geometric triangle parallel

Question

$DC$ is parallel to $AB$.

Find the value of $BE$ and $DC$.

I've tried too many times but still can't figure it out.

Answer 1

Hint: In both cases we can use the law of cosines. This says that if you have a triangle with side lengths $a$, $b$, and $c$, with the angle opposite the side of length $c$ as $\theta$, then you have the relation

$$c^2=a^2+b^2-2ab\cos\theta$$

Notice that in each situation, you have $3$ out of the $4$ variables in this formula, so you can solve for the last one.

Alternatively, if you wanted to, you can use the law of sines for the first question. The law of sines states that if you have a triangle with side lengths $a$, $b$, and angles $\theta_a$ and $\theta_b$ such that $\theta_a$ is opposite the side of length $a$ and $\theta_b$ is opposite the side of length $b$, the you have the relation $$\frac{a}{\sin\theta_a}=\frac{b}{\sin\theta_b}$$