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Grade: 12

                        

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

4 years ago

Answers : (4)

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
4 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
3 years ago
Venkata Rakesh Y
9 Points
							
3sinP+4cosQ = 6 ----> (1)
3cosP+4sinQ = 1 -----> (2)
(1)²+(2)² =>   (3sinP+4cosQ)²+(3cosP+4sinQ)² = 36+1
                       9sin²P + 16cos²Q + 24sinPcosQ + 9cos²P + 16 sin²Q + 24sinQcosP = 37
                       9(sin²P+cos²P) + 16(cos²Q+sin²Q) + 24(sinPcosQ+cosPsinQ) = 37
                       9(1)+16(1)+24sin(P+Q) = 37
                       25+24sin(P+Q) = 37
                       24sin(P+Q) = 12
                        sin(P+Q) = ½
                       P+Q = 30° or 150° 
(1)=>  3sinP+4cosQ = 6
           3sinP = (6 – 4cosQ)
            sinP = (6 –  4cosQ)/3
            cosQ max value is 1
            -cosQ min value is -1
            6-4cosQ min value is 6-4 = 2
            sinP >= 2/3 >= 0.67
             As SinP >= 2/3 > 1/2
                  P must be greater than 30° (Because Sine is an increasing function in [0° ,90° ] )
         If P+Q = 30° ( P,Q > 0 because they are angles of a triangle)
            then P has a maximum value of less than 30°
            But from the above P must be greater than 30°
            Hence P+Q is not 30° .
            Thus P+Q = 150°
4 months ago
Kushagra Madhukar
askIITians Faculty
605 Points
							
Dear student,
Please find the solution to your problem below.
 
3sinP+4cosQ = 6
3cosP+4sinQ = 1
Square and add the equations, we have,
                       (3sinP+4cosQ)²+(3cosP+4sinQ)² = 36+1
                       9sin²P + 16cos²Q + 24sinPcosQ + 9cos²P + 16 sin²Q + 24sinQcosP = 37
                       9(sin²P+cos²P) + 16(cos²Q+sin²Q) + 24(sinPcosQ+cosPsinQ) = 37
                       9(1)+16(1)+24sin(P+Q) = 37
                       25+24sin(P+Q) = 37
                       24sin(P+Q) = 12
                        sin(P+Q) = ½
                       P+Q = 30° or 150° 
 Now,  3sinP + 4cosQ = 6
           3sinP = (6 – 4cosQ)
            sinP = (6 –  4cosQ)/3
            cosQ max value is 1
            -cosQ min value is -1
            6 – 4cosQ min value is 6 – 4 = 2
            sinP >= 2/3 >= 0.67
             As SinP >= 2/3 > 1/2
                  P must be greater than 30° (Because Sine is an increasing function in [0° ,90° ] and sin 45o = ½ )
But, for that P  must be greater than 45o, in which case Q will become negative, which is not possible in case of a triangle.
 
Hence, P + Q = 150o
Hence, R = 180o – (P + Q) = 30o
(Kudos to Venkata Rakesh for getting at the right answer)
 
Thanks and regards,
Kushagra
4 months ago
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