標題:
Geometry
發問:
AB//CD. AD intersects CB at M. AB = 5 cm. CB = 8 cm. AD = 10 cm. Angle ABC = 40 degree. Find AM and CM.(Ans. correct to 3 sig. fig.)
最佳解答:
let AM = x and BM = y then MD = 10 - x and CM = 8 - y Since AM / MD = BM / CM, x / (10 - x) = y / (8 - y) 8x - xy = 10y - xy y = 4x / 5 In triangle ABM, by cosine formula, x^2 = 5^2 + y^2 - 2(5)(y)cos 40 x^2 = 5^2 + (4x / 5)^2 - 2(5)(4x / 5)cos 40 x^2 = 25 + 16x^2 / 25 - (8cos 40)x 25x^2 = 625 + 16x^2 - (200cos 40)x 9x^2 + (200cos 40)x - 625 = 0 x = 3.40 or -20.4(rejected) y = 4(3.40) / 5 = 2.72 therefore, AM = 3.40 cm and CM = 8 - 2.72 = 5.28 cm
其他解答:
I found the following method workable: Through A draw a line // to DC and cutting DC produced at E. Apply cosine rule to triangle ADE to find the length of EC.....
Geometry
發問:
AB//CD. AD intersects CB at M. AB = 5 cm. CB = 8 cm. AD = 10 cm. Angle ABC = 40 degree. Find AM and CM.(Ans. correct to 3 sig. fig.)
最佳解答:
let AM = x and BM = y then MD = 10 - x and CM = 8 - y Since AM / MD = BM / CM, x / (10 - x) = y / (8 - y) 8x - xy = 10y - xy y = 4x / 5 In triangle ABM, by cosine formula, x^2 = 5^2 + y^2 - 2(5)(y)cos 40 x^2 = 5^2 + (4x / 5)^2 - 2(5)(4x / 5)cos 40 x^2 = 25 + 16x^2 / 25 - (8cos 40)x 25x^2 = 625 + 16x^2 - (200cos 40)x 9x^2 + (200cos 40)x - 625 = 0 x = 3.40 or -20.4(rejected) y = 4(3.40) / 5 = 2.72 therefore, AM = 3.40 cm and CM = 8 - 2.72 = 5.28 cm
其他解答:
I found the following method workable: Through A draw a line // to DC and cutting DC produced at E. Apply cosine rule to triangle ADE to find the length of EC.....
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