Trig problems need help!?

how can the following be done.

1. From a point on level ground between two telephone poles, cables are attached to the top of each pole. One cable is 74.8 m long, and the other is 66.7 m long. If the angle of intersection between the two cables is 103.6 degrees,find the distace between the poles. answer is 111.3 metres

2. A triangular piece of land is bounded by 186m of freeway fence, 135m of brick wall, and 215m of concreate median. FInd the angle between the fence and the median.

Law of Cosines
a^2 = b^2 + c^2 - 2bc cosA
or
a^2 = 74.8^2 + 66.7^2 - 2(74.8)(66.7)cos103.6

same, but solving for A instead of a.
135^2 = 186^2 + 215^2 - 2(186)(215)cosA
135^2 - 186^2 - 215^2 = -2(186)(215)cosA
divide by -2(186)(215) and use inverse cos

1. We can draw a picture for this problem. We are told that there are 2 phone poles of equal length. This means that the 2 poles are parallel and are always equally far apart. Then we can make a point in the center of the 2 poles at the bottom and draw 2 lines to the tops of the poles. We can label them 74.8 and 66.7. Then we can label the angle 103.6 degrees. Drawing a line from the top of the telephone pole to the other creates a triangle. We now have to use the law of cosines.
c^2 = a^2 + b^2 - 2*a*b*cos(angle c)
c^2 = 74.8^2 + 66.7^2 - 2*74.8*66.7*cos(103.6)
c = 111.3
2. Again we can use the law of cosines. The fence is 186 and the median is 215. These are the 2 sides that are adjacent to angle c. We have a c side too, it's the side opposite of the angle we need to measure. 135^2 = 186^2 + 215^2 - 2*186*215*cos(x). x = 38.14 degrees