In a quadrilateral ABCD, measuredangle A = fr | vedclass

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(B) Let $|\vec{AC}| = k$. Then $|\vec{AD}| = 3k$ and $|\vec{AB}| = 5k$. Since $\vec{AC}$ bisects $\angle A = \frac{2\pi}{3}$, the angle between $\vec{

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$\cos^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{7}}\right)$

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(B) Let $|\vec{AC}| = k$. Then $|\vec{AD}| = 3k$ and $|\vec{AB}| = 5k$. Since $\vec{AC}$ bisects $\angle A = \frac{2\pi}{3}$, the angle between $\vec{AB}$ and $\vec{AC}$ is $\frac{\pi}{3}$, and the angle between $\vec{AD}$ and $\vec{AC}$ is $\frac{\pi}{3}$. Let $\hat{u}$ and $\hat{v}$ be unit vectors along $\vec{AB}$ and $\vec{AD}$ respectively. Then $\vec{AC} = \frac{|\vec{AC}|}{2\cos(\pi/6)} (\hat{u} + \hat{v}) = \frac{k}{2(\sqrt{3}/2)} (\hat{u} + \hat{v}) = \frac{k}{\sqrt{3}} (\hat{u} + \hat{v})$. Thus, $\vec{AC} = \frac{1}{\sqrt{3}} (\vec{AB}/5 + \vec{AD}/3) 	imes k$ is not correct; rather $\vec{AC} = \frac{k}{2\cos(\pi/6)} (\frac{\vec{AB}}{5k} + \frac{\vec{AD}}{3k}) = \frac{1}{\sqrt{3}} (\frac{\vec{AB}}{5} + \frac{\vec{AD}}{3})$. We have $\vec{BC} = \vec{AC} - \vec{AB} = \frac{1}{5\sqrt{3}} \vec{AB} + \frac{1}{3\sqrt{3}} \vec{AD} - \vec{AB} = \frac{1-5\sqrt{3}}{5\sqrt{3}} \vec{AB} + \frac{1}{3\sqrt{3}} \vec{AD}$. Using the dot product formula $\cos 	heta = \frac{\vec{AB} \cdot \vec{BC}}{|\vec{AB}| |\vec{BC}|}$, we calculate the angle to be $\cos^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{7}}ight)$.

In a quadrilateral $ABCD$, $\measuredangle A = \frac{2\pi}{3}$ and $\vec{AC}$ is the bisector of angle $A$. If $15|\vec{AC}| = 5|\vec{AD}| = 3|\vec{AB}|$, then the angle between $\vec{AB}$ and $\vec{BC}$ is

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