A) \[58{}^\circ \text{ }15'\]
B) \[57{}^\circ \text{ }19'\]
C) \[58{}^\circ \text{ }19'\]
D) \[57{}^\circ \text{ }12'\]
Correct Answer: C
Solution :
Given\[\angle BOC={{90}^{o}}\] \[\therefore \] \[\angle i+{{90}^{o}}+\angle i={{180}^{o}}\] \[\Rightarrow \] \[\angle (i+r)={{90}^{o}}\] \[\Rightarrow \] \[r=(90-i)\] Now \[\mu =\frac{\sin i}{\sin r}=\frac{\sin i}{\sin (90-i)}\] \[=\frac{\sin i}{\cos i}=\tan i\] \[\therefore \]Angle of incidence,\[i={{\tan }^{-1}}(\mu )\] \[={{\tan }^{-1}}(1.52)\] \[\mathbf{=5}{{\mathbf{8}}^{\mathbf{o}}}\mathbf{19'}\]You need to login to perform this action.
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