My neice asked me for help with her national 5 homework. (National 5 is a new qualification in Scotland roughly equivalent to the old standard grade, so I would have thought quite easy). I have made a crude MS paint of the problem and linked it below.
Basically, with the information given, you need to work out the length QR.
Now, I have solved it, but I had split the triangle into two right angle triangles, create simultaneous equations to work out the length of the perpendicular line I drew to make the triangles, and the lengths that PR is split into by this line, and then finally Pythagoras to work out the length QR.
I don't remember problems this difficult when doing my Standard Grade, so am I missing something bleeding obvious, or is it really this difficult?
(The way I did it isn't difficult per se, just much more so than I remember problems at this level being)
I am no maths wiz, just the best she has in her immediate circle of friends/family.
First use the geometric information to get the inner angles of the triangle: At $P$ it is $85^\circ$, at $R$ it is $180-125-(360-340)=35^\circ$, so at $Q$ it must be $180-85-35=60^\circ$.Now use the sine law: $\sin(60)/30=\sin(85)/x$, so $x=30\cdot \sin(85)/\sin(60)$.