3 years ago
Find, correct to the nearest degree, the three angles of the triangle with the given vertices.
A (1, 0,-1), B (4, -3,0), C (1, 5 ,5)
∠ CAB=____ °
∠ ABC=____ °
∠ BCF=____ °
/angleCAB = ?
/angleABC = ?
/angleBCA = ?
1 Answers
Best Answer
chuckwebb9 Staff answered 3 years ago
chuckwebb9 Staff answered 3 years ago
Step 1
Find theree angles.
Given vertices: А (1, 0, -1), В (4, -3, 0), С (1, 5, 5).
Distance formula:
d=sqrt((x_(2) - x_(1))^2 + (y_(2) - y_(1))^2 + (z_(2) - z_(1))^2)
Step 2
a be the distance between B and C.
a = sqrt((x_(2) - x_(1))^(2) + (y_(2) - y_(1))^(2) + (z_(2) - z_(1)0^(2))
= sqrt((1 - 4)^(2) + (5 - (-3))^(2) + (5 - 0)^(2))
=sqrt((-3)^(2) + (8)^(2) + (5)^(2))
= sqrt((-3)^(2) + (8)^(2) + (5)^(2))
7sqrt(2)
Step 3
b be the distance between A and C.
b = sqrt((x_(2) - x_(1))^(2) + (y_(2) - y_(1))^(2) + (z_(2) - z_(1)^(2))
= sqrt((1-1)^(2) + (5 - 0)^(2) + (5 - (-1))^(2))
(SK= sqrt((0)^(2) + (5)^(2) + (6)^(2))
= sqrt(61)
Step 4
c be the distance between A and B.
c = sqrt((x_(2) - x_(1))^(2) + (y_(2) - y_(1))^(2) + (z_(2) - z_(1)^(2))
= sqrt((4-1)^(2) + (-3 - 0)^(2) + (0 - (-1))^(2))
= sqrt((3)^(2) + (-3)^(2) + (1)^(2))
= sqrt(19)
Step 5
Apply cosine law. To get an angle A.
cos A=(b^(2)+c^(2)-a^(2))/(2bc)
=((sqrt(61))^(2)+(sqrt(19))^(2)-(7sqrt(2))^(2))/(2*sqrt(61)*sqrt(19))
=-0.2644
A=cos^(-1)(-0.2644)
A=105.33°