A) \[60{}^\circ \]
B) \[30{}^\circ \]
C) \[52\frac{1}{2}{}^\circ \]
D) \[67\frac{1}{2}{}^\circ \]
Correct Answer: C
Solution :
sin2 \[\left( A+B-C \right)=1\] = sin \[90{}^\circ \] \[\Rightarrow 2\left( A+B-C \right)=90{}^\circ \] \[\Rightarrow A+\text{ }B-C=45{}^\circ \] .....(i) \[\tan \,(B+C-A)=\sqrt{3}\] = tan \[60{}^\circ \] \[\Rightarrow B+C-A=60{}^\circ \] ..... (ii) On adding both the equations, \[A+B-C+B+C-A\] \[=45{}^\circ +60{}^\circ \] \[\Rightarrow 2B=105{}^\circ \Rightarrow B=52\frac{1{}^\circ }{2}\]You need to login to perform this action.
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