Question

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

Answer 1

summary: You have triangle ABC, angle C = 10°, AC=90 and BC = 100
using the cosine law:
AB^2 = 90^2 + 100^2 - 2(90)(100)cos10fl

do the math

Answer 2

AB = AC + CB = -90 + 100Ft[10o]
AB = -90 + 100*cos10+100*sin10
AB = -90 + 98.5+17.4i = 8.5 + 17.4i = 19.4 Ft.