in the first place. By making sure that the same letter is placed for the angle and opposite side, with the side being in lower case and the angle in upper case, solving is just a case of plugging in values and then rearranging for an unknown.

Example

Find the unknown angle x.

Sine rule example 2

Here we have the angle of 76^\circ as well as its opposite side length 8.4. We also have the side length corresponding to the unknown angle , therefore we can put our values into the equation \frac{a}{sinA}=\frac{b}{sinB}. Labelling the angle x and side length 5.7 as a and A and the angle 76^\circ and side length 8.4 as b and B we get:

\frac{5.7}{sin(x)}=\frac{8.4}{sin(76)}

The right-hand side can then be solved and rearranged to find sin(x)=0.66. Then by taking the inverse of sin to isolate the x term we get:

x=sin^{-1}(0.66)=41.30^\circ

The cosine rule

The cosine rule uses the same labelling of the triangle as we used for the sine rule where opposite pairs of angles and sides are given the same letter but the side is lower case and the angle is in upper case.

a^2=b^2+c^2-2bc \times cosA

When using the sine rule, we have to know both the side length and angle of at least one pair (so we had to have both a and A for example) to be able to work out another unknown. Things are a little different with the cosine rule as we can find an unknown side length without knowing any complete pair of opposites. We do, however, need to know both of the other side lengths and the corresponding angle to the side we wish to work out. This is clear from the formula given above.

The two situations where the cosine is useful:

  • When we know two sides and one angle (but no pairs) and wish to find the side length opposite the known angle.
  • When we know all three side lengths and want to find one of the angles.

As with the previous rule, it helps massively to label each triangle and assign the letters a, b, c, A, B and C to sides and angles. It will also help us to label the single angle that we are working with as A since this will then be the same as the letter in the formula to avoid confusion.

As with the sine rule, you should put a lot of focus on labelling the triangle correctly as this is where the majority of mistakes are made in these questions. Once the triangle has been labelled it is really just a case of plugging different values into the equation and then finding the unknown that we are looking for.

Example

Find the side length x in the following diagram:

Cosine rule example

We begin by labelling the different sides and angles of the triangle. Remember, we said that it is easiest to call the single known angle A as this is what it is in the cosine formula. This will then make x=a. Listing what we know, we find:

LetterabcA
Amount in this casex6cm5.6cm58^\circ

From this we can substitute in the values which we know into the cosine formula:

a^2=b^2+c^2-2bc \times cosA
x^2=6^2+5.6^2-2(6)(5.6) \times cos(58)

Calculating the value of x in this case then gives us the answer:

x=5.63cm

Example

Find the unknown angle marked as x in the following diagram:

Cosine rule example 2

In this example we have all of the side lengths of the triangle but none of the angle sizes. Since we are looking to find the angle x we need to rename this as since the cosine rule has only one angle which is named A. This then makes the side opposite (3.8cm) the side of . The other two sides can then be b and c but which is which is of no relevance.

Substituting these into the cosine equation we find:

a^2=b^2+c^2-2bc \times cosA
3.8^2=5.9^2+6.3^2-2(5.9)(6.3) \times cosA

Now we can calculate the different squares in this equation and rearrange to get cosA as the subject of the formula, isolated to one side. This gives:

cosA=\frac{5.9^2+6.3^2-3.8^2}{2(5.9)(6.3)}=\frac{143}{177}

Then by finding the inverse of cos for each side we get the value for the angle x:

x=cos^{-1}(0.81)=36^\circ

Which rule to use

Now that you have seen both of the sine and cosine rules and how they relate to triangles of any size, we can try to decide which is the best to use in certain situations.

You should look to use the sine rule always; however, if one of the following situations occur, use the cosine rule:

We know all three side lengths and want to know one particular angle.

We know two sides and one angle and want to know the side opposite the known angle.

These two situations are fairly easy to spot (especially the first) and when they occur we should immediately use the cosine rule. Even if you are not very sure about which to use, just label the triangle up correctly (using the A for the unknown angle or a if you are looking for an unknown side) and then try using both formulae. You may not find the unknown immediately if you have used the wrong one for the situation but it cannot hurt to use both if you cannot decide which is needed.

Solving a triangle

If you are asked to ‘solve a triangle’ you are being told to find all of the unknown angles and sides. This is obviously done using all of the techniques which we have already learned in this lesson and ones previous. By using the sine and cosine rules, angles in a triangle and other trigonometry skills we have looked at, you should now be able to do this given quite limited information on the shape to begin with.

There are six different facts that we can know about a triangle: its three side lengths and its three angles. To enable us to calculate all of these we need to be told only three, with at least one being a side length. From this rather limited information we can completely solve a triangle for all of its parts.

Practical applications

In your exam you will probably be asked a question which relates to a real-life situation and requires you to make use of the techniques that we have learnt in this course to solve a triangle. However, this may not always be clear and a small story may be used in order to illustrate the question. These can involve different things like vectors and bearings and you need to be able to spot the pieces of information that are needed.

By drawing a diagram of a worded question you can easily keep tabs on the angles and sides that you know and convert the words to a picture using shapes. We will work through an example of this sort now:

Example

A ship travels on a bearing of 135^\circ for 5 miles and then changes course to travel due south for 3.5 miles. How far is the ship from its initial location?

This will be a lot easier is we take each piece of information and map the ship’s course out with all known angles and side lengths included:

Trigonometry applications

We have been asked to find the distance that the ship is from its start point. This must be some unknown side length across from the 135^\circ angle which we have already added to the diagram. By drawing out the problem we can construct a triangle and label the side which we need to find as x:

Trigonometry applications example

Clearly, we need to make use of the cosine rule here as we have two known sides and an unknown side with the opposite angle known. Labelling the unknown x as aas 135^\circ and b and c as 5 and 3.5 respectively, we can add the value into the formula for the cosine rule:

a^2=b^2+c^2-2bc \times cosA
x^2=5^2+3.5^2-2(5)(3.5) \times cos135

This then gives us a value for x as 7.87 miles

Therefore, we have taken a worded question and found the correct answer simply by drawing out each small piece of information given and then making use of the formulae and techniques which we have learnt in geometry.

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