TS EAMCET · Maths · Vector Algebra
In a quadrilateral \(\mathrm{ABCD}, \measuredangle \mathrm{A}=\frac{2 \pi}{3}\) and AC is the bisector of angle \(\lfloor\mathrm{A}\). If \(15|\stackrel{\rightharpoonup}{\mathrm{AC}}|=5|\stackrel{\rightharpoonup}{\mathrm{AD}}|=3|\stackrel{\rightharpoonup}{\mathrm{AB}}|\), then angle between \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{BC}}\) is
- A \(\operatorname{Cos}^{-1}\left(\frac{\sqrt{3}}{\sqrt{7}}\right)\)
- B \(\operatorname{Cos}^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{7}}\right)\)
- C \(\operatorname{Cos}^{-1}\left(\frac{4\sqrt{3}}{5\sqrt{7}}\right)\)
- D \(\operatorname{Cos}^{-1}\left(\frac{3\sqrt{3}}{4\sqrt{7}}\right)\)
Answer & Solution
Correct Answer
(B) \(\operatorname{Cos}^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{7}}\right)\)
Step-by-step Solution
Detailed explanation
\(\angle \mathrm{CAB} = \frac{1}{2} \angle \mathrm{A} = \frac{1}{2} \cdot \frac{2\pi}{3} = \frac{\pi}{3}\) Let \(|\stackrel{\rightharpoonup}{\mathrm{AC}}|=k\). Then \(|\stackrel{\rightharpoonup}{\mathrm{AB}}|=5k\) and \(|\stackrel{\rightharpoonup}{\mathrm{AD}}|=3k\). In…
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