Diketahui sin A=5/13    dan tan B=-24/7 denagn A sudut lancip dan B sudut tumpul.

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tentukan nilai dari :
a. sin A cos B — cos A sin B
b. cos A cos B + Sin A sin B
c tan A + tan B
  1-tan A tan B

Diketahui sin A=5/13    dan tan B=-24/7 denagn A sudut lancip dan B sudut tumpul.

Jawaban Terkonfirmasi

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