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,

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

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

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

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

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Applying cosines law,

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

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

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

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

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

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

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