Hello davyboy123 Originally Posted by

**davyboy123** I have attached the image of the problem to look at.

a). Apply the Sine rule to triangle QRS to determine the length of QS

b). Apply the cosine rule to triangle QPS to determine the length of QP [IMG]file:///C:/Documents%20and%20Settings/hunter/Desktop/ICA1%20-%20SPECIMEN-5.jpg[/IMG]

I'm not quite sure where your problem is here. $\displaystyle \angle QSR =180-(15+30)=135^o$. So in $\displaystyle \triangle QSR$:$\displaystyle \frac{QS}{\sin 30^o}= \frac{15}{\sin 135^o}$

...etc

And $\displaystyle \angle QSP =180-135= 45^o$. So in $\displaystyle \triangle QSP$:$\displaystyle QP^2 = QS^2+PS^2-2.QS.PS.\cos45^o$

... etc

Can you complete it now?

Grandad