Physics Two Dimensional Motion Question 133
Question: If $ \overrightarrow{A}=\overrightarrow{B}-\overrightarrow{C} $ , then, the angle between $ \overrightarrow{A} $ and $ \overrightarrow{B} $ is
Options:
A) $ \text{ta}{{\text{n}}^{-1}}\frac{B^{2}+A^{2}-C^{2}}{2AB} $ B) $ {{\sin }^{-1}}\frac{B^{2}+A^{2}-C^{2}}{2AB} $ C) $ {{\cos }^{-1}}\frac{A^{2}+B^{2}-C^{2}}{2AB} $ D) $ {{\sec }^{-1}}\frac{A^{2}+B^{2}-C^{2}}{2AB} $
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Answer:
Correct Answer: C
Solution:
[c] Given, $ \vec{A}=\vec{B}-\vec{C} $ $ \
Therefore $ $ \vec{C}=\vec{B}-\vec{A} $ If $ \theta $ is the angle between $ \vec{A} $ and $ \vec{B} $ , then $ C^{2}=B^{2}+A^{2}-2AB,cos\theta $ $ \
Therefore \cos \theta =\frac{B^{2}+A^{2}-C^{2}}{2AB} $
BETA
