Applied maths question

aarond280

can anyone tell me how they got sin 􀀁 = 30 in question 5 (b) and where they got the r thanks heres the link http://www.examinations.ie/archive/markingschemes/2009/LC020ALP000EV.pdf

MathsManiac

In the diagram given in the question, imagine drawing a line from the centre of B to the point of tangency. This gives a right-angled triangle, where the side opposite alpha is r, and the hyopeneuse is 2r, (taking r to be the radius of the spheres)

aarond280

MathsManiac wrote: »

In the diagram given in the question, imagine drawing a line from the centre of B to the point of tangency. This gives a right-angled triangle, where the side opposite alpha is r, and the hyopeneuse is 2r, (taking r to be the radius of the spheres)

oh right and is there any other way of doing that withouth using the radius

MathsManiac

Not sure why you would want to avoid referring to the radius - it makes an appearance and then immediately gets cancelled out, so it's not exactly getting in the way!

I guess you could imagine extending the given diagram to draw a pattern showing seven identical circles - six of them arranged round one, (like a flower!).

You could then use that pattern to argue that line shown in the original diagram makes an angle of 30 degrees with the twelve o'clock line, by rotational / reflective symmatry, as it's a twelfth of the way around the pattern. Might be hard to call it a proof, but it's enough for "show", I'd say.

Other than that, I can't think of an obvious way.