How do you find the exact value of the third side given $triangle ABC$ , $AB=2sqrt2, BC=4, mangleB=(3pi)/4$ ?

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Answer 1 · 2017-02-13T11:12:05

$|AC| = 2 sqrt{ 10 }$

The Law of the Cossins

$|AC|^2 = |AB|^2 + |BC|^2 - 2 * |AB| * |BC| * cos hat B$

$|AC|^2 = 8 + 16 + 2 * 2 sqrt 2 * 4 * cos (180º - 135º)$

$|AC| = sqrt{ 24 + 16 sqrt 2 * sqrt 2 / 2}$

$|AC| = sqrt{ 40 }$

How do you find the exact value of the third side given triangle ABCtriangle ABC, AB=2sqrt2, BC=4, mangleB=(3pi)/4AB=2sqrt2, BC=4, mangleB=(3pi)/4?