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High school mathematics compulsory 5 exam
According to sine theorem, b/a=sinB/sinA.

Therefore, cosA/cosB=sinB/sinA, that is, sinAcosA=sinBcosB and sin2A=sin2B.

From b/a=4/3, a and b are not equal, that is, a and b are not equal.

So 2A+2B=π, A+B=π/2, that is, △ABC is a right triangle and C=π/2.

Let a=3k and b=4k.

A2+B2 = (3k) 2+(4k) 2 = 25k2 = C2 =100, then k 2 = 4 and k=2.

a=6、b=8 .

Let the inscribed circle radius r of △ABC.

△ area of △ABC = (1/2) r (a+b+c) =12r = (1/2) ab = 24, r=2.