Here's the question:
I tried to find right triangles, and I drew them in. However, I'm not sure on what to do next?
Using trigonometry to find length AB?
\(\begin{array}{|rcll|} \hline \cos(\alpha) &=& \dfrac{18}{70-x} \qquad (1)\\\\ \cos(\alpha) &=& \dfrac{18+24}{70} = \dfrac{42}{70} \qquad (2) \\\\ \cos(\alpha) =\dfrac{18}{70-x} &=& \dfrac{42}{70} \\\\ \dfrac{18}{70-x} &=& \dfrac{42}{70} \\\\ \dfrac{70-x}{18} &=& \dfrac{70}{42} \\\\ 70-x &=& \dfrac{70\cdot 18}{42} \\\\ x &=& 70- \dfrac{70\cdot 18}{42} \\\\ x &=& 70- 30 \\ \mathbf{x} & \mathbf{=}& \mathbf{40\ \text{cm}} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline 24^2+AB^2 &=& x^2 \quad & \quad x=40 \qquad \text{Pythagoras theorem} \\ 24^2+AB^2 &=& 40^2 \\ AB^2 &=& 40^2 - 24^2 \\ AB^2 &=& 1024 \quad & \quad 1024 = 32^2 \\ AB^2 &=& 32^2 \\ \mathbf{AB} & \mathbf{=}& \mathbf{32\ \text{cm}} \\ \hline \end{array}\)
Here's another way using similar triangles
Call the center of the larger circle, C and the center of the smaller circle, D
And call the point formed by the tangent and the radius at the "top" of the smaller circle, E
Note that the vertical angles formed by the intersection of the tangent lines are equal...and if we connect the centers of each circle, these angles will be bisected and equal
So...we have triangles ABC and EBD, angle BAC = angle BED
And angles ABC and EBD are also equal
Therefore.by AA congruency...triangle ABC is similar to triangle EBD
Then AC / ED = BC/BD
Then 24 / 18 = BC /BD
Then 4/3 = BC/ BD
Then CD has 7 equal parts
And each equal part is 70/7 = 10
So...BC is 4 of these = 40
So...by the Pythagorean Theorem, √ [ BC^2 - AC^2 ] = BC
So....√[40^2 - 24^2 ] = √1024 = 32 cm = BC