Question

How do you find the exact value of the third side given `triangle RST`, RS=9, `ST=9sqrt3`, `mangleS=(5pi)/6`?

Answer 1

`=9sqrt(7)`

Explanation:

You would apply the cosine theorem to calculate the third side:

`bar(RT)=sqrt(bar(RS)^2+bar(ST)^2-2*bar(RS)*bar(ST)*cos hatS)`

`=sqrt(9^2+(9sqrt(3))^2-2*9*9sqrt3*cos(5/6pi))`

`=sqrt(81+243-cancel162^81sqrt(3)*(-sqrt(3)/cancel2)`

`=sqrt(324+243)=sqrt(567)=9sqrt(7)`