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posted by Anonymous 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

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

2966.76

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