¿Pregunta de trigonometría?
¿Cómo hallo el área de un triángulo en el plano cartesiano si me dan sus vértices y me dicen que emplee el seno de un ángulo?
(Dice hallar el triangulo cuyos vértices son A(1,-3)B(3,3) y C(6,-1) empleando el seno del ángulo BAC
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Verified answer
Hola
Pendiente AB
tan(aAB) = (3 - (-3))/(3 - 1) = 6/2 = 3
aAB = 71.56º
Pendiente AC
tan(aAC) = (-1 - (-3))/(6 - 1) = 2/5
aAC = 21.80º
A = aAB - aAC = 49.76º
AB = raiz( (3 - (-3))^2 + (3 - 1)^2 )
AB = raiz( (6)^2 + (2)^2 )
AB = raiz( 36 + 4 )
AB = raiz(40)
AB = 6.32
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AC = raiz( (-1 - (-3))^2 + (6 - 1)^2 )
AC = raiz( (2)^2 + (5)^2 )
AC = raiz( 4 + 25 )
AC = raiz(29)
AC = 5.38
*************
Area = (1/2) * AB * AC * sen(A)
Area = (1/2) * 6.32 * 5.38 * sen(49.76º)
Area = 12.98
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En forma exacta
Area = (1/2) * Abs
|...1...-3....1..|
|...3....3....1..|
|...6...-1....1..|
Area = (1/2) *
Abs [ ( (1) (3) (1) + (3) (-1)(1) + (6) (-3) (1) ) -
- ( (1) (3) (6) + (1) (-1) (1) + (1) (-3) (3) ) ]
Area = (1/2) * Abs[ ( 3 - 3 - 18 ) - ( 18 - 1 - 9 ) ]
Area = (1/2) Abs( -18 - 8)
Area = (1/2) Abs( -26)
Area = (1/2) * 26
Area = 13
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