Geometric Proofs I am not sure where I went wrong in this question. I am meant to find the length of the belt that is around the pulley.

I have also attached my working out.

Thank you!!
 A: Consider the following diagram

The length of an arc of is equal to the angle times the radius. In the following, all angles are in radians.


*

*Length of long arc: $\alpha R$

*Length of short arc: $(2\pi-\alpha) r$

*Length $L$ of two straight lines: consider the striped triangle. We can write:


$$\tan \beta=\frac{L}{R-r} \implies L=(R-r)\tan\left(\pi-\frac{\alpha}{2}\right)$$
Therefore, the total length is
$$\alpha R + (2\pi-\alpha) r + 2(R-r)\tan\left(\pi-\frac{\alpha}{2}\right)$$
which, in your case ($\alpha=3.3161255$, $R=8$ cm, $r=7$ cm), gives $70.16$ cm.
A: Refer to the figure:
$\hspace{1cm}$
The total length of the belt:
$$L=2AE+l_{AC}+l_{DE}=2\cdot \tan 85^\circ+\frac{14\pi}{360^\circ}\cdot 170^\circ+\frac{16\pi}{360^\circ}\cdot 190^\circ\approx 70.16.$$
A: There are actually at least two things that don't quite "add up" here.
One is your interpretation of the apparent intent of the problem;
the other is the apparent intent of the problem itself.
It's a bit difficult to draw the diagram to scale (for more than one reason),
but here is a schematic showing two ways we can pass a belt around two pulleys of unequal size:

In any case, the belt travels from the large pulley to the small one along a line tangent to both pulleys, and travels back along another line tangent to both pulleys.
Since the pulleys are of unequal size, the two parts of the belt between the pulleys cannot be parallel.
So if the large pulley has its center at $F$ and the belt is tangent to it at $D$ and $E$, the tangent lines along which the belt travels will intersect at a point.
I have labeled this point $P$.
Now there are two places we can think of putting the smaller pulley.
One place is between the tangent lines on the far left of the diagram,
putting the center of the pulley at the point labeled $B',$
so that the crossing point $P$ is between the pulleys.
In that case the belt crosses over itself in the space between the pulleys.
The other place is between $P$ and the larger pulley,
putting the center of the smaller pulley at the point labeled $B.$
In that case the belt runs from the top of one pulley to the top of the other, and back again from the bottom of that pulley to the bottom of the first pulley.
Looking at the photograph of the printed diagram that came with the question,
I believe the intent of the author was that the belt should travel from the top of one pulley to the top of the other along the straight line segment near the top of the diagram, and then back from the bottom of one pulley to the bottom of the other along the straight line segment near the bottom of the diagram. 
That is, we are meant to put the center of the smaller pulley at a point $B$ between the larger pulley and the point $P$ where the tangent lines intersect.
In your calculations, you found that the smaller angle
$\angle DFE = 170^\circ$ and that the angle made by the tangent lines at $P$ is $\angle DPE = 10^\circ.$
Therefore by symmetry $\angle DPF = \angle EPF = 5^\circ$
and  $\angle DFP = \angle EFP = 85^\circ.$
So far this all agrees with the apparently intended interpretation of the problem.
Where you disagreed with the author's apparent interpretation is that you put the two pulleys on opposite sides of $P.$
That is, you put the smaller pulley at the point labeled $B'$ in the diagram above rather than at the point labeled $B$ where I believe you were intended to put it.
This results in a much longer belt, because your belt does not merely need to reach from $E$ to $A$ and from $D$ to $C$ but must also reach from $A$ to $C'$ and from $C$ to $A'.$
What I meant originally when I said, "You went wrong in the orientation of angle $\angle ABC,$"
is that if you had put the smaller pulley at $B,$
the smaller angle $\angle ABC$ would have faced toward the left.
Instead, you put it at $B'$, where the smaller angle $\angle A'B'C'$ faces toward the right.
And indeed the angles $\angle A'B'F$ and $\angle C'B'F$ are each $85$ degrees,
but you were intended to make
$\angle ABF = \angle CBF = 95^\circ.$
(Note that according to the right angles at $A$ and $E,$
the radius $AB$ should be parallel to $EF.$)

But you are not the only one whose interpretation of the problem might be questioned. The original author's apparent intent is questionable as well, because the diagram shown not only is not drawn to scale;
it cannot be drawn to scale.
In order for the outer tangents of two circles of diamters $7$ and $8$ to make an angle of $10^\circ,$ the circles must overlap,
contrary to what is shown in the diagram.
To put it another way, if I had constructed my diagram so that the circle around $F$ had radius $8$ and so that the angle $\angle DPE$ was actually $10^\circ,$ 
if you then put a circle of radius $7$ tangent to the lines $DP$ and $EP$ so that the center of that circle was at a point $B$ between $P$ and $F$, the circle around $B$ would overlap the circle around $F.$
It is my guess that the author of the problem did not think of this, since the circles could easily be made non-overlapping (as shown in the diagram) by a different choice of numerical values.
In that sense, I think it may be fair to say that the author also "went wrong" in the solution to this problem.
