# Thread: Applied pure trig question

1. ## Applied pure trig question

A belt ABCDEFA of length L passes round two wheels, centres O_1 and O_2. The parts ABC and DEF of the belts are in contact with the wheels and parts AF and CD are straight. If O_1A = a and O_2F = 3a and the angle AO_1B = theta radians show that:

4atan (theta) = 4a(theta) + L - 6a*pi

Next, find the shortest length belt in terms of a.

Im pretty stuck on this one. Any ideas??
I am looking at the equation to see what its saying but can't see where tan(theta) comes from - Obviously a triangle is drawn and the Opposite and Adjacent sides are used some how.
The arc AC on the small wheel is 2a(2*pi - (theta)) but I'm just kinda stuck

2. Hello dojo
Originally Posted by dojo
A belt ABCDEFA of length L passes round two wheels, centres O_1 and O_2. The parts ABC and DEF of the belts are in contact with the wheels and parts AF and CD are straight. If O_1A = a and O_2F = 3a and the angle AO_1B = theta radians show that:

4atan (theta) = 4a(theta) + L - 6a*pi

Next, find the shortest length belt in terms of a.

Im pretty stuck on this one. Any ideas??
I am looking at the equation to see what its saying but can't see where tan(theta) comes from - Obviously a triangle is drawn and the Opposite and Adjacent sides are used some how.
The arc AC on the small wheel is 2a(2*pi - (theta)) but I'm just kinda stuck
What you don't say - but what I deduce from your comment about the arc $AC$ - is that $BO_1O_2E$ is a straight line.

So look at the attached diagram. The key fact that you seem to be missing is that when we draw $O_1G$ parallel to $AF,\; O_1O_2G$ is a right-angled triangle with $\angle O_1O_2G = \theta$ and $O_2G = 2a$. So $O_1G = AF = 2a\tan\theta = CD$.

So now add the arc lengths $ABC$ and $D EF$ to $AF + CD\; (=4a\tan\theta)$, equate the sum to $L$, and the result follows.

The shortest belt possible is when $O_1O_2 = a + 3a = 4a$. This gives, from $\triangle O_1O_2G, \;\cos\theta = \tfrac12$. So with this value of $\theta$, solve the equation for $L$ and you're there.

Can you complete it now?