A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
Correct Answer: C
Solution :
Let\[O\]is the point on centre and\[P\]is the point on circumference. Therefore, angle\[QOR\]is double the angle\[QPR\]. So, it is sufficient to find the angle\[QOR\]. Now, slope of\[OQ\], \[{{m}_{1}}=\frac{4-0}{3-0}=\frac{4}{3}\] Slope of\[OR\], \[{{m}_{2}}=\frac{3-0}{-4-0}=-\frac{3}{4}\] Again, \[{{m}_{1}}{{m}_{2}}=-1\] Therefore, \[\angle QOR={{90}^{o}}\] Which implies that\[\angle QPR=\frac{{{90}^{o}}}{2}={{45}^{o}}\].You need to login to perform this action.
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