What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
Someone , can help me with this problem? (Smirk)
What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
Re: What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
Re: What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
Yes sir.. I really need to have a graph or diagram for that. And of course a solution. Can you help me sir?
Re: What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
no ... you make the sketch and set up the problem for a solution.
Re: What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
How to make that sketch sir? Is it in the editor? or I need to upload an image?
Re: What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
Quote:
Originally Posted by
Adikz 
Someone , can help me with this problem? (Smirk)
What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
Study this web page.
You can work out this problem from the given.
Re: What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
Quote:
Originally Posted by
Adikz How to make that sketch sir? Is it in the editor? or I need to upload an image?
With pencil and paper! We're not saying that you should draw a picture here so that we can understand it, we are saying that you should draw a picture so you can understand it. You don't see that you will have an iscosceles triangle with two sides of length 21 and base of length 32? And you are asked to find the measure of the 'vertex' angle- the one opposite the side of length 32. As skeeter suggested initially, use the cosine law. Do you know what that is?
Re: What is the central angle subtended by a 32 m chord on a circle of a radius 21 m?
Code:
* * *
A * 32 * B
♥---------------♥
* * * *
21 * * 21
* * θ * *
* ♥ *
* O *
* *
* *
* *
* * *
$\displaystyle \text{Law of Cosines: }\:\cos\theta \:=\:\frac{21^2 + 21^2 - 32^2}{2(21)(21)} \:=\:-\frac{142}{882}$
$\displaystyle \theta \;=\;\cos^{\text{-}1}\!\left(\text{-}\tfrac{142}{882}\right) \;=\;99.26481291^o \;\approx\;99.3^o $
Like this? (Giggle)
