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標題:

F. 4 A. Maths(compound angle)[20 points!!] urgent!

發問:

1.It is given that sin A= 3/5 and cos(A+B)= -7/25, where A and B are acute angles. Without solving for A and B, find the value of sin B. 2.It is given that tanA=40/9 and tan(A-B)= 5/12. Without solving for A and B, find the values of (a) tand B, and (b) tan(2A-B) I need steps please.

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最佳解答:

1) sinA = 3/5 所以對邊係3 斜邊係5 鄰邊= 開方 (5^2 - 3^2) = 4 so cosA = 4 / 5 cos(A + B) = -7 / 25 cosAcosB - sinAsinB = -7 / 25 4/5 * cosB - 3 / 5 sinB = -7 / 25 20cosB - 15sinB = -7 (全條式乘25) ------第一式 因為 cos (A + B) = -7/25 用番相同方法.. 計到對邊係24 因為兩隻都係acute angle so 0<180 SO sin (A+B) > 0 so sin(A + B) = 24/25 sin(A + B) = 24/25 sinAcosB + sinBcosA = 24/25 3/5 * cosB + 4/5sinB = 24/25 15cosB + 20sinB = 24 15cosB = 24-20sinB cosB = (24-20sinB)/15 put in to 一式 20 * (24 - 20sinB)/15 - 15sinB = -7 (96 - 80sinB)/3 - 15sinB = -7 96 - 80sinB - 45sinB = -21 (全條式乘3) 96 - 125sinB = -21 sinB = 117 / 125 2)a) tanA = 40/9 tan(A-B) = 5/12 tan(A-B) = (tanA - tanB) / (1+tanAtanB) = 5/12 (40/9 - tanB) / (1+ 40/9 * tanB) = 5/12 (代個tanA = 40/9入去) 12 * (40/9 - tanB) = 5* (1+ 40/9 * tanB) 480 / 9 - 12tanB = 5 + 200tanB / 9 480 - 108tanB = 45 + 200tanB (全式乘9) 435 = 308tanB tanB = 435 / 308 2)b) tan2A = 2tanA / (1- tanA ^2 ) = (2 * 40/9) / ( 1 - (40/9)^2) = - 720 / 1519 (禁計數機) tan(2A-B) = (tan2A - tanB) / (1+tan2AtanB) = (-720/1519 - 435/308) / (1 + -720 / 1519 * 435/308) = - 5.71 (3sig.fig.)

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