Asked by Anonymous
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 and 70 feet, respectively. How far apart are the houses?
feet
feet
Answers
Answered by
helper
Triangle ACB, C = 70 deg, a = 70, b = 50
find side c
Use Law of Cosines--Case III
(Given two sides and the included angle)
c^2 = a^2 + b^2 - 2ab cos C
c^2 = 70^2 + 50^2 - 2(70)(50) cos 70
Solve for c
find side c
Use Law of Cosines--Case III
(Given two sides and the included angle)
c^2 = a^2 + b^2 - 2ab cos C
c^2 = 70^2 + 50^2 - 2(70)(50) cos 70
Solve for c
Answered by
Anonymous
2966.76
Answered by
a
x^2 = 7400-((7000*(cos(70)) = 5005.858997
square root of 5005.858997 =
70.75209535 ft
square root of 5005.858997 =
70.75209535 ft
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.