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In a triangle PQR if 3sinP+4cosQ=6 and 4sinQ+3cosP=1 then angle R is equal to
square both the equations and add both them.u will get 25+ 24(sinpcosq+cospsinq)=37 24(sin(p+q))=12 sin(p+q)=0.5 hence, p+q=30 or 150if p+q=30,then sin p3sin p+4cosqwhich is incorrect according to the given questionhence p+q=150and R=30 DEGREES.HOPE THIS IS CLEAR
If you solve the given equations we get sin(p+q) = ½therefore the principal value = 30 degreeGeneral solutiions = npi+(-1)^n (principal value)p+q = 30, 150, …....If p+q =30 then r = 150 This is impossible because of one of the triangle must be 90 degree in trigonometry. If p+q = 150 then r = 30 This is possible because of the angle is 30, 60, 90Therefore the correct answer is r = 30 degree Ayyanar, Puducherry
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