Problem statement: Given a triangle with side lengths 4 and 6, their corresponding opposite angles have a 1:2 ratio. Find the length of the third side.
I solved the problem in 2 ways and got as an answer {5} on one of them and {4; 5} on the other. Can someone explain to me what's going on?
On
Hint:
The sine rule has ambiguous case . See http://en.wikipedia.org/wiki/Law_of_sines#The_ambiguous_case.
The difference between the two ways is how to use the rules of cosines. You may want to check if each solution is sufficient.
Suppose that $BC=4$. Let $D$ be the midpoint of the side $AB$. Since $\angle{BCD}=\angle{DBC}=x$, one has to have $BD=CD$. However, this does not hold because $$BD=6/2=3,\ \ \ CD=\sqrt{3^2+4^2}=5.$$ This is a contradiction. Hence, $BC\not =4$.