A triangular lot sits at the corner of two streets that intersect at an angle of 64°. One street side of the lot is 115 ft, and the other 128 ft. How long is the back of the lot?

Answer 1

This problem can best be solved using the Law of Cosines:

c² = a² + b² - (2 a b) cos(C)

Let

a = 115

b = 128

C = 64^o

Then

c² = 115² + 128² - 2(115)(128)cos(64^o) = 16703.35343852956

c = 129 (to the precision allowed by the given measurements)

For those unfamiliar with the Law of Cosines, simply break up the problem into two right triangles and use the standard definitions of Trig functions:

Starting at the far end of the 115 ft stretch, draw a line to the point on the 128 ft span such that the new line and the 128 ft span are perpendicular. Designate the length of this new segment as y, and the segment from the right angle to the 64^o intersection angle as x. We now have two right triangles, and from one of them we can see that

x = 115 cos(64^o)

y = 115 sin(64⁰)

All that is left to do is to use the Pythagorean formula on the other right triangle to compute its hypotenuse, d, (the right triangle having y as one leg and 128-x as the other):

d² = y² + (128-x)²

Upon substitution, we get the same value of d.

Answer 2

Use the Law Cosines

d² = 115² + 128² - 2(115)(128) cos 64°