A) lies between \[\sqrt{2}\] and 1
B) lies between 2 and \[\sqrt{2}\]
C) is less than 1
D) is greater than 2
Correct Answer: B
Solution :
\[\mu =\frac{\sin \left( \frac{A+\delta m}{2} \right)}{\sin A/2}\] \[=\frac{\sin \left( \frac{A+A}{2} \right)}{\sin A/2}\] In given situation value of A varies from 0 to \[90{}^\circ \]. So, \[{{\mu }_{\min }}=2\cos \frac{{{90}^{o}}}{2}=\sqrt{2}\]and\[{{\mu }_{\max }}=2\cos {{0}^{o}}=2\]You need to login to perform this action.
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