Help with two geometry questions

Question

anonymous · Answer

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yttrium · Answer

There's a technique on finding the sides of special triangles (30-60-90) easier.
Always remember that the side opposite 30 degrees is half of the hypotenuse. And the side opposite 60 degrees is square root of 3 multiplied by the side opposite 30 degrees. Did you get it??

yttrium · Answer

@AdeyNyx ?

anonymous · Answer

using sinQ and tan Q procedure much easy =.@Yttrium - sorry  i didnt understood ur way dude.

yttrium · Answer

Ooops. That's actually the same. :)

yttrium · Answer

Let me give you an example.
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yttrium · Answer

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let say side CA is 4. Using trigonometric functions, we can conclude that side AB = 2sqrt3 and side BC = 2. But even without that we can also find it measure. Just knowing the the side opposite 30 deg is half of the hypotenuse. Therefore, side BC  = 2. And the side opposite 60 degrees is sqrt of 3 times the length of the side opposite 30. Hence side AB = side BC * sqrt of 3 which is AB = 2sqrt3. Hope you understand my point now. :3