已知三角形ABC中.sin(A-B)=1/2.tanA*tanB=2+根号3.且边长AB=12(1)求cos(A-B);(2)求角C的大小;(3)求三角形ABC的面积别说到此为止
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已知三角形ABC中.sin(A-B)=1/2.tanA*tanB=2+根号3.且边长AB=12(1)求cos(A-B);(2)求角C的大小;(3)求三角形ABC的面积别说到此为止
已知三角形ABC中.sin(A-B)=1/2.tanA*tanB=2+根号3.且边长AB=12
(1)求cos(A-B);
(2)求角C的大小;
(3)求三角形ABC的面积
别说到此为止
已知三角形ABC中.sin(A-B)=1/2.tanA*tanB=2+根号3.且边长AB=12(1)求cos(A-B);(2)求角C的大小;(3)求三角形ABC的面积别说到此为止
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(1)
sin(A-B)=1/2
A-B=π/6或5π/6
∵tanA*tanB=2+√3>0
∴0<A<π/2,0<B<π/2,
-π/2<A-B<π/2,
∴A-B=π/6
∴cos(A-B)=√3/2;
(2)
A-B=π/6
A=B+π/6
2+√3=tanA*tanB
=sinA*sinB/(...
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(1)
sin(A-B)=1/2
A-B=π/6或5π/6
∵tanA*tanB=2+√3>0
∴0<A<π/2,0<B<π/2,
-π/2<A-B<π/2,
∴A-B=π/6
∴cos(A-B)=√3/2;
(2)
A-B=π/6
A=B+π/6
2+√3=tanA*tanB
=sinA*sinB/(cosA*cosB)
=tan(B+π/6)*tanB
={(√3/3+tanB)/[1-(√3/3)tanB]}*tanB
(3+2√3)-(2+√3)tanB=√3(tanB)^2+tanB
(tanB)^2+(1+√3)tanB-(2+√3)=0
tanB=1或-2-√3(因0<B<π/2,所以后者舍去)
B=π/4
A=B+π/6=5π/12
C=π-π/4-5π/12=π/3;
(3)
AB/sinC=BC/sinA
BC=ABsinA/sinC
=12(√2/2)/(√3/2)
=4√6
s=(1/2)AB*BCsinB
=(1/2)12*4√6*(√2/2)
=24√3
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