How do you find the three angles of the triangle with the given vertices: A(1,0) B(4,6) C(-3,5)?

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Answer 1 · 2018-03-25T05:30:57

#color(brown)("The three angles of the triangle are " #

#color(blue)(hat A = 65.22^@, hat B = 55.31^@, hat C = 59.47^@#

Given : # A(1,0), B(4,6), C(-3,5)#

Using distance formula,

#d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2#

#a = sqrt((4+3)^2 + (6-5)^2) = sqrt50#

#b = sqrt((1+3)^2 + (0-5)^2) = sqrt41#

#c = sqrt((1-4)^2 + (0-6)^2) = sqrt45#

Applying cosines law,

#cos A = (b2 + c^2 - a^2) / (2 b c) #

#cos A = (41 + 45 - 50) / (2 sqrt 41 sqrt 45) ~~ 0.4191#

#hat A = cos ^(-1) 0.4191 = 65.22^@#

#cos B = (50 + 45 - 41) / (2 sqrt 45 sqrt 50) ~~ 0.5692#

#hat B = cos ^(-1) 0.5692= 55.31^@#

#cos C = (50 + 41 - 45) / (2 sqrt 50 sqrt 41) ~~ 0.508#

#hat C = cos ^(-1) 0.508 = 59.47^@#