A) \[105{}^\circ \]
B) \[60{}^\circ \]
C) \[45{}^\circ \]
D) \[75{}^\circ \]
E) \[90{}^\circ \]
Correct Answer: D
Solution :
\[\therefore \]Angles A, B, C are in AP \[\therefore \] \[B=\frac{A+C}{2}\] \[\Rightarrow \] \[2B=A+C\] \[\therefore \] \[A+B+C=180{}^\circ \] \[\Rightarrow \] \[3B=180{}^\circ \] \[\Rightarrow \] \[B=60{}^\circ \] Now, \[sin\text{ }B:sin\text{ }C=\sqrt{3}:\sqrt{2}\] \[\Rightarrow \] \[\sin 60{}^\circ :\sin C=\sqrt{3}:\sqrt{2}\] \[\Rightarrow \] \[\frac{\sqrt{3}}{2}\times \frac{1}{\sin C}=\frac{\sqrt{3}}{2}\] \[\Rightarrow \] \[\sin C=\frac{1}{\sqrt{2}}=\sin 45{}^\circ \] \[\Rightarrow \] \[C=\frac{1}{\sqrt{2}}=\sin 45{}^\circ \] \[\Rightarrow \] \[C=45{}^\circ \] \[\therefore \] \[\angle A=180{}^\circ -(\angle B+\angle C)\] \[=180{}^\circ -(60{}^\circ +45{}^\circ )\] \[=75{}^\circ \]
You need to login to perform this action.
You will be redirected in
3 sec
