A) \[{{a}^{2}}\]
B) \[1\]
C) \[2a\]
D) \[a\]
Correct Answer: D
Solution :
[d] \[PQ=\sqrt{{{(0-a\sin \phi )}^{2}}+{{(-a\cos \phi -0)}^{2}}}\] |
\[=\sqrt{{{a}^{2}}{{\sin }^{2}}\phi +{{a}^{2}}{{\cos }^{2}}\phi }=a\sqrt{{{\sin }^{2}}\phi +{{\cos }^{2}}\phi }=a\sqrt{1}=a\] |
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