Question

Geometry Find the value of the angle $\theta$in degrees rounded to the nearest tenth of a degree.

Short answer

The answer is $\theta =38.9°.$

Step-by-step solution

Step 1. Given information

We have two circles small circle have radius r1 = 2 and bigger circle have radius r2 = 4.

These two circle has two common tangents meeting at some point at an angle $\theta$.

Given:

$$\begin{gathered} PS=2 \\ QC=4 \\ \angle POS=\angle QOC=\frac{\theta }{2}.(SinceitsymmetricaboutthelinedrawnfromO). \\ SC=2+4=6, \\ And\angle OPS=\angle OQC=90°. \\ (Sincethe\tan gentisperpendiculartothecircle.) \end{gathered}$$

Step 2. Triangles and its properties

$$\begin{gathered} In∆POS, \\ \sin \left(\frac{\theta }{2}\right)=\frac{PS}{OS}. \\ In∆QOC, \\ \sin \left(\frac{\theta }{2}\right)=\frac{QC}{OC}=\frac{QC}{OS+SC}. \end{gathered}$$

Step 3. Solving the equations from step 2

$$\begin{gathered} \sin \left(\frac{\theta }{2}\right)=\frac{PS}{OS}=\frac{2}{OS}, \\ Now, \\ \sin \left(\frac{\theta }{2}\right)=\frac{QC}{OS+SC}, \\ \frac{2}{OS}=\frac{4}{OS+6}, \\ SolvingforOSweget, \\ 2OS+12=4OS, \\ 2OS=12, \\ OS=6, \\ Nowfromaboveequation, \\ \sin \left(\frac{\theta }{2}\right)=\frac{2}{OS}=\frac{2}{6}=\frac{1}{3}, \\ \frac{\theta }{2}={\sin }^{-1}\left(\frac{1}{3}\right), \\ \theta =2{\sin }^{-1}\left(\frac{1}{3}\right), \\ \theta =38.9424....° \\ \theta \cong 38.9°. \end{gathered}$$