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

AKASH
21 Points
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 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
pavithra Ayyanar
21 Points
3 years ago
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
Venkata Rakesh Y
9 Points
11 months 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°
11 months ago
Dear student,

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