math Question~~~

a)AD = CD (equilateral triangle sides)
DE = DE (common)
ㄥADE = ㄥCDE = 60° (equilateral triangle angles)△ADE ≡ △CDE (S.A.S.)So AE = EC
bi)The required angle = ㄥACE ,
let AC = 1 , then AE = CE = sin 60° = √3 / 2cos ㄥACE
= (AC² + CE² - AE²) / (2 AC CE)
= 1 / ( 2 (√3 / 2) )
= √3 / 3ㄥACE = 54.7°ii)The required angle = ㄥAEC ,sin(ㄥAEC/2) = (AC/2) / AE = (1/2) / (√3 / 2) = √3 / 3ㄥAEC/2 = 35.264...ㄥAEC = 70.5°


2011-10-10 23:11:04 補充：
Alternatively of bii) :

ㄥAEC = 180 - 2ㄥACE = 180 - 2*54.7356 = 70.5°