標題:

Angle of triangle

發問:

For triangle ABC, angle A is 65 degree. M is a point inside triangle ABC such that angle AMB = 105 dgree , angle AMC = 130 degree and angle MAC = 29 degree. Find angle MCB.

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http://img87.imageshack.us/img87/3937/tria.jpg 圖片參考:http://img87.imageshack.us/img87/3937/tria.jpg Angle BAM = Angle A – Angle MAC = 65 – 29 = 36 Angle MCA = 180 – 130 – 29 = 21 Angle BMC = 360 – 105 – 130 = 125 Applying sine rules: Consider triangle ABM, x / sin 36 = c / sin 105 … (1) Consider triangle BCM, x / sin Y = a / sin 125 … (2) Consider triangle ABC, a/ sin 65 = c / sin(21 + Y) c / a = sin(21 + Y) / sin 65 … (3) (1)/(2) = > sinY / sin 36 = c sin 125 / a sin 105 Use (3) => sinY / sin 36 = (sin 125 / sin 105)[sin(21 + Y) / sin 65] sinY sin 65 = (sin 36)(sin 125 / sin 105)sin(21 + Y) 0.9063sinY = 0.4985sin(21 + Y) = 0.4985(sinYcos21 + sin21cosY) 0.9063sinY – 0.4654sinY = 0.1786cosY 0.4409sinY = 0.1786cosY sinY / cosY = 0.1786/0.4409 tanY = 0.4051 Y = 22.05

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