Angle of a sector

H

harith

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Nov 23, 2013
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12
Hello, I tried solving this problem but can't find any solutions to it. Any help will be appreciated.Sector.jpg
 
Denis said:
Harith, WHAT did you try? WHAT does "can't find any solutions" mean?
Looks to me like you're trying to get your homework done...
Click to expand...

I am not even schooling, got an interview next week so am going through the papers. I tried with the sine rule before. But now I know how to do it. Thanks for the reply. :)
 
angle
harith said:
Hello, I tried solving this problem but can't find any solutions to it. Any help will be appreciated.View attachment 3624
Click to expand...
Suppose that \(\displaystyle M \) is the midpoint of \(\displaystyle \overline {AB} \), then \(\displaystyle \Delta OMA \) is a right triangle.

So \(\displaystyle m\left( {\angle AOB} \right) = 2 \arcsin \left( {\frac{8}{{15}}} \right) \)
 
Last edited by a moderator:
pka said:
Suppose that \(\displaystyle M \) is the midpoint of \(\displaystyle \overline {AB} \), then \(\displaystyle \Delta OMA \) is a right triangle.

So \(\displaystyle m\left( {\angle AOB} \right) = 2 \arcsin \left( {\frac{8}{{15}}} \right) \)
Click to expand...
... the TeX is no rendering,
 
Thanks, by the way if I have 2 sides given and the triangle is not a right angle, can I use cosine rule or sine rule?

I know its a silly question, personally i think the answer is 'no' but still just confirming.
 
Renderer on?

angle

pka said:
Suppose that \(\displaystyle M \) is the midpoint of \(\displaystyle \overline {AB} \), then \(\displaystyle \Delta OMA \) is a right triangle.

So \(\displaystyle m\left( {\angle AOB} \right) = 2 \arcsin \left( {\frac{8}{{15}}} \right) \)
Click to expand...

Your statement seems to render from here.

Suppose that \(\displaystyle M \) is the midpoint of \(\displaystyle \overline {AB} \), then \(\displaystyle \Delta OMA \) is a right triangle.

So \(\displaystyle m\left( {\angle AOB} \right) = 2 \arcsin \left( {\frac{8}{{15}}} \right) \)


P.S> Isn't the fact that O is the center of the circle suggestive? If one knew the length of all three sides of the triangle the cos law would supply an included angle and the sin law would provide the other angles. But even if all of that is true it would still be the longer solution route, so ...
 
Last edited by a moderator:
pka said:
... the TeX is no rendering
Click to expand...

This is because your LaTex coding contains a reserved system word (angle).

I'll fix the issue, by inserting the word 'angle' at the top of the post.

For additional information, please see items #16 and #17 HERE.


Cheers ~ Mark
 
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