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) \[{{\sec }^{-1}}\frac{{{A}^{2}}+{{B}^{2}}-{{C}^{2}}}{2AB}\]
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 \[{{C}^{2}}={{B}^{2}}+{{A}^{2}}-2AB\,cos\theta \] |
| \[\therefore \cos \theta =\frac{{{B}^{2}}+{{A}^{2}}-{{C}^{2}}}{2AB}\] |
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