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- Dec 13th 2008, 12:16 AMhelloyingangles problem
thanks

- Dec 13th 2008, 12:45 AMandreas
(a)

$\displaystyle \angle{DBC}=28$ from this we know $\displaystyle \angle{BCD}=180-2*28=124$. $\displaystyle X+X+\angle{ACB}=180$ and

$\displaystyle X+\angle{ACB}=124 $ solving these equations you get $\displaystyle X=56$ and $\displaystyle \angle{ACB}=180 - 2*56=68$. You also know that $\displaystyle \angle{ABC}=2*28=56$. Now, what is $\displaystyle \angle{BAC}$ ? It is $\displaystyle 180 -68-56=56$.

(b) and (c) are much more easier questions, try to solve them. If you do not succeed ask here again. Good luck!