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

Question

AKASH · Answer

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 · Answer

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 Ayyanar, Puducherry

Venkata Rakesh Y · Answer

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 °

Kushagra Madhukar · Answer

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 45 o = ½ ) But, for that P must be greater than 45 o , in which case Q will become negative, which is not possible in case of a triangle. Hence, P + Q = 150 o Hence, R = 180 o – (P + Q) = 30 o (Kudos to Venkata Rakesh for getting at the right answer) Thanks and regards, Kushagra

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

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

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