Can some please check my work and see if I did this the right way or not? Thank you!Attachment 29704

Printable View

- Nov 9th 2013, 04:36 PMsakonpure6Clock Question - Check my work please!
Can some please check my work and see if I did this the right way or not? Thank you!Attachment 29704

- Nov 9th 2013, 05:24 PMSlipEternalRe: Clock Question - Check my work please!
The second hand does not represent the radius of the clock. Its length runs along the hypotenuse of the right triangle that runs from the tip of the second hand to the tip of the hour hand, along the hour hand, and then finally along the second hand. The angle $\displaystyle \theta$ does not equal $\displaystyle \dfrac{2}{3}$. An angle of $\displaystyle \dfrac{2}{3}$ is approximately 10% of the circumference of the circle. The angle you want is $\displaystyle \dfrac{2}{3}$ of $\displaystyle \dfrac{\pi}{2}$. In other words, $\displaystyle \theta = \dfrac{\pi}{3}$. And $\displaystyle \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$.

- Nov 9th 2013, 06:21 PMsakonpure6Re: Clock Question - Check my work please!
- Nov 9th 2013, 07:04 PMSlipEternalRe: Clock Question - Check my work please!
The angle between the hour hand and the minute hand is a right angle. That angle is $\displaystyle \dfrac{\pi}{2}$. Two thirds of that is $\displaystyle \dfrac{\pi}{3}$.

- Nov 10th 2013, 10:35 AMsakonpure6Re: Clock Question - Check my work please!
Thank you for this approach! It really slipped me!

- Nov 10th 2013, 10:37 AMSlipEternalRe: Clock Question - Check my work please!
Also, I am using radians. If you are using degrees, then a right angle is 90 degrees, so 2/3 of that would be 60 degrees.