Question

在△ABC中,已知A=45°,COSB=4/5,（1）求COSC的值,（2）若BC=10,D为AB中点,求CD的长

Answer 1

1

A=45°

sinA=cosB=√2/2

cosB=4/5

sinB=√(1-cos^2B)=3/5

cosC=cos(180-(A+B))=-cos(A+B)=-cosAcosB+sinAsinB

=-√2/2*4/5+√2/2*3/5

=-√2/10

2

BC/sinA=AC/sinB

AC=sinB/sinA*BC=3/5/√2/2 *10=6√2

AB^2=BC^2+AC^2-2*BC*ACcosC=10^2+(6√2)^2-2*10*6√2*(-√2/10)=196

AB=13

BD=13/2

CD^2=BC^2+BD^2-2BC*BD*cosB=10^2+(13/2)^2-2*10*13/2*4/5=153/4

CD=√153/2