Posted by **Anonymous** on Sunday, February 13, 2011 at 5:58pm.

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?

feet

- college Trig -
**helper**, Sunday, February 13, 2011 at 6:11pm
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

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