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,...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
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![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,...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](/uploads/image/z/1241224-16-4.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2Ca+b+c%E5%88%86%E5%88%AB%E6%98%AF%E8%A7%92A+B+C+%E7%9A%84%E5%AF%B9%E8%BE%B9%2CcosA%E7%AD%89%E4%BA%8E5%E5%88%86%E4%B9%8B%E6%A0%B9%E5%8F%B75.tanB%3D3.%EF%BC%881%EF%BC%89%E6%B1%82%E8%A7%92C%E7%9A%84%E5%80%BC%3F%282%29%E5%BD%93a%3D4%E6%97%B6%2C...%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2Ca+b+c%E5%88%86%E5%88%AB%E6%98%AF%E8%A7%92A+B+C+%E7%9A%84%E5%AF%B9%E8%BE%B9%2CcosA%E7%AD%89%E4%BA%8E5%E5%88%86%E4%B9%8B%E6%A0%B9%E5%8F%B75.tanB%3D3.%EF%BC%881%EF%BC%89%E6%B1%82%E8%A7%92C%E7%9A%84%E5%80%BC%3F%282%29%E5%BD%93a%3D4%E6%97%B6%2C%E6%B1%82%E4%B8%89%E8%A7%92)
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,...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
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,...
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?
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,...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
(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
(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) = -co...
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(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。
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