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12 April, 05:25

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 20° , and then walks off the distance to each house, 90 feet and 100 feet, respectively. How far apart are the houses?

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  1. 12 April, 05:48
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    You can use trig combinations or (what i'd do) use pythagorus

    AB = sqrt[ (100^2) + (90^2) ] = 134.5
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