Question

(本小題滿分14分)
已知a、b、c是△ABC中A、B、C的對(duì)邊, 關(guān)于x的方程b (x 2 + 1 ) + c (x 2– 1 ) –2ax =" 0" 有兩個(gè)相等的實(shí)根, 且sinCcosA – cosCsinA="0," 試判定△ABC的形狀.

Answer 1

等腰直角三角形

∵(b + c)x 2 –2ax + (b – c ) = 0有相等實(shí)根， 
∴⊿=" 4a" 2 – 4( b + c )(b – c) = 0，                                    3分
∴ a 2 + c 2 – b 2 = 0， 
∴ B = 90° .                                                       3分
又sinCcosA – cosCsinA="0" ,
得 sin (C – A) =" 0,"                                                 3分
∵–< C – A <  .                                               2分
∴ A = C.
∴△ABC是B為直角的等腰直角三角形.                             3分