Surveying+a+Property

= **Dakota Findley** =

**Aby Sobotka Briner**
= = = **Your friend has an old property of which she would like to find the boundaries. The deed is quite old, so she knows where the front two corners of the property are and that the between them is 220 feet. The shape is an irregular Quadrilateral and the other three sides have distances of 525 feet, 248 feet, and 542 feet in a clockwise direction from the front left corner. More importantly, it is known that the front left corner of the property is a right angle. Construct a blueprint of the property so that she can locate all of her boundary lines and corners.** =



Work:

1. First, I found AC. AC = 525^2 + 220^2 AC = √324025

AC = 569.23
2. Next, I found m<A.

sin(A) = sin(90) 220 569.23 Cross Multiply...

569.23*sin(A) = 220 569.23 569.23

sin-1(.3865) = A

A = 22.74 degrees
3. Lastly, I solved for m<C. m<C = 180-22.74+90

C = 67.26 degrees


1. First, I found m<A.

m<CAD = 542^2 = 248^2 + 569.23^2 - 2*248*569.23*cos(A) 293764 = 385526.79 - 282338.08*cosA -385526.79 -385526.79

-91762.79 = -282338.08*cos(A) -282338.08 -282338.08

.325 = cos(A)

cos-1(.325) = A

A = **71.03 degrees**
2. Next, I found m<C.

m<ACD = 248^2 = 542^2 + 569.23^2 - 2*542*569.23*cosC 61504 = 617786.79 - 617045.32*cosC -617786.79 -617786.79

-556282.79 = -617045.32*cosC -617045.32 -617045.32

.9015 = cosC

cos-1(.9015) = C

C = 25.64 degrees
3. Lastly, I solved for m<D.

m<ADC = 180 - 25.64 + 71.03

D = 83.33 degrees


FINAL ANGLES:

m<A = 22.74 + 71.03

m<B = 90 degrees
m<C = 67.26 + 25.64