Hi,

All variables given in blue are known values. I am trying to calculate the angle X in terms of the variables I know. How can I do this?

http://img100.imageshack.us/img100/2673/trih.jpg

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- Oct 21st 2012, 08:52 AMCorpsecreateFinding an angle within this triangle
Hi,

All variables given in blue are known values. I am trying to calculate the angle X in terms of the variables I know. How can I do this?

http://img100.imageshack.us/img100/2673/trih.jpg - Oct 21st 2012, 09:05 AMSorobanRe: Finding an angle within this triangle
Hello, Corpsecreate!

Quote:

All variables given in blue are known values.

I am trying to calculate the angle X in terms of the variables I know.

http://img100.imageshack.us/img100/2673/trih.jpg

The sum of the (interior) angles of a triangle is always equal to 180^{o}.

Hence: .$\displaystyle A + B + (C + X) \;=\;180$

Got it?

- Oct 21st 2012, 09:07 AMBobPRe: Finding an angle within this triangle
You can calculate the angle between the side$\displaystyle b$ and the horizontal, add $\displaystyle X$ to that and you have the angle $\displaystyle C.$

- Oct 21st 2012, 06:09 PMCorpsecreateRe: Finding an angle within this triangle
Angle $\displaystyle C$ is the entire interior angle, not just the angle up to the horizontal. $\displaystyle A + B + C$ will give $\displaystyle 180$, $\displaystyle A + B + (C + X)$ will give a number larger than $\displaystyle 180$.

The angle side $\displaystyle b$ makes with the horizontal is $\displaystyle A + phi$, I dont see how adding $\displaystyle X$ to this angle would give me angle $\displaystyle C$?

NVM: Question has been solved! - Oct 22nd 2012, 12:53 AMBobPRe: Finding an angle within this triangleQuote:

The angle side $\displaystyle b $ makes with the horizontal is $\displaystyle A + phi$, I dont see how adding $\displaystyle X$ to this angle would give me angle $\displaystyle C$?

NVM: Question has been solved!

This will equal the angle below the horizontal at $\displaystyle C.$ Add $\displaystyle X$ and you get $\displaystyle C.$

BTW., some labels on the diagram would have helped.