A) \[\text{si}{{\text{n}}^{-\text{1}}}\text{ }\left( \text{tan r} \right)\]
B) \[{{\sin }^{-1}}(\tan i)\]
C) \[\text{si}{{\text{n}}^{-1}}\text{ }\left( \text{tan r}' \right)\]
D) \[{{\sin }^{-1}}(\tan (i+r)\]
Correct Answer: B
Solution :
From figure,\[i=r\]and\[r=90-r\] \[\mu =\frac{\sin r'}{\sin r}=\frac{\sin (90-r)}{\sin r}\] \[=\frac{\cos r}{\sin r}=\frac{1}{\tan r}\] We know, \[\mu =\frac{1}{\sin C}\] \[\therefore \] \[\frac{1}{\sin C}=\frac{1}{\tan r}\] \[\Rightarrow \] \[\sin C=\tan r\] \[\therefore \] \[C={{\sin }^{-1}}(\tan r)={{\sin }^{-1}}(\tan i)\]You need to login to perform this action.
You will be redirected in
3 sec