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EXAMEN DE TRIGONOMETRIA
Sabiendo que sen = 1/3 y que se encuentra en el cuadrante 0 < < /2
Calcular:
La anchura de una calle es de 20 metros y desde el centro de la misma se ven los ángulos más altos de dos edificios, situados cada uno en un extremo de la calle. Calcula la altura de estos 2 edificios
Clacular:
Cos 1125º
Tg (-120)
Cosec 5/6
Sen 300º
Simplifica:
Demuestra el teorema del seno
Calcula los lados y los ángulos que faltan del siguiente triángulo sabiendo que el ángulo C=120º; y que su lado a=4, y el lado c=4"3
SOLUCIONES:
1.
Cos = "8/3
Tg (180 - ) = -1/"8
Cos (/2 - ) sen =1/3
Cosec (2 - ) = -3
2.
La altura de los edificios es igual a 10 metros.
Se realiza por un sistema, en el que se utiliza la tangente, para así poder despejar el seno
3.
Cos 1125º = "2/2
Tg (-120) = "3
Cosec 5/6 = 2
Sen 300º = -"3/2
4.
La solución es igual a cero
Es teoría, al que representar un triángulo equilátero y en él hallar los senos hasta llegar a una igualdad. Se encuentra en todos los libros de texto
5.
El lado b es igual 4
El ángulo A es igual a 30º
El ángulo B es igual a 30º
Cos
1 - sen
"
1 + sen
cos
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Enviado por: El remitente no desea revelar su nombre Idioma: castellanoPaís: España