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 (A to C) and 70 feet (C to B), respectively. How far apart are the houses? (answer to 1 decimal place) in feet

Question

Mathematics

Dominique Dalton

13 August, 16:04

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 (A to C) and 70 feet (C to B), respectively. How far apart are the houses? (answer to 1 decimal place) in feet

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Answers (1)

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Marquez · Answer 1

|AB|^2 = |AC|^2 + |CB|^2 - 2|AC||CB|cos70° = 50^2 + 70^2 - 2 (50) (70) cos 70° = 2,500 + 4,900 - 7,000cos 70° = 7,400 - 2,394.14 = 5,005.86

|AB| = sqrt (5,005.86) = 70.8 feet.

Therefore, the houses are 70.8 feet apart.