Can someone show how to solve the triangles?

You can solve them by using the law of cos, then the law of sin
First: use the law of the cos to find the side that corresponds to the angle given: for example, #10 (the unknown side)^2= (a given side)^2+(the other given side)^2-2(a given side)(the other given side)(Cos(the degree of the corresponding angle of the unknown side) 
Official formula: c^2=a^2+b^2-2abCosC
d=((3)^2+(4)^2-2(3)(4)Cos(30degree))^1/2
d=(9+16-24((3)^1/2)/2)^1/2(simplify)
d=2.053(to the 3rd decimal place, as reference)

Secondly, use the law of sin to find out the remaining sides
Sin(One angle)/corresponding side=Sin(another angle)/corresponding side=Sin(the third angle)/corresponding side, officially: Sin(A)/a=Sin(B)/b=Sin(C)/c
in this case: Sin(30degree)/2.053=SinB/3(find the non obtuse angle using the law of sin, because it can not deal with an obtuse angle)
3xSin(30)/2.053=SinB 
Sin^-1(3xSin(30degree)/2.053)=B
71.766=A=B
46.94(degree)=B

Thirdly, fin the third, larger angle by subtraction
46.94+30+C=180
C=180-46.94-30
C=103.06

Don’t worry about it

Can you provide the tape diagram?