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

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4 years ago

```							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
```
4 years ago
```							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
```
3 years ago
```							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 605 Points
```							Dear student,Please find the solution to your problem below. 3sinP+4cosQ = 63cosP+4sinQ = 1Square 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 = 150oHence, R = 180o – (P + Q) = 30o(Kudos to Venkata Rakesh for getting at the right answer) Thanks and regards,Kushagra
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4 months ago
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