EJEMPLO :

• En el siguiente triángulo cuyos vértices son lospuntos  de coordenads A (-4,8) B(4,4) Y C (-2,2) . Hallar los ángulos interiores

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mAB = Y1-Y2/X1-X2                                mAC= 8-2/-4+2= -6/2 = -3

mAB= 8-4/-4-4= 4/ -8 = -1/2                     mBC= 4-2/4+2 = 2/6 =1/3.

Tg Θ A = m2-m1/1+m2.m1                        Tg Θ  C =   m2-m1/1+m2.m1                                                    

Tg Θ= -1/2-(-3)/1+(-1/3)(-3)                      Tg Θ = (-3 - 1/3)/ (1-3(1/3))

Tg Θ =-1/2+3 / 1+3/2                                  Tg Θ= (-9-1/3)/(1-3/5)

Tg Θ  = (-1+6/2)/(2+3/2)=5/5= 1                 Tg Θ = (-10/3)/(0/3) = -30/0 = infinito

    Tg Θ a = 1                                                      

Θ A = inv tg 1

Θ A  = 45º

Tg Θ B= m2-m1/1+m2.m1                            Comprobación:

Tg Θ = 1/3-(-1/2)/1+1/3(-1/2)                       A+B+C =180º

Tg Θ =(1/3+1/2)/(1-1/6)                               45º +45º +90º = 180º

Tg Θ = (5/6)/(5/6)                                               180º  = 180º

Tg Θ = 5/5 

Tg Θ  = 1  

Θ = Inv Tg 1

Θ = 45º