find the side lengths.
By the law of since, the angle measures are in the same order as the side lengths.
Triangle LM has vertices K(3,2) L(-1,5) and M(-3,-7) Write the angles in order from the least to greatest measure
2 answers
Method 1:
use the slope of the 3 lines to find the angles the lines make with the x-axis.
Then subtract the angles to determine the angle between the lines
method 2:
use the formula tanØ = |(m1-m2)/(1+m1m2)|
where Ø is the angle between lines with slope m1 and m2
(this is really the same as method2)
method 3:
find the lengths of each line, then use the cosine law to find the angle between one pair to find one angle, then the sine law to find a second angle, then the sum of the 3 angles of a triangle to find the 3rd angle
I will find the smallest angle using all 3 methods
let Ø be the angle between LM and KM
method1:
slope of LM = 12/2 = 6, so LM makes an angle of 80.5° with the x-axis
slope of KM = 3/2 , so KM makes an angle of 56.3° with the x-axis
Ø = 80.5 - 56.3 = 24.2°
.... repeat for the other 2 angles
Method 2
tanØ = |(6-1.5)/(1 + 6(1.5)|
= 4.5/10 = 9/20
Ø = arctan(9/20) = 24.2°
etc
method 3
using distance formulas:
LM = √148 , LK = 5 , KM = √117
5^2 = √148^2 + √117^2 - 2√148√117cosØ
cosØ = (148 + 117 - 25)/(2√117√148) = .91192..
Ø = arccos(.9112..) = 24.2°
etc
use the slope of the 3 lines to find the angles the lines make with the x-axis.
Then subtract the angles to determine the angle between the lines
method 2:
use the formula tanØ = |(m1-m2)/(1+m1m2)|
where Ø is the angle between lines with slope m1 and m2
(this is really the same as method2)
method 3:
find the lengths of each line, then use the cosine law to find the angle between one pair to find one angle, then the sine law to find a second angle, then the sum of the 3 angles of a triangle to find the 3rd angle
I will find the smallest angle using all 3 methods
let Ø be the angle between LM and KM
method1:
slope of LM = 12/2 = 6, so LM makes an angle of 80.5° with the x-axis
slope of KM = 3/2 , so KM makes an angle of 56.3° with the x-axis
Ø = 80.5 - 56.3 = 24.2°
.... repeat for the other 2 angles
Method 2
tanØ = |(6-1.5)/(1 + 6(1.5)|
= 4.5/10 = 9/20
Ø = arctan(9/20) = 24.2°
etc
method 3
using distance formulas:
LM = √148 , LK = 5 , KM = √117
5^2 = √148^2 + √117^2 - 2√148√117cosØ
cosØ = (148 + 117 - 25)/(2√117√148) = .91192..
Ø = arccos(.9112..) = 24.2°
etc