Question

已知在三角形abc中，sinA(sinB+cosB)-sinC=0,sinB+cosC=0,求A,B的值

Answer 1

解：由已知sinA(sinB+cosB)– sinC = 0，所以sinA(sinB+cosB)= sinC = sin(A + B) = sinAcosB + cosAsinB，移项可得sinAsinB – cosAsinB = 0，所以sinA – cosA= 0，所以sinA = cosA，因此cosA ≠ 0，所以tanA = 1，可得A = kπ + arctan1 = kπ + π/4，k∈Z，而A是三角形ABC的内角，所以只能是k = 0，对应的A = π/4，所以C = π – A –B = 3π/4 – B，代入sinB + cosC = 0可得sinB + cos(3π/4 – B) = 0，展开得sinB + cos(3π/4)cosB + sin(3π/4)sinB = 0，即(1 + √2/2)sinB – (√2/2)cosB = 0，所以(1 + √2/2)sinB = (√2/2)cosB，进而tanB = (√2/2)/(1 + √2/2) = 1/(√2 + 1) = √2– 1，所以B = kπ + arctan(√2 – 1) = kπ + π/8，k∈Z，而B是三角形ABC的内角，所以只能是k = 0，对应的B = π/8 ；
综上所述，A = π/4，B = π/8 。