MathematicsMedium19×since 2002Q5026In a ΔPQR, \Delta PQR,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu}ΔPQR, If 3 sin P+4 cos Q=63{\mkern 1mu} \sin {\mkern 1mu} P + 4{\mkern 1mu} \cos {\mkern 1mu} Q = 63sinP+4cosQ=6 and 4sinQ+3cosP=1,4\sin Q + 3\cos P = 1,4sinQ+3cosP=1, then the angle R is equal to :A5π6{{5\pi } \over 6}65πBπ6{{\pi } \over 6}6πCπ4{{\pi } \over 4}4πD3π4{{3\pi } \over 4}43πCheck answerSkip