question_answer

A triangle \[ABC\] satisfies the relation \[2\sec 4C+{{\sin }^{2}}2A+\]\[\sqrt{\sin B}=0\] and a point \[P\] is taken on the longest side of the triangle such that it divides the side in the ratio \[1:3\]. Let \[Q\] and \[R\] be the circum centre and orthocentre of A ABC. If \[QR:QP:RP=1:\alpha :\beta \], then the value of\[{{\alpha }^{2}}+{{\beta }^{2}}\].

A) \[9\]                                     

B) \[8\]

C) \[6\]                                     

D) \[7\]

Correct Answer: A

Solution :

\[2\sec 4C+{{\sin }^{2}}2A+\sqrt{\sin B}=0\] \[A={{45}^{o}},\,\,B={{90}^{o}}\]and\[C={{45}^{o}}\] Let\[AQ=a\], then\[BP=\frac{a}{2}\], \[PQ=\frac{a}{2}\]and\[QR=a\] \[\therefore \]\[PR=\sqrt{{{a}^{2}}+\frac{{{a}^{2}}}{4}}=\frac{\sqrt{5}a}{2}\] \[\therefore \]\[1:\alpha :\beta =\frac{a}{2}:a:\frac{\sqrt{5}a}{2}=1:2:\sqrt{5}\] \[\therefore \]\[\alpha =2\]and\[\beta =\sqrt{5}\]\[\therefore \]\[{{\alpha }^{2}}+{{\beta }^{2}}=9\]