Solve a math problem of trigonometric function in high school, which is urgent online.

1, certificate:

sinAcosB+cosAsinB=3/5

sinAcosB-cosAsinB= 1/5

So sinAcosB=2/5, cosAsinB= 1/5.

(sinacosb)/(cosasinb) = tana/tanb = 2, that is, tanA=2tanB, and the certificate is completed.

2. Solution:

c=AB=3，sinC=sin(A+B)=3/5

From sine theorem, there is b=csinB/sinC.

Draw a picture by yourself with h=bsinA=csinAsinB/sinC.

Cos(A+B)=-4/5, that is, sinAsinB-cosAcosB=4/5.

SinAcosB=2/5, cosAsinB= 1/5, so sinAsinBcosAcosB=2/25.

Let sinAsinB=x and cosacosb = y.

X-y=4/5，xy=2/25。

At the same time, take the positive root and get x = (square root of 2+6) /5, that is, sinasinb = (square root of 2+6) /5.

Therefore, h = csinasinb/sinc = 3 * (square root of 2+6)/5/(3/5) = square root of 2+6.