It is given that in △ABC，a：b = 4：5. Based on this，Sam made the deductions below：
From the sine formula，we have
sinA：sinB = 4：5
Hence，sin²A：sin²B = 16：25
(1-sin²A)：(1-sin²B) = (1-16)：(1-25)
cos²A：cos²B = -15：-24 = 5：8
Therefore，cosA：cosB = √5：√8
(a) His final conclusion that‘cosA：cosB = √5：√8’is wrong. Explain.
(b) If it is known in addition that a：b：c = 4：5：6，can you find the ratio
cosA：cosB：cosC?
f4 maths (Trigonometric Formulas)
2006-10-17 1:33 am
回答 (2)
2006-10-17 1:47 am
✔ 最佳答案
(a)sin²A：sin²B = 16：25
(1-sin²A)：(1-sin²B) = (1-16)：(1-25)
this step is wrong, so his final conclusion that‘cosA：cosB = √5：√8’is wrong
(b)
a：b：c = 4：5：6
set a=4k, b=5k, c=6k
cosA
=(b^2+c^2-a^2)/2bc
=(25+36-16)/60
=3/4
cosB
=(a^2+c^2-b^2)/2ac
=(16+36-25)/48
=27/48
=9/16
cosC
=(a^2+b^2-c^2)/2ab
=(16+25-36)/40
=5/40
=1/8
cosA：cosB：cosC
=3/4: 9/16: 1/8
=12:9:2
2006-10-17 1:45 am
(a)sin²A：sin²B = 16：25 doesn't means (1-sin²A)：(1-sin²B) = (1-16)：(1-25)
it means (1-sin²A)：(1-sin²B) = (1-16x)：(1-25x) where 16x=sin²A
(b) a：b：c = 4：5：6 therefore a=4x, b=5x,c=6x
cosA=(b^2+c^2-a^2)/2bc=((5x)^2+(6x)^2-(4x)^2)/2(5x)(6x)=3/4
cosB=(a^2+c^2-b^2)/2ac=9/16
cosC=(a^2+b^2-c^2)/2ab=1/8
cosA：cosB：cosC=(3/4) : (9/16): (1/8)
=((3/4)*16) :((9/16)*16) : ((1/8)*16)
=12:9:2
it means (1-sin²A)：(1-sin²B) = (1-16x)：(1-25x) where 16x=sin²A
(b) a：b：c = 4：5：6 therefore a=4x, b=5x,c=6x
cosA=(b^2+c^2-a^2)/2bc=((5x)^2+(6x)^2-(4x)^2)/2(5x)(6x)=3/4
cosB=(a^2+c^2-b^2)/2ac=9/16
cosC=(a^2+b^2-c^2)/2ab=1/8
cosA：cosB：cosC=(3/4) : (9/16): (1/8)
=((3/4)*16) :((9/16)*16) : ((1/8)*16)
=12:9:2
收錄日期: 2021-04-12 22:51:49
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