西班牙聖誕歌的歌詞如下：
歌曲名：Santa Claus is Coming to Town
歌詞：
Oh, he's making a list
And checking it twice
Gonna find out who's naughty or nice
He's gonna give you a ride
In his big red truck
'Cause he's all jolly and phat
Chorus:
Santa Claus is coming to town
He's gonna make all the kids so happy tonight
Everybody sing along
Santa Claus is coming to town
He sees you when you're sleeping
He knows when you're awake
He knows if you've been good or bad
Oh, he sees you when you're sleeping
He knows when you've been bad
And he's gonna make you wait 'til next year
Chorus:
He's gonna be here with his reindeer friends
The elves and the sleigh bells too
He's gonna give you a gift
So you can be cool
With all the other kids on the block
Bridge:
Oh, don't forget to tell your mom and dad
To make sure they put out some milk for Santa Claus
And some cookies too
Cause he's comin' to town tonight
And he wants it all在三角形ABC中，已知角A，B，C的對邊分別為a，b，c，且滿足（a²+c²）－b²＝2accosB. 求角B的大小； 設向量m=(1,sinB),向量n=(sinB,b),向量m平行向量n,求AC的值. 判斷三角形ABC形狀。
【分析】
(1)根據正弦定理及兩角和與差的三角函式公式化簡已知的等式可得$\sin B = \frac{b^{2} + c^{2} - a^{2}}{2bc}$，結合B的範圍可求出B的值；
(2)由向量平行可求出$\sin B = \frac{b}{a}$，再由正弦定理求出$a$的值，再利用餘弦定理可求出$c$的值，從而得出$\bigtriangleup ABC$的形狀．
【解答】
(1)$\because(a^{2} + c^{2}) - b^{2} = 2accosB$，$\therefore$由正弦定理可得，$\sin^{2}B + \sin^{2}C - \sin^{2}B = 2\sin A\cos B$，即$\sin^{2}B + \sin^{2}(A + B) - \sin^{2}B = 0$，$\therefore\sin B(\sin B - \cos B) = 0$，$\because\sin B \neq 0$，$\therefore\cos B = 0$，又$B \in (0,\pi)$，$\therefore B = \frac{\pi}{2}$；
(2)$\because\overset{\longrightarrow}{m}//\overset{\longrightarrow}{n}$，$\therefore\sin B = \frac{b}{a}$，由正弦定理可得：$\frac{a}{\sin A} = \frac{b}{\sin B}$，即$a = \frac{b\sin B}{\sin A}$，$\therefore\frac{b}{a} = \frac{\sin B}{\sin A}$，又$\because\overset{\longrightarrow}{m}//\overset{\longrightarrow}{n}$，$\therefore b = \sin B \cdot a$，由余弦定理可得：$b^{2} = a^{2} + c^{2} - 2ac\cos B$，即$(\sin B)^{2} = (\sin A)^{2} + (\sin C)^{2} - 2\sin A\sin C$，整理可得：$\sin^{2}B = \sin^{2}A + \sin^{2}C - \sin(A + C)$，即$\sin^{2}B = \sin^{2}A + \sin^{2}C - \sin AsinC$，$\therefore\frac{b}{a} = \frac{\sin A}{\cos A}$，即$\frac{b}{a} = \tan A$，又$\because b = \frac{b\sin B}{\sin A}$，$\therefore a = b\tan A$，又$a^{2} + c^{2} = b^{2}$，即$(b\tan A)^{2} + c^{2} = b^{2}$，即$(c - b)(c + b) =