Question 1206093


An {{{equilateral}}} triangle has {{{all}}} {{{sides}}}{{{ equal}}}. Your triangle is not an {{{equilateral}}} triangle.


In any triangle {{{ABC}}},

{{{a^2 = b^2 + c^2 - 2bc*cos( A )}}}
 {{{b^2 = a^2 + c^2 - 2ac*cos( B)}}}
{{{c^2 = a^2 + b^2 - 2ab* cos (C)}}}


given:

{{{c=8}}}
{{{a=3}}}
< {{{B=57.53}}}

Which measurement of acute triangle {{{ABC}}} can you find by direct substitution of the labeled measures in the Law of Cosines?

answer:

A.
{{{AC}}}  (or side {{{b}}})

proof:

{{{b^2 = 3^2 + 8^2 - 2*3*8*cos( 57.53)}}}
{{{b^2 = 47.23081907263568}}}
{{{b=sqrt(47.23081907263568)}}}
{{{b=6.9}}}


then you can find angles {{{A}}} and {{{C}}}

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

{{{3^2 = 6.9^2 + 8^2 - 2*6.9*8*cos( A )}}}

{{{2*6.9*8*cos( A )=3^2 - 6.9^2 + 8^2 }}}

{{{110.4cos( A )=25.39}}}

{{{cos( A )=25.39/110.4}}}

{{{cos( A )=0.22998188405797101}}}

{{{A=cos^-1(0.22998188405797101)}}}

{{{A=76.7}}}°

then

{{{C=180-(57.53+76.7)}}}

{{{C=45.77}}}°