In the given figure, \[\mathbf{PS}=\mathbf{SQ}\], then \[\mathbf{sin}\

1.	In the given figure, \[\mathbf{PS}=\mathbf{SQ}\], then \[\mathbf{sin}\theta =\]?
A.	\[\frac{4{{Q}^{2}}-3{{p}^{2}}}{p}\]
B.	\[\frac{p}{4{{Q}^{2}}-3{{p}^{2}}}\]
C.	\[\frac{\sqrt{4{{Q}^{2}}-3{{p}^{2}}}}{p}\]
D.	1
Answer» C. \[\frac{\sqrt{4{{Q}^{2}}-3{{p}^{2}}}}{p}\]

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In the given figure, \[\mathbf{PS}=\mathbf{SQ}\], then \[\mathbf{sin}\theta =\]?

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