Law of sine and cosine

Ihatdogs10 said:
I can use cos because the given is 2 sides and 1 included angle?
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Then why are you waiting around - calculate the lengths of those blue lines.

Then find all the angles of those triangles.
 
Hi I try to solve the first line did I get it right?
 

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Oh, I am sorry I just think of using it so I need to use the formula from #11 right?
Sorry I did not process everything yesterday
 
nasi112 said:
divide by [MATH]4[/MATH] means I want quarter the line of AB

you can do it like this also
first angle has a length of [MATH]0.25[/MATH] x [MATH]1912.87[/MATH]
second angle has a length of [MATH]0.5[/MATH] x [MATH]1912.87[/MATH]
third angle has a length of [MATH]0.75[/MATH] x [MATH]1912.87[/MATH]
Click to expand...
Did I get it right
 

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The length of the first blue line on the left [MATH]= 1073.92[/MATH]
Now, you can use Laws of Sines

[MATH]\frac{\sin \theta}{\frac{1912.87}{4}} = \frac{\sin 63.32^{o}}{1073.92}[/MATH]
 
You had two mistakes to find the blue line.

The first mistake, you have used one of the length as [MATH]0.25[/MATH]. We don't have any lines of this length.

The second mistake, you have used the angle [MATH]\cos 78.72^{o}[/MATH] to find the blue line. No, It is the wrong angle.
You should use [MATH]\cos 63.32^{o}[/MATH].
 
To find the blue line use this

[MATH]a^2 = b^2 + c^2 - 2bc \cos \theta[/MATH]
 
Hi the formula that you have given I get 1742
 

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you have used [MATH]1912[/MATH], you should use [MATH]\frac{1912}{4}[/MATH]
 
hello, i am confused because i get 961.9 instead of 1073. how did you do it?
 
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