Matematika

Pertanyaan

diketahui sin A=5/13    dan tan B=-24/7 denagn A sudut lancip dan B sudut tumpul.
tentukan nilai dari :
a. sin A cos B -- cos A sin B
b. cos A cos B + Sin A sin B
tan A + tan B
  1-tan A tan B

1 Jawaban

  • sinA = 5/13, cosA = 12/13, tanA = 5/12
    tanB = -24/7, sinB = 24/25, cosB = -24/25

    a.
    sinA cosB - cosA sinB
    = 5/13 (-24/25) - 12/13 24/25
    = -120/325 - 288/325
    = -408/325

    b.
    cosA cosB + sinA sinB
    = 12/13 (-24/25) + 5/13 24/25
    = -288/325 + 120/325
    = -168/325

    c.
    (tanA + tanB) / (1 - tanA tanB)
    = (5/12 + (-24/7)) / (1 - 5/12 (-24/7))
    = (-253/84) / (1 - (-120/84))
    = (-253/84) / (1 + 120/84)
    = (-253/84) / (204/84)
    = -253/84 * 84/204
    = -21252/17136

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