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Grade: 12
        In a triangle PQR if 3sinP+4cosQ=6 and 4sinQ+3cosP=1 then angle R is equal to
3 years ago

Answers : (2)

AKASH
21 Points
							
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 150
if p+q=30,
then sin p
3sin p+4cosq
which is incorrect according to the given question
hence p+q=150
and R=30 DEGREES.
HOPE THIS IS CLEAR
3 years ago
pavithra Ayyanar
21 Points
							
If you solve the given equations we get sin(p+q) = ½
therefore the principal value = 30 degree
General 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, 90
Therefore the correct answer is r = 30 degree
 
  1. Ayyanar, Puducherry
2 years ago
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