For the first attachment the instructions are:
solve for x
for the second attachment the instructions are:
Solve for <B if possible. If 2 solutions find both.
This phrase will help you solve every trig problem; SOH CAH TOA
S=sin = O (opposite length)/H (hypotenuse)
C=cos = A (adjacent length)/H (hypotenuse)
T= tan = O (opposite length)/A (adjacent length)
Its better if you learn then just get answers, especially when your problems are fairly easy and very important to know.
So to help you start, lets do the first one.
The triangle is a 30, 60, 90 triangle where you are given the hypothesis is 1.
So using the 60 degree angle, you can say cos(60)=y/1, because cos(X)=a/h ^^^^^ABOVE^^^^
so y=cos60
you can do the same thing to solve for x; the sin(60)= x/1, because sin(X)=O/h; ^^^^ABOVE^^^^
so x=sin(60)
So you know, in a 30 60 90 triangle, the short leg (opp. the 60) is x(sqrt3)/2, the hypothesis is 2x, and the long leg (opp. the 30) is x.
Now that you know how to do them, you can do the rest by simply following the SOHCAHTOA phrase, and just PLUGGIN AND CHUGGIN BABY!
I know I have to learn this, but right now I need to hand a paper in by tomorrow. I really don't know what to do, it's getting late. The last three triangles need the ambiguous case, I don't know how to do that. If you don't have time to answer them for me, I understand.
Thanks for all your help.