# Thread: High School Math Problem Involving Circles, Radii and Tangent (Image Included)

1. ## High School Math Problem Involving Circles, Radii and Tangent (Image Included)

I've been trying to solve this problem for a while now, but I still cannot do it. I understand that AB has a length of 5cm, and so if BC is "x" then AC = 5 + x. From there, I'm stuck. Please can somebody show me how to do it. Thanks!

2. well first you need to find the angle at which the line AB makes with the horizontal (angle of ABX)

from the picture i attach, angle of ABX can be found by:
tan ABX = (AE - BD) / ED
tan ABX = (4-1)/4+1)
tan ABX = 3/5 = 0.6
ABX = 30.9637 degrees

from observation this will also be the angle of ACE:

sin ACE = AE / AC
sin 30.9637 = (radius of larger circle) / AC
AC = radius of larger circle/sin 30.9637
AC = 7.774603526 cm

3. Thanks! But why are the the two angles equal?

4. imagine moving the line BX down perpendicularly to the vertical until it intercepts line EDC(above line EDC). Since line AC is straight and line BX is parallel to line EDC the angles will also be the same.

5. Hmm. I've got the mark scheme here, and it has a different answer.

I couldn't quite understand the working, that's why I came here.

It says:

BC/1 = (BC + 5)/4
4x = x + 5
x = (5/3)
AC = 6 + (2/3)

6. ahhh i see my mistake now sorrie >.<

it is suppose to be

sin ABX = 4-1/4+1 = 0.6

again using observation angle ACE = angle ABX

sin ACE = radius of larger circle/AC
0.6 = 4/AC
AC = 4/0.6

7. But doesn't AB = XB = ED? So why don't you get the same angle?

8. lol thats where i made my mistake. i mistakenly assumed that AB = ED however it is not actually equal because the circle is not drawn up to scale. the smaller circle is actually very much smaller.

from the attachment AY is not equal in length to EZ. The only possibility for AB = ED is when both circles are equal in size(same radius)

9. Ok, that makes sense now.

But, I still don't understand the mark scheme working:
How did they get:

BC/1 = (BC + 5)/4

and

4x = x + 5

10. Also, surely AB has to be larger than XB, as it is the hypotenuse. By looking at the picture, BX's length is the two radii added together, plus a bigger gap. I'm confused.

11. $BDC$ and $AEC$ are similar.

So,
$\frac{|BC|}{|AC|} = \frac{|BD|}{|AE|}$

$\frac{1+x}{6+x}=\frac{1}{4}$

$x = \frac{2}{3}$

$|AC| = 6 + x = 6 + \frac{2}{3} = \frac{20}{3}$

12. well they are using the same method but more vaguely.

consider the triangle BCD and triangle ACE. obviously we can see that triangle BCD is the smaller version of triangle ACE.

from the attachment,
for the smaller triangle,
cos theta = BD/BC ====(1)

for the larger triangle,
cos theta = AE/(AB+BC) ======(2)

from (1) and (2)

BD/BC = AE/(AB+BC)
BC/BD = (AB+BC)/AE

let BC = x
x/1 = ((4+1)+x)/4
4x = 5+x
3x = 5
x=5/3 = BC

now AC = AB + BC = 4+1+5/3
=5+ 3/3 + 2/3
=6 + 2/3

13. Do you mean Sin Theta, not Cos Theta?

Ok, I understand how to work out the answer now using the similar triangles method. But, using the trig method, I can't understand how AB is bigger than BX (the horizontal).

14. well you already got the thinking right. the triangle ABX has a hypotenuse of AB and hypotenuse are always longer than any other side length of the triangle.

For exampleattachment)
sin theta = 8/10 = 0.8 ===> means that the line OB is a fraction of line AB
sin alpha = 6/10 = 0.6 =====> means that line OA is a fraction of line AB
this is true for all triangles

in our case A=A, B=B,O=X.
cos ABX = BX/AB ====> line BX is a fraction of length of line AB(BX<AB)

15. I'm still confused. I appreciate you for all trying to help me, and I can understand all the methods of working this out.

But I still don't understand why AB is larger then BX (the horizontal).

I can't understand your diagram that well. D:

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