Question 278179
If the sides of a triangle are AB=16, BC=16, and AC=23 
what is each degree of all three angles.
<pre><font side = 4 color = "indigo"><b>
Let {{{a=BC}}}, {{{b=AC}}}, {{{c=AB}}}

{{{drawing(200,200,-2,18,-2,18, triangle(0,0,16.53125,15.99117799,16,0),
locate(0,0,A),locate(16,0,B),locate(16.5,17.5,C), locate(8,0,a=16),
locate(14.5,8,c=16),locate(5,10,b=23) )}}}

We use the law of cosines:

{{{cos(A)=(b^2+c^2-a^2)/(2bc)}}}

{{{cos(A)=(23^2+16^2-16^2)/(2*23*16)}}}

{{{cos(A)=529/736}}}

{{{cos(A)=.71875}}}

{{{A="44.04862567°"}}} or {{{A="44°2'55.052''"}}}

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

{{{cos(B)=(16^2+16^2-23^2)/(2*16*16)}}}

{{{cos(B)=-17/512}}}

{{{cos(B)=-.033203125}}}

{{{B="91.90274865°"}}} or {{{B="91°54'9.895''"}}}
 
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Since ABC is isosceles, we know that angle C=angle A,
or we can calculate it as a check with the law of cosines:

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

{{{cos(C)=(16^2+23^2-16^2)/(2*16*23)}}}

{{{cos(C)=529/736}}}

{{{cos(C)=.71875}}}

{{{C="44.04862567°"}}} or {{{A="44°2'55.052''"}}}

Edwin</pre>