Question 1138026




<img src="http://upload.wikimedia.org/wikipedia/commons/thumb/1/14/Triangle.Isosceles.svg/200px-Triangle.Isosceles.svg.png">


 let the base angles be {{{alpha}}} and {{{beta}}}=>{{{alpha = beta}}}

given: {{{a=10}}}, {{{b=10}}}, and {{{c=4}}}

let altitude to base {{{c}}} be {{{h}}}, then sides {{{a}}},{{{h}}}, and{{{ c/2 }}}form right triangle where {{{a}}} is hypothenuse,{{{ h}}} and{{{ c/2}}} legs

since {{{a=10}}} and {{{c/2=4/2=2}}}, we have

{{{h=sqrt(10^2-(2)^2)}}}

{{{h=sqrt(100-4)}}}

{{{h=sqrt(96)}}}


using the sine rule:

{{{sin(alpha)=h/10}}}

{{{sin(alpha)=sqrt(96)/10}}}

{{{alpha=sin^-1(sqrt(96)/10)}}}

{{{alpha=78.5}}}°

then

{{{beta=78.5}}}°

and third angle {{{gamma=180-(78.5+78.5)}}}°=>{{{gamma=180-157 }}}°=>{{{gamma=23}}}°


or, you can do it using the cosine rule:

{{{a=10}}}, {{{b=10}}}, and{{{ c=4}}}

{{{a^2=c^2+b^2-2cb*cos(alpha)}}}


{{{2cb*cos(alpha) =c^2+b^2-a^2}}}

{{{cos(alpha) =(c^2+b^2-a^2)/(2cb)}}}

{{{cos(alpha) =(4^2+10^2-10^2)/(2*4*10)}}}

{{{cos(alpha) =16/80}}}
{{{cos(alpha) =1/5}}}

{{{alpha=cos^-1(1/5)}}}

{{{alpha=78.5}}}° => {{{beta=78.5}}}°