三角形ABC的内角A,B,C的对边为a,b,c,若cosA=4/5,cosC=5/13,a=1,则b=
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发布时间:2024-02-12 16:45
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时间:2024-02-14 23:06
sinA = 3/5, sinC = 12/13
sinB = sin(A+C)
= sinAcosC + cosAsinC
= 3/5 * 5/13 + 4/5 * 12/13
= 63/65
又因为:a/sinA = b/sinB
所以:b = a/sinA * sinB
= a/(3/5) * 63/65
= 5/3 * 63/65
= 21/13