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why?

yes you're correct sorry, typo. Definitely got it in the exam though!

theres a solution in the Q16 thread.

This is for 16b) gradient OT = tan theta, therefore gradient tp = -1/tan theta... point T is (cos theta, sin theta) and use equation of a line formula

sack your tutor mate, its definitely k>32.

DAMN, i got that quadratic but never thought to use the discriminant.

I don't think your way would work, because that vertical line is only in one dimension, i.e. only has 'height', and no 'depth' or 'width', but by putting it on the other view you are saying that it does have width or depth.

You can disregard the vertical line in the bottom view, because since it's vertical, it is already its true length. so just work out the angled section, then add it to the vertical section, like this:

carrotsticks alone.

Draw a scale diagram of the forces you know and the ones you need to find out, using the angles provided. Measure the unknown lines and scale to work out the force...

you basically had it there, except use perpendicular distance formula = distance from point formula

Decent paper, but I have no idea if i nailed it or stuffed up completely. Anyway, NO MORE ENGLISH EVER!!!!!!!

no its quite easy, you just have to project them both to the same plane. i'm doing a pic, just wait... you pivot x around until it is horizontal, then project it straight down until it intersects the horizontal line of the top of the beam. Then measure 'y', the blue line, to get...

I haven't got two on one day, but i've got four consecutive next week, which sucks big time.

I got 4 pages for all of them except for the comprehension which was 3 and the creative which was 1.5.

...^

passage of time, not moments in time

actually, the extended response wasn't too bad, but i spent too long on the short answers, and ran out of time for the creative, which was fairly retarded. and, "There are currently 57 users browsing this thread. (23 members and 34 guests)"

yep he was right.

3 hours 30 minutes!!!