## FT1 – Pythagoras

**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

**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)}=2.81km (2 d.p.)

## FT3 – Trigonometry – Finding Angles

**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

**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

**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 -2**a** 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: