A) \[10{}^\circ \]
B) \[30{}^\circ \]
C) \[45{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: D
Solution :
[d] Given \[\left( \frac{{{\sin }^{2}}x+1}{\sin x} \right)=\frac{7}{2\sqrt{3}}\] \[\Rightarrow 2\sqrt{3}{{\sin }^{2}}x\ -7sinx+2\sqrt{3}=0\] \[\operatorname{sinx}=\frac{7\pm \sqrt{49-48}}{4\sqrt{3}}\] \[=\frac{7\pm \sqrt{1}}{4\sqrt{3}}\] \[=\frac{8}{4\sqrt{3}},\frac{6}{4\sqrt{3}}\] \[=\frac{2}{\sqrt{3}},\frac{\sqrt{3}}{2}\] Clearly, \[x=60{}^\circ \]
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