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# Using trigonometry to find length AB?

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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?

Guest Aug 1, 2018
#1
+20033
+1

Using trigonometry to find length AB?

$$\begin{array}{|rcll|} \hline \cos(\alpha) &=& \dfrac{18}{70-x} \qquad (1)\\\\ \cos(\alpha) &=& \dfrac{18+24}{70} = \dfrac{42}{70} \qquad (2) \\\\ \cos(\alpha) =\dfrac{18}{70-x} &=& \dfrac{42}{70} \\\\ \dfrac{18}{70-x} &=& \dfrac{42}{70} \\\\ \dfrac{70-x}{18} &=& \dfrac{70}{42} \\\\ 70-x &=& \dfrac{70\cdot 18}{42} \\\\ x &=& 70- \dfrac{70\cdot 18}{42} \\\\ x &=& 70- 30 \\ \mathbf{x} & \mathbf{=}& \mathbf{40\ \text{cm}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline 24^2+AB^2 &=& x^2 \quad & \quad x=40 \qquad \text{Pythagoras theorem} \\ 24^2+AB^2 &=& 40^2 \\ AB^2 &=& 40^2 - 24^2 \\ AB^2 &=& 1024 \quad & \quad 1024 = 32^2 \\ AB^2 &=& 32^2 \\ \mathbf{AB} & \mathbf{=}& \mathbf{32\ \text{cm}} \\ \hline \end{array}$$

heureka  Aug 1, 2018
edited by heureka  Aug 1, 2018
#2
+90055
+1

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

CPhill  Aug 1, 2018
edited by CPhill  Aug 1, 2018
edited by CPhill  Aug 1, 2018

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