question_answer

The\[\Delta PQR\]is inscribed in the circle\[{{x}^{2}}+{{y}^{2}}=25.\] If\[Q\]and\[R\]have coordinates (3, 4) and (-4, 3), respectively. Then,\[\angle QPR\]is equal to

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\[\angle QPR=\frac{{{90}^{o}}}{2}={{45}^{o}}\].