6. Trigonometría Tri C4S21 Àngulo de cercha (Lenguaje algebraico) Una cercha con forma de triángulo isósceles está formada por un pendolón cuya altura equivale  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»#«/mo»«mi»L«/mi»«mn»1«/mn»«/math» metros. Si la cuchilla tiene una dimensión igual a  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»#«/mo»«mi»L«/mi»«mn»2«/mn»«/math» metros, determine el ángulo formado entre las cuchillas y la base.

 

 

Observaciones:

-Ingresa las cifras sin unidad de medida

-Utiliza punto para separar las cifras decimales

-Utiliza hasta tres cifras después del punto en el resultado

-Expresa el resultado utilizando el sìmbolo de grado (º)

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iiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKACiiigAooooAKKKKAP/9k= Revisemos el desarrollo del problema.

Si consideramos que:

-Altura pendolón: «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»a«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mo»#«/mo»«mi»L«/mi»«mn»1«/mn»«mo»§#160;«/mo»«mi»m«/mi»«/math»

-Longitud cuchilla: «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»b«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mo»#«/mo»«mi»L«/mi»«mn»2«/mn»«mo»§#160;«/mo»«mi»m«/mi»«/math»

Por teorema del seno se tiene que:

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»s«/mi»«mi»e«/mi»«mi»n«/mi»«mo»(«/mo»«mi»§#945;«/mi»«mo»)«/mo»«/mrow»«mi»a«/mi»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mi»s«/mi»«mi»e«/mi»«mi»n«/mi»«mo»(«/mo»«mn»90«/mn»«mo»§#176;«/mo»«mo»)«/mo»«/mrow»«mi»b«/mi»«/mfrac»«/math»

Es decir:

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»s«/mi»«mi»e«/mi»«mi»n«/mi»«mo»(«/mo»«mi»§#945;«/mi»«mo»)«/mo»«/mrow»«mrow»«mo»#«/mo»«mi»L«/mi»«mn»1«/mn»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mi»s«/mi»«mi»e«/mi»«mi»n«/mi»«mo»(«/mo»«mn»90«/mn»«mo»§#176;«/mo»«mo»)«/mo»«/mrow»«mrow»«mo»#«/mo»«mi»L«/mi»«mn»2«/mn»«/mrow»«/mfrac»«/math»

 

Despejando «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»§#945;«/mi»«/math» se tiene que:

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»§#945;«/mi»«mo»=«/mo»«mi»a«/mi»«mi»r«/mi»«mi»c«/mi»«mi»s«/mi»«mi»e«/mi»«mi»n«/mi»«mfenced»«mfrac»«mrow»«mo»#«/mo»«mi»L«/mi»«mn»1«/mn»«mo»§#183;«/mo»«mi»s«/mi»«mi»e«/mi»«mi»n«/mi»«mo»(«/mo»«mn»90«/mn»«mo»§#176;«/mo»«mo»)«/mo»«/mrow»«mrow»«mo»#«/mo»«mi»L«/mi»«mn»2«/mn»«/mrow»«/mfrac»«/mfenced»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»§#945;«/mi»«mo»=«/mo»«mo»#«/mo»«mi»s«/mi»«mi»o«/mi»«mi»l«/mi»«/math»

 

En tu calculadora «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»a«/mi»«mi»r«/mi»«mi»c«/mi»«mi»s«/mi»«mi»e«/mi»«mi»n«/mi»«mo»(«/mo»«mi»c«/mi»«mi»a«/mi»«mi»n«/mi»«mi»t«/mi»«mi»i«/mi»«mi»d«/mi»«mi»a«/mi»«mi»d«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/math» es equivalente a «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mi»e«/mi»«msup»«mi»n«/mi»«mrow»«mo»-«/mo»«mn»1«/mn»«/mrow»«/msup»«mo»(«/mo»«mi»c«/mi»«mi»a«/mi»«mi»n«/mi»«mi»t«/mi»«mi»i«/mi»«mi»d«/mi»«mi»a«/mi»«mi»d«/mi»«mo»)«/mo»«/math»

 

Reporta y/o califica esta pregunta:

 1. Copia este código: Tri. C2. S21.

 2. Presiona este enlace, pega el código y completa el formulario

 

 

]]> 1 .3333333 0 0 Ángulo=#sol]]> Muy bien!!!

]]> *]]> Al parecer, cometiste un error.

]]> - Santiago Sur 24/6

En la pregunta sugiero espaciar un poco el dibujo de los textos. Corregido

En la observación:

  • Solo el título con negrita Corregido
  • No debe decir hasta 3 cifras decimales, porque no acepta 1 o 2 decimales y además debe decir si se aproxima o no. Corregido

En la retroalimentación:

  • ¿Por qué asume que el ángulo es recto? Porque la altura en un triángulo isósceles con esa forma genera un ángulo recto
  • También se puede  hacer sin utilizar el teorema del seno.

 

 

 

02/06/2016   Fatima Toro Lagos

Realizado-Pendiente: Sugiero que tal vez en el encabezado podría decir: "En una cercha con forma de un triángulo isósceles....."

Realizado-Pendiente: En las observaciones dice utiliza coma para separar los decimales, y el consenso es utilizar punto.

Realizado-Pendiente: Donde se ingresa la respuesta, sugiero dejar fijo "Ángulo = ".

Realizado-Pendiente: Dice "Observación" en vez de "Observaciones"

Pendiente: Las imágenes no siempre se cargan, por lo que la figura debe hacerse en wiris, allí se pueden etiquetar todos los segmentos y

ángulos. (acuerdo)

Además si la dibujas en wiris, puedes agregar el pendolón para que quede más claro, pues son término que los estudiantes no conocen.

 

Marco Antonio:

Sugiero colocar en vez de sen

 

 

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