Solution: 1 In a triangle, a = x cm, b = 2 cm, and b = 45 degrees. According to the sine theorem, a/sinA=b/sinB=c/sinC=2R(R is the radius of the circumscribed circle), and there is x/sinA=b/sinB, and x is obtained by substituting it into the number.

If b=c=2, then c = 45 and x=2.

If b=a=2, then x=2.

2. Because sinC=sin(A+B), you know this, so SINC = SIN (A+B) = COSA SINB on the right of Sina COSA.

In the problem, tell 2cosBsinA=sinC, substitute sinC= sinAcosB+cosAsinB to get cosBsinA= sinBcosA, translate, cosBsinA- sinBcosA=0, that is, sin(A-B)=0, then A=B, so it is an isosceles triangle.

3. According to the sine theorem, a/sinA=b/sinB=c/sinC=2R(R is the radius of the circumscribed circle), we can know that the angle of the largest edge pair is the largest, so the angle corresponding to x+2 is obtuse. Using cosine theorem, we can get cosα=[x? +(x+ 1)？ -(x+2)？ ]/[2x (x+ 1)] < 0, the solution is-1 < x < 3,

According to the conditions of triangle, x+(x+ 1) > x+2, and x > 1,

So 1 < x < 3.

4. Assuming that AC and BD intersect at point O, the sum of the internal angles of the triangle is 180, and other angles can be obtained: DAC=30, CBD=60, AOB = 105. From the sine theorem, it can be concluded that AD= root number 3, △BOC= root number 2, and then △

In triangle AOB, AB can be obtained by cosine theorem or cos∠AOB, and finally AB= root number 5.

5. From c/sinC=2R, we can get c = 2r sinc = 2× 2× sin60 = 2 * radical number 3.

CosC=(a？ +b？ -c？ ) /2ab, from a:b=3:4, let a=3x and b=4x, and substitute for the solution, x=2* radical number 39/ 13, so a=3x =6* radical number 39/ 13, and b=4x =8* radical.