Question 252654
Given triangle ABC with a=12,b=10 and c=8 find the angles in degrees and minutes
<pre><font size = 4 color = "indigo"><b>
{{{cos(A) = (b^2+c^2-a^2)/(2bc)}}}

{{{cos(A) = (10^2+8^2-12^2)/(2*10*8)}}}

{{{cos(A) = (100+64-144)/(160)}}}

{{{cos(A) = 20/160=1/8=0.125}}}

{{{A = "82.81924422°" = "82°49'"}}} 

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{{{cos(B) = (a^2+c^2-b^2)/(2ac)}}}

{{{cos(B) = (12^2+8^2-10^2)/(2*12*8)}}}

{{{cos(B) = (144+64-100)/(192)}}}

{{{cos(B) = 108/192=9/16=0.5625}}}

{{{B = "55.77113367°" = "55°46'"}}}

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{{{cos(C) = (a^2+b^2-c^2)/(2ab)}}}

{{{cos(C) = (12^2+10^2-8^2)/(2*12*10)}}}

{{{cos(C) = (144+100-64)/240}}}

{{{cos(C) = 180/240=3/4=.75}}}

{{{C = "41.40962211°" = "41°25'"}}}

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Checking:

 82° 49'
 55° 46'
 41° 25'
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178°120' = 180°

Edwin</pre>