Question

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

• A. $\cos^{-1}(\frac{\sqrt{3}}{\sqrt{7}})$
• B. $\cos^{-1}(\frac{3\sqrt{3}}{2\sqrt{7}})$
• C. $\cos^{-1}(\frac{4\sqrt{3}}{5\sqrt{7}})$
• D. $\cos^{-1}(\frac{3\sqrt{3}}{4\sqrt{7}})$

Hint:

When finding the angle between vectors representing sides of a triangle, be careful about the direction of vectors. The angle 'at' a vertex, say B, is the angle between $\vec{BA}$ and $\vec{BC}$. The angle between $\vec{AB}$ and $\vec{BC}$ is the exterior angle at B. Exam questions often ask for the interior angle even if vectors point outwards.

Answer 1

The Correct Option is B

Step 1: Set up a coordinate system and define vectors.
Let $A$ be the origin $(0,0)$, and place $\vec{AB}$ along the positive x-axis. From the given ratios, let $15|AC| = 5|AD| = 3|AB| = k$. Then $|AB| = k/3$, $|AD| = k/5$, $|AC| = k/15$. Choose $k = 15$, giving: \[ |AB| = 5, |AD| = 3, |AC| = 1. \] Then $\vec{AB} = 5\vec{i}$. Angle at $A$ is $\angle A = 2\pi/3 = 120^\circ$, and $AC$ is the bisector, so \[ \angle BAC = \angle CAD = 60^\circ. \] Hence, $\vec{AC} = 1(\cos 60^\circ \vec{i} + \sin 60^\circ \vec{j}) = \frac{1}{2}\vec{i} + \frac{\sqrt{3}}{2}\vec{j}$.
Step 2: Find $\vec{BC$.}
\[ \vec{BC} = \vec{AC} - \vec{AB} = \left(\frac{1}{2}-5\right)\vec{i} + \frac{\sqrt{3}}{2}\vec{j} = -\frac{9}{2}\vec{i} + \frac{\sqrt{3}}{2}\vec{j}. \]
Step 3: Find the angle between $\vec{AB$ and $\vec{BC}$.}
The angle $\theta$ between vectors is given by: \[ \cos\theta = \frac{\vec{AB}\cdot\vec{BC}}{|\vec{AB}||\vec{BC}|}. \] Compute the dot product: \[ \vec{AB}\cdot\vec{BC} = (5\vec{i}) \cdot \left(-\frac{9}{2}\vec{i} + \frac{\sqrt{3}}{2}\vec{j}\right) = -\frac{45}{2}. \] Magnitudes: \[ |\vec{AB}| = 5, |\vec{BC}| = \sqrt{\left(-\frac{9}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{21}. \] Thus, \[ \cos\theta = \frac{-45/2}{5\sqrt{21}} = -\frac{9}{2\sqrt{21}}. \]
Step 4: Adjust for acute angle.
The question options are positive, implying the acute angle between the vectors. So take \[ |\cos\theta| = \frac{9}{2\sqrt{21}}. \] Rationalize to match the options: \[ \frac{9}{2\sqrt{21}} = \frac{9}{2\sqrt{3}\sqrt{7}} = \frac{3\sqrt{3}}{2\sqrt{7}}. \]
Step 5: Conclusion.
Hence, the angle between $\vec{AB}$ and $\vec{BC}$ is: \[ \boxed{\cos^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{7}}\right)}. \]