# Getting different answers using different methods in a geometrical problem

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?

• The second solution is the direct way of using the cosine law to find the POSITIVE value a side. The first one is an indirect (or even can be thought of as an improper) way of using the cosine law. This is because an extraneous root has been generated from the quadratic. Further testing (like @mathlove ‘s work) is needed to find out which one (4 or 5) is the invalid value.
– Mick
Commented May 19, 2015 at 14:06

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, $BC\not =4$.
• @user1113314: In the first, you used $AC^2=BA^2+\color{red}{BC}^2-2\cdot BA\cdot \color{red}{BC}\cdot\cos{\angle{ABC}}$. In the second, you used $\color{red}{BC}^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos{\angle{BAC}}$. See also Mick's comment. Commented May 19, 2015 at 17:08