In triangle ABC, A=40degree B=60degree, and its area is 80cm^2
(a) Find the ration a:b:c in terms of sines of angles
(b) Which side has the longest length
(c) Hence find the length of the longest side
I just wanna derive the answer of (c) , thank you
Trigonometry F.4 ( not A. Maths )
2007-04-20 1:58 am
回答 (2)
2007-04-20 2:51 am
✔ 最佳答案
(a) Find the ration a:b:c in terms of sines of anglesa/sin40 = b/sin60 = c/sin80
so, a:b:c = sin40 : sin60 : sin80
(b) Which side has the longest length
c
(c) Hence find the length of the
area = 80 = 0.5 (ksin40)(ksin60)sin8 0 = 0.5*k^2*sin40sin60sin80
160/sin40sin60sin80 = k^2
k = 17.1 cm
So, longest side = 17.1 * sin 80 = 16.8 cm
2007-04-20 2:37 am
(a) a/sin40 = b/sin60 = c/sin80, where a=BC, b=AC, c=AB
(b) sin40< sin60 < sin80, therefore, a>b>c.
i.e. BC (a) is the longest.
(c) Area = 80 = 0.5 * a * b * sin80
a^2 = 160sin40/(sin80sin60) (b= asin60/sin40)
a = 10.98 cm
(b) sin40< sin60 < sin80, therefore, a>b>c.
i.e. BC (a) is the longest.
(c) Area = 80 = 0.5 * a * b * sin80
a^2 = 160sin40/(sin80sin60) (b= asin60/sin40)
a = 10.98 cm
參考: me
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