A) \[{{180}^{{}^\circ }}-3A\]
B) \[{{180}^{{}^\circ }}-2A\]
C) \[{{90}^{{}^\circ }}-A\]
D) \[{{180}^{{}^\circ }}+2A\]
Correct Answer: B
Solution :
As, we know that \[\mu =\frac{\sin \left( \frac{A+{{D}_{m}}}{2} \right)}{\sin \frac{A}{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\cot \frac{A}{2}=\frac{\sin \left( \frac{A+{{D}_{m}}}{2} \right)}{\sin \frac{A}{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left( \frac{A+{{D}_{m}}}{2} \right)}{\sin \frac{A}{2}}\] \[\sin \left( \frac{\pi }{2}-\frac{A}{2} \right)=\sin \left( \frac{A+{{D}_{m}}}{2} \right)\] \[\Rightarrow \,\,\,\,\,\frac{\pi }{2}-\frac{A}{2}=\frac{A}{2}+\frac{{{D}_{m}}}{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{D}_{m}}=\pi -2A\] \[{{D}_{m}}={{180}^{{}^\circ }}-2A\]
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