Finding Side AB and Angle ABC of a Park Problem

[ur5pointos2sl]

Ok my math is a little rusty. I ran across a problem that states:
"A park is being considered in a space between a small river and a highway as a rest stop for travelers. Boundary BC is perpendicular to the highway and boundary AD makes an angle of 75 Deg. with the highway. BC= 160.0 m , AD= 270.0 m, and the boundary along the highway = 190.0m long. What are the length of side AB and the magnitude of angle ABC?"

I have attached a file that is almost identical to the one in the book. I need help figuring out where to start this problem(not an answer please). Any help would be appreciated.

Thanks

 

[tiny-tim]

Welcome to PF!

"A park is being considered in a space between a small river and a highway as a rest stop for travelers. Boundary BC is perpendicular to the highway and boundary AD makes an angle of 75 Deg. with the highway. BC= 160.0 m , AD= 270.0 m, and the boundary along the highway = 190.0m long. What are the length of side AB and the magnitude of angle ABC?"

Hi ur5pointos2sl ! Welcome to PF!

I can't see your picture file yet, but I think the best way is probably to use x and y coordinates, starting with C as the origin.

 

[ur5pointos2sl]

My problem is that I can't seem to use law of sine/cosine because there is always 2 things to be solved for. I think I am missing something.

I put C at the origin and looked at it that way. If i separate the figure into two triangles that doesn't seem to lead me anywhere either. I can solve for one but then the other I am completely stuck on.

 

[Dick]

You don't need coordinates. DO split the quadrilateral into two triangles. Solve the right triangle first. Find length of BD and the two angles. Then you have side-angle-side on the remaining triangle. Solve it.

 

[ur5pointos2sl]

You don't need coordinates. DO split the quadrilateral into two triangles. Solve the right triangle first. Find length of BD and the two angles. Then you have side-angle-side on the remaining triangle. Solve it.

Thank you. That is exactly what I did last night except I solved for the right triangle then used law of cosine to solve for the other. I finally realized my answer was off because I was rounding to one decimal place.