To find the distance from the house at A to the house at B, a surveyor measures the angle ACB, which is found to be 70°, and then walks off the distance to each house, 50 feet and 70 feet, respectively. How far apart are the houses?

Question

feet

Answer 1

Triangle ACB, C = 70 deg, a = 70, b = 50
find side c

Use Law of Cosines--Case III
(Given two sides and the included angle)

c^2 = a^2 + b^2 - 2ab cos C
c^2 = 70^2 + 50^2 - 2(70)(50) cos 70

Solve for c

Answer 2

2966.76

Answer 3

x^2 = 7400-((7000*(cos(70)) = 5005.858997
square root of 5005.858997 =

70.75209535 ft