Question

数学题求？！

数学题求？！数学题求解在三角形abc中角abc的对边分别为abc，sina+sinb/sinc=cosa+cosb／cosc 求角c， 若a=2 b=3求abc面积

Answer 1

(sinA+sinB)/sinC=(cosA+cosB)/cosC

(cosA+cosB)sinC=(sinA+sinB)cosC

cosAsinC+cosBsinC=sinAcosC+sinBcosC

sinAcosC-cosAsinC=sinCcosB-cosCsinB

sin(A-C)=sin(C-B)

A、B、C为三角形内角，A、C、B成等差数列

2C=A+B

A+B+C=180°

3C=180°

C=60°

1. 面积1/2*absinc=0.5*2*3*3^1/2=(3*3^1/2)/2

Answer 2

解：
(1)
由正弦定理、余弦定理得：
(a+b)/c=[(b²+c²-a²)/(2bc) +(a²+c²-b²)/(2ac)]/[(a²+b²-c²)/(2ab)]
整理，得：a³+b³-ac²-bc²=0
(a+b)(a²-ab+b²)-c²(a+b)=0
(a+b)(a²+b²-c²-ab)=0
a+b恒>0，因此只有a²+b²-c²-ab=0
a²+b²-c²=ab
由余弦定理得：cosC=(a²+b²-c²)/(2ab)=ab/(2ab)=½
C为三角形内角，C=π/3
(2)
S△ABC=½absinC
=½·2·3·sin(π/3)
=½·2·3·(√3/2)
=3√3/2

Answer 3

设A(x1.y1),B(x2,y2),x1+x2=2*2=4,y1+y2=2x1^2/16+y1^2/4=1x2^2/16+y2^2/4=1相减得到:(x1-x2)(x1+x2)/16+(y1+y2)(y1-y2)/4=0即K(AB)=(y1-y2)/(x1-x2)=-(4/16)/(2/4)=-1/2那么AB方程是y-1=-1/2(x-2),即有y=-x/2+2代入到椭圆方程中有:x^2/16+(x^2/4+4-2x)/4=12x^2-8x=02x(x-4)=0x1=0,x2=4y1=2,y2=0AB=根号(2^2+4^2)=2根号5