Would appreciate any help on the following problem, Thank You!

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- May 22nd 2010, 05:21 AMmathguy20Help with Special Triangle Problem
Would appreciate any help on the following problem, Thank You!

- May 22nd 2010, 05:48 AMskeeter
given $\displaystyle AB = 2$

BC is the hypotenuse of the 45-45-90 triangle ...

$\displaystyle BC = (AB)\sqrt{2} = 2\sqrt{2}$

BC is also the longer leg of a 30-60-90 triangle ... BC is the short leg.

$\displaystyle BD = \frac{2\sqrt{2}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}$

from this point, you should be able to finish up. - May 22nd 2010, 12:10 PMmathguy20
- May 27th 2010, 04:03 AMmathguy20
bump.... Anybody able to confirm if my previous answer was correct?

Thanks!