TrigonometryFind, correct to the nearest degree, the three angles of the triangle with the given vertices.
3 months 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 = ?
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Best Answer
chuckwebb9 Staff answered 3 months 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°
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