在三角形ABC中，a b c分别是角A B C 的对边，cosA等于5分之根号5.tanB=3.（1）求角C的值？(2)当a=4时，...

Question

在三角形ABC中，a b c分别是角A B C 的对边，cosA等于5分之根号5.tanB=3.（1）求角C的值？(2)当a=4时，求三角形ABC的面积？

Answer 1

(1)
cosA = √5/5 > 0, A为锐角; sinA = √(1 - cos²A) = √(1 - 1/5) = 2√5/5
tanB = 3 > 0, B为锐角; sinB/√(1 - sin²B) = 3, sinB = 3√10/10, cosB = sinB/tanB = √10/10
cosC = cos[π-(A+B)] = -cos(A+B) = -cosA*cosB + sinA*sinB
= -(√5/5)(√10/10) + (2√5/5)(3√10/10)
= -√2/10 + 3√2/5
= √2/2
C = π/4

(2)
由正弦定理, R为三角形外接圆的半径, a/sinA = c/sinC = 2R
c = asinC/sinA = 4sin(π/4)/(2√5/5)
= √10
S = (1/2)acsinB
= (1/2)*4*√10*3√10/10
= 6

Answer 2

解：（1）因为cosA = √5/5 > 0, A为锐角，所以 sinA = √(1 - cos²A) = √(1 - 1/5) = 2/√5，
又 tanB = 3 > 0, B为锐角，所以cosB=1/√(1+ tan^2B )=1/√10，sinB= tanBcosB= 3/√10，
cosC = cos[π-(A+B)] = -cos(A+B) = -cosAcosB + sinAsinB
= -(√5/5)(√10/10) + (2√5/5)(3√10/10)= -√2/10 + 3√2/5= √2/2
C = π/4
（2）S=(1/2)*a^2sinBsinC/sin(B+C)=(1/2)*16*(3/√10)*(√2/2)/(2/√5)=6。
答：C = π/4；三角形ABC的面积为6。