$DC$ is parallel to $AB$.
Find the value of $BE$ and $DC$.
I've tried too many times but still can't figure it out.
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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 $$\frac{a}{\sin\theta_a}=\frac{b}{\sin\theta_b}$$
Peter Woolfitt
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But my teacher didn't mention about any use of sin or cosine.. I forgot how he did it, but the reason why I didn't write it down is because I wanted to figure it myself and I somehow did.. Question: Is it possible to just take the right triangle which is and rotate it by 90 degrees so that the E is upward, C is downward and D is the point that's to the right. So that would be LABE = LCED? – aaaapluss Jan 25 '15 at 17:40
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@aaaapluss Unfortunately, we can't do that. Even though both triangles have a $30^\circ$ angle, the other angles aren't the same – Peter Woolfitt Jan 25 '15 at 18:05
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Well that's strange because this chapter doesn't even show how to use the cos and sin. – aaaapluss Jan 25 '15 at 18:22