Re: [問題] 請問一題數學
: 三角函數
bc(√3 /2) = 15a----I
ab*sinC = 35c----II
a*sinB = 35-----III
c*sinB = 15-----IV
by III & IV --> a/c = 7/3 --> a : c = 7 : 3
take a = 7c/3 into I ---> b√3 /2 = 35 --> b = 70/√3
II---> (7c/3)*(70/√3)sinC = 35c
---> (7/3)*(2/√3)sinC = 1 --> sinC = 3√3/14
sin(B+C) = √3 /2 ---> sinBcosC + sinCcosB = 13*sinB/14 + (cosB)*3√3/14
---> 7√3 = 13sinB + (cosB)*3√3
---> [7√3 - 13sinB]^2 = 27(1 - ((sinB)^2)
we can get sinB = 5√3/14 or 4√3/7(not a solution, check by yourself ^_^)
70^2 /3 = 58c^2 /9 - (14c^2/3)*(11/14) = 58c^2 /9 - 11c^2 /3 --> solve c ^_^
and then you can solve the area by yourself ^_^
好啦我有事如果你沒解出來我晚點再解
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※ 編輯: j0958322080 (140.115.223.6), 11/01/2015 12:29:54
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11/01 12:43, , 1F
11/01 12:43, 1F
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