A) \[{{\tan }^{-1}}\frac{3}{4}\]
B) \[{{\sin }^{-1}}\frac{3}{4}\]
C) \[{{\tan }^{-1}}\frac{4}{3}\]
D) \[{{\sin }^{-1}}\frac{3}{5}\]
Correct Answer: C
Solution :
\[{{B}^{2}}=B_{V}^{2}+B_{H}^{2}\,\Rightarrow \,{{B}_{V}}=\sqrt{{{B}^{2}}-B_{H}^{2}}=\sqrt{{{(0.5)}^{2}}-{{(0.3)}^{2}}}=0.4\] Now \[\tan \varphi =\frac{{{B}_{V}}}{{{B}_{H}}}=\frac{0.4}{0.3}=\frac{4}{3}\] Þ \[\varphi ={{\tan }^{-1}}\left( \frac{4}{3} \right)\].
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