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?
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
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.
To find the distance between the houses, you can use the law of cosines. The law of cosines states that in any triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of their lengths and the cosine of the included angle.
In this case, let's denote the distance from point A to the unknown distance between the houses as x. The distances from point B to point C and from point A to point C are given as 90 feet and 100 feet, respectively. The angle ACB is given as 10 degrees.
Applying the law of cosines, we have:
x^2 = 90^2 + 100^2 - 2(90)(100)cos(10)
Now we can solve this equation to find the value of x:
x^2 = 8100 + 10000 - 18000cos(10)
x^2 = 18179.71
Taking the square root of both sides:
x ≈ 134.92 feet
Therefore, the houses are approximately 134.92 feet apart.