Question

(理)已知a、b、c是△ABC三边长,关于x的方程ax²-2x-b=0(a＞c＞b)的两根之差的平方等于4,△ABC的面积S=,c=7.

(1)求角C;

(2)求a、b的值.

(文)在△ABC中,sinA+cosA=,AC=2,AB=3.

(1)求tanA的值;

(2)求△ABC的面积.

Answer 1

(理)解:(1)设x₁、x₂为方程ax²x-b=0的两根,

则x₁+x₂=,

x₁·x₂=.                                                                 

∴(x₁-x₂)²=(x₁+x₂)²-4x₁x₂==4.

∴a²+b²-c²=ab.                                                              

又cosC=,

∴cosC=.∴C=60°.                                                        

(2)由S=absinC=10,

∴ab=40.                                                               ① 

由余弦定理c²=a²+b²-2abcosC,

即c²=(a+b)²-2ab(1+cos60°).

∴7²=(a+b)²-2×40×(1+).

∴a+b=13.                                                             ② 

由①②,得a=8,b=5.                                                         

(文)解:(1)∵sinA+cosA=cos(A-45°)=,

∴cos(A-45°)=.                                                          

又∵0°＜A＜180°,∴A-45°=60°,

故A=105°.                                                                

∴tanA=tan(45°+60°)=.                                      

(2)∵sinA=sin(45°+60°)

=sin45°cos60°+cos45°sin60°

=,                                                              

∴S_△ABC=AB·AC·sinA

=×2×3×

=().