I dont really understand what question 'a' is asking, can anyone explain it to me?
Thanks
<QPR is just the angle created by the segment PQ and PR(the lower left angle of the triangle), and they want you to find the cosine of that angle
They tell you that sin(<QPR)=$\displaystyle \frac{2}{3}$ so using this, and the identity $\displaystyle sin^2(x)+cos^2(x)=1$ you can get the answer
$\displaystyle \left(\frac{2}{3}\right)^2+cos^2(x)=1$
$\displaystyle \frac{4}{9}+cos^2(x)=1$
$\displaystyle cos^2(x)=1-\frac{4}{9}=\frac{5}{9}$ and take the sqrt of both sides to get
$\displaystyle cos(x)=\pm\frac{\sqrt{5}}{3}$ and if you go by the picture, the angle is acute, so the angle must lie in the first quadrant, so we can ignore the negative value
You are asked to calculate the $\displaystyle \cos(\angle(QPR))$
1. You know already $\displaystyle \sin(\angle(QPR))=\dfrac23$
2. You know that $\displaystyle (\sin(\alpha))^2+(\cos(\alpha))^2=1$
3. Therefore:
$\displaystyle (\cos(\angle(QPR)))^2+(\sin(\angle(QPR)))^2=1$
$\displaystyle (\cos(\angle(QPR)))^2+\dfrac49=1$
$\displaystyle (\cos(\angle(QPR)))^2=\dfrac59$
$\displaystyle \cos(\angle(QPR))=\dfrac13 \sqrt{5}$
Thanks, didn't realise I had to use that identity;
Can you also help me with part 'b', to find the side QR I would have to use te cosine rule (?), do I use the answer from part 'a'?
so I find the inverse cosine of $\displaystyle \frac{\sqrt{5}}{3} $ = 41.8.
Is this the angle I use?
That isn't necessary here. As you've stated you have to use the cosine rule:
$\displaystyle QR=\sqrt{7^2+(3\sqrt{5})^2-2 \cdot 7 \cdot 3\sqrt{5} \cdot \cos(\angle(QPR))}$
$\displaystyle QR=\sqrt{49+45-42 \cdot \sqrt{5} \cdot \dfrac13 \sqrt{5}}$
$\displaystyle QR=\sqrt{49+45- 70 } = \sqrt{24}=2\sqrt{6}$