A) \[40{}^\circ \]
B) \[42{}^\circ \]
C) \[45{}^\circ \]
D) \[48{}^\circ \]
Correct Answer: B
Solution :
| [b] We have, \[r=15cm\] and \[l=\frac{1}{2}(22)=11cm\] |
| We know that \[l=2\pi r\left( \frac{\theta }{360{}^\circ } \right)\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\theta =\frac{11\times 360{}^\circ }{2\times \frac{22}{7}\times 15}=\frac{90{}^\circ \times 7}{15}=6{}^\circ \times 7=42{}^\circ \] |
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