Trigonometry & Pythagoras Answers | Higher Tier | MME


HT1 – Pythagoras

Question: Below is a right-angled triangle. Work out the exact length of the side marked x, giving your answer in surd form.

Answer: To do this we will need to use Pythagoras theorem a^2+b^2=c^2

We then add the numbers we know into the equation. This gives us


Next, we need to rearrange to make x the subject.

x^2=13^2 - 9^2

Next if we then calculate the right-hand side:

x^2=169 - 81


Finally, we square root both sides

x=\sqrt{88} or x=2\sqrt{22}

HT2 – Trigonometry

Question: Find the size of the angle marked q to 1 d.p.

Answer: The two sides we’re concerned with are the hypotenuse and the opposite (to the angle) – O and H. Therefore, we want the ‘SOH’ part of ‘SOHCAHTOA’, so will be using sin. We have O=13,H=15, and the angle is q, so we get


Then, to get q, we have to apply the inverse sin function: \sin^{-1} to both sides. It cancels out the sin on the left-hand side, and we get


Finally, putting this into the calculator we get

q=60.0735...=60.1\degree\text{ (1dp)}

HT3 – Trig & Pythagoras in 3D

Question: Shape ABCDEFG is a cuboid. Find the length of side FC, marked in red, to 3sf.

Answer: Firstly, the shape is a cuboid, which means it is full of right-angles. Every corner is a right-angle, in fact, which means our rules mentioned on the card will come in useful.

So, we need to find FC. Let’s first we consider the triangle FCH on its own.

Because we’re in a cuboid, this is a right-angled triangle. We know that side CH is 5cm, but otherwise we have no information about this triangle.

We need an extra side-length or angle for us to be able to find the length of FC. If we find the length of FH, for example, then we can use Pythagoras to find FC.

It is very useful then to notice that FH is also a hypotenuse of a different triangle…

To find side-length FH, we’re going to now consider the triangle FEH. Fortunately, we do know a little more about this triangle.

As before, this is also a right-angled triangle, but the difference is we know one of the side-lengths as well as an angle. This is enough information for us to use SOHCAHTOA to find the length of FH.

FE is adjacent to the angle, and we’re looking for the hypotenuse FH, so we will use \cos.

\cos(26) = \dfrac{\text{Adjacent}}{\text{Hypotenuse}} = \dfrac{9}{FH}

FH \times \cos(26) = 9

FH = 9 \div \cos(26) = 10.013… \text{cm}

Now we know FH, our first triangle, FCH, looks like this:

Firstly, note that the length of FH is expressed as 10.013… cm because when you’re putting it into your calculator, you should use the exact answer (by utilising the ANS button) and not a rounded-off decimal.

Now, we know two side-lengths of this triangle, we can use Pythagoras’ theorem to find the third, FC, which is the answer to the whole question.

(FC)^2 = 5^2 + (10.013…)^2

FC = \sqrt{5^2 + (10.013…)^2} = 11.2 \text{cm (3s.f.)}

HT4 – The Sine Rule

Question: Find the obtuse angle marked x to 2 s.f.

Answer: Here we can use the ‘flipped-over’ version of the sine rule to make things easier:

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

 Then, as usual, we label the triangle. Making sure to label A as the unknown, 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 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.

HT5 – The Cosine Rule

Question: Find the length of the side marked x to 2 significant figures.

Answer: Firstly, we need appropriately label the sides of this triangle. Firstly, we set a=x, and therefore we get that A=19, since it is the angle opposite. It doesn’t matter how we label the other two sides, so here we’ll let b=86 and c=65.

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

Now, subbing these values into the cosine rule equation, we get


Then, taking the square root, and putting it into the calculator, we get

x=\sqrt{7,396+4,225-11,180\cos(19)}=32\text{cm (2sf)}

HT6 – Trigonometry Common Values

Question: Without using a calculator, write down the exact value of \sin(w).

Answer: This question requires a bit less work but a bit more thought.

Since two sides of this triangle are the same length, it must be an isosceles triangle. In an isosceles triangle, we must have two angles the same – specifically the two angles that aren’t given to us (we can’t have one angle be the same as the right-angle, because then the sum of the angles in the triangle would go above 180\degree).

If those two angles are the same, the other one must also be w. Then, because angles in a triangle sum to 180\degree, we get


Subtract 90 from both sides to get


Then, dividing both sides by 2 gives us: w=45. Now we know the size of w, from the values of sin that we memorised we get that


ALTERNATIVE METHOD: According to SOHCAHTOA, the sin of w must be equal to the opposite side divided by the hypotenuse. The opposite side is given to us: 2, but the hypotenuse is not. However, we can find it using Pythagoras, since this is a right-angled triangle.

The hypotenuse is c, and then a and b are both 2, so the equation a^2+b^2=c^2 becomes


Square rooting both sides, we get


At this point you can simplify the surd and make it into 2\sqrt{2}, but you don’t have to. Then, now we know the hypotenuse, we get


This is an exact value so is the correct answer. If you simplified the surd, you’ll notice that there is a 2 on top and bottom that can cancel, and so we get


Which is in the familiar form. However, if you didn’t do this last step, your answer was still correct.

HT7 – Graphs of Trigonometric Functions

Question: On the same axes, plot graphs of y=\sin{\left(x\right)} and y=\cos(x) between -180^o and 180^o

Answer: If you can’t remember their shapes, check a few points. So, we have that:

\cos(0)=1,\,\,\text{ and }\,\,\cos(90)=0

Which is enough to start of the pattern of the cos graph. Similarly, we have:

\sin(0)=0,\,\,\text{ and }\,\,\sin(90)=1

Which is enough to start the pattern of the sin graph. If you aren’t sure, just try more values. The resulting graph looks like:

HT8 – Vectors

Question: In the diagram, we have vectors \vec{AE}=3a-2b and \vec{DC}=2a+4b. E and B are the midpoints of AD and AC respectively. Find an expression for \vec{EB} in terms of a and b and state whether or not it is parallel to \vec{DC}.

Answer: There are a lot of steps here, so take your time to read through it and make sure you understand.

We will find \overrightarrow{EB} by doing


The first vector is straightforward, because we know \overrightarrow{AE}, and that is just the same vector in the opposite direction. So, we get


Now we need \overrightarrow{AB}. Since B is the midpoint of AC (given in the question), we must have that \overrightarrow{AB}=\frac{1}{2} \overrightarrow{AC}. Therefore, looking at the diagram, we get that


We’re given the second part of this, \overrightarrow{DC}=2\mathbf{a}+4\mathbf{b}, and since E is the midpoint of AD, we can also work out the first part:


Now, at long last, we have everything we need and can go back through our work, filling in the gaps. Now we have \overrightarrow{AD}, we get that

\overrightarrow{AB}=\dfrac{1}{2}\left(6\mathbf{a}-4\mathbf{b}+2\mathbf{a} +4\mathbf{b}\right)=\dfrac{1}{2}\left(8\mathbf{a}\right)=4\mathbf{a}

Therefore, finally we have that


If \overrightarrow{EB} and \overrightarrow{DC} are parallel, then one must be a multiple of the other. Well, if we multiply \overrightarrow{EB} by 2 then we get


Therefore, we’ve shown that 2\overrightarrow{EB}=\overrightarrow{DC}, and thus the two lines must be parallel.