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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so we got a ABC tri
AC =50 ;BC = 70 and angle c=70 degrees
law of cosines

(AB)^2 = 50^2 + 70^2 -2(50)(70) cos(70)

google gives me:
sqrt((50^2) + (70^2) - (2 * 50 * 70 * cos(70))) = 54.4680234

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