A) 11.9°
B) 16.0°
C) 15.3°
D) 9.11°
Correct Answer: A
Solution :
Total deviation = 0 \[{{\delta }_{1}}+{{\delta }_{2}}+{{\delta }_{3}}+{{\delta }_{4}}+{{\delta }_{5}}=({{\mu }_{1}}-1)\,{{A}_{1}}-({{\mu }_{2}}-1)\,{{A}_{2}}\] \[+({{\mu }_{3}}-1)\,{{A}_{3}}-({{\mu }_{4}}-1)\,{{A}_{4}}+({{\mu }_{5}}-1)\,{{A}_{5}}=0\] \[\Rightarrow \]\[2\times {{A}_{2}}(1.6-1)=3\,(1.53-1)\,9\] \[\Rightarrow \]\[{{A}_{2}}=3\left( \frac{0.53\times 9}{1.2} \right)={{11.9}^{o}}\]
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