Trigonometry & Pythagoras Answers | Foundation Tier | MME

# TRIGONOMETRY & PYTHAGORAS FOUNDATION ANSWERS

## FT1 – Pythagoras

Question: Below is a right-angled triangle. Work out the length of the side marked $x$. Give your answer to two decimal places.

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

$9^2+x^2=13^2$

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:

$x^2=169 - 81$

$x^2=88$

Finally, we square root both sides

$x=9.38$

Which gives the final answer to be $x=9.38$ (2d.p.)

## FT2 – Trigonometry – Finding Lengths

Question: Find the length of the side marked $z$ (2 d.p.)

Answer: Here we’re dealing with the opposite and adjacent sides, so we need a formula with both of those in. This means we’ll have to use $\tan(x)=\frac{O}{A}$.

First we rearrange the formula: we’re after the adjacent so we want to make A the subject.

$A\times \tan(x)=O$

$A=\dfrac{O}{\tan(x)}$

Now we’ve made A the subject, we just need to plug in the values given.

$z=\dfrac{3.6}{\tan(52)}=2.81$km (2 d.p.)

## FT3 – Trigonometry – Finding Angles

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

$\sin(q)=\dfrac{O}{H}=\dfrac{13}{15}$

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

$q=\sin^{-1}\left(\dfrac{13}{15}\right)$

Finally, putting this into the calculator we get

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

## FT4 – Trigonometry Common Values

Question: Without using a calculator, determine the size of angle $w$.

Answers: In this question we’re given the lengths of the opposite and adjacent sides, so we can apply the formula $\tan(w)=\frac{O}{A}$.

We can substitute our values straight in:

$\tan(w)=\dfrac{2}{2}=1$

Now, if you’ve remembered the trig values from this card, you’ll know that the value of $w$ which gives us $\tan(w)=1$ is $45\degree$. So $w=45\degree$

## FT5 – Vectors

Question: Express the vector a as a column vector. Draw the vector –2a beside it.

Answers: The vector shown in the diagram goes 2 steps to the left and 1 step downwards. That’s the same as saying it goes 1 steps in the negative y direction and 2 step in the negative x direction. So the vertical part of the column vector is -1, and the horizontal part of its column vector is -2. We write the column vector with the horizontal part at the top, and the vertical part at the bottom, so it looks like this:

Let $\mathbf{a}=\begin{pmatrix}-2\\-1\end{pmatrix}$

The vector -2a is what we get when we multiply a by -2. This doubles the size of it and switches its direction, so it looks like this on the grid: