Quadrilateral $ABCD$ has $AB = BC = CD$, $\angle ABC = 70^\circ$ and $\angle BCD = 170^\circ$. What is the measure of $\angle BAD$? $\mathrm{(A)}\ 75\qquad$ $\mathrm{(B)}\ 80\qquad$ $\mathrm{(C)}\ 85\qquad$ $\mathrm{(D)}\ 90\qquad$ $\mathrm{(E)}\ 95$
Solution
Put $B=(0,0)$, $C=(1,0)$ and $AB=BC=CD=1$. Then $A=(\cos70^\circ,\sin70^\circ)$ and, since $\angle BCD=170^\circ$, $D=(1+\cos10^\circ,\sin10^\circ)$.
$\overrightarrow{AD}=(1+\cos10^\circ-\cos70^\circ,\ \sin10^\circ-\sin70^\circ)=(1+\sin40^\circ,\ -\cos40^\circ)=(1+\cos50^\circ,\ -\sin50^\circ)=2\cos25^\circ\,(\cos25^\circ,\ -\sin25^\circ)$, so $\overrightarrow{AD}$ points in the direction $-25^\circ$.
$\overrightarrow{AB}=(-\cos70^\circ,-\sin70^\circ)$ points in the direction $250^\circ=-110^\circ$.
