Página — Matematica 9° Secundaria (pág. 95)

Observando la figura deducimos:

\tan45^{\circ}=\frac{cateto opuesto}{cateto adyacente}=\frac{1}{1}=1

Luego, tan 45^{\circ}=1

f)

\cot45^{\circ}=\frac{\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\c_{\c_{\mathrm{\quad}}}}{\c_{\mathrm{\quad}}}}}{\cfrac{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\quad}}{\cfrac{\c_{\quad}}{\cfrac{\c_{\quad}}{\c_{\quad}}}}}}}}}}=\frac{1}{\cfrac{\cfrac{\cfrac{\cfrac{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}}{\cfrac{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}}{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}}{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}}{\cfrac{\cfrac{\c_{\mathrm{\quad}}}{\cfrac{\c_{\mathrm{\quad}}}}{\c_{\cfrac{\mathrm{\quad}}}{\c_{\mathrm{\quad}}}}}}}}}}=\frac{1}{\cfrac{\cfrac{\cfrac{\c_{\mathrm{\quad}}}{\c_{\mathrm{\quad}}}}{\cfrac{\c_{\mathrm{\quad}}}{\c_{\mathrm{\quad}}}}}}}}=\frac{\cfrac{\c_{\mathrm{\quad}}}{\c_{\mathrm{\quad}}}}}}=\frac{\c_{\mathrm{\quad}}}}{\c_{\mathrm{\quad}}\c_{\mathrm{}\quad}}}=\frac{\c_{\mathrm{}\quad}}{\c_{\mathrm}}}={\frac{\c_{\quad}}{\c_\mathrm{}}}{\c_{\mathrm}}}={\frac{\c_{\quad}}{\c_}{\mathrm}}{\frac{\c_}{\mathrm}{\frac{\c_}{}\frac{\c_{}}{\mathrm}{\c_}{\mathrm}}{\frac{\c_}{\frac{\c_}{}}{\frac{\c_}{\mathrm}{\frac}{\c_}{\frac{}\c_{}}{\frac{\c_}{\mathrm}{\frac}{\c_}{}}{\frac{\c_}{\frac}{\frac{\c_}{\frac}{\c_}{}{\frac\frac{\c_}{}{\frac}{\c_}{\frac}{\frac{\c_}{}}{\frac\frac{\c_}{}{\frac}{\frac{\c_}}{\frac{\frac}{\c_}{}{\frac}{\frac\frac{\c_}{}{\frac}{\frac}{\frac{\c_}{\frac}{\frac}{\frac{\c_}{}}{\frac{\frac\frac}{\frac{\c_}{}}{\frac{\frac}{\frac\frac{\c_}{}}{\frac{\frac}{\frac{\frac\frac}{\frac{\frac}{\frac}{\frac{\frac\frac{\c_}{}}{\frac}{\frac{\frac\frac{\frac}{\frac}{\frac{\frac}{\frac\frac{\frac}{\frac}{\frac{\frac\frac{}\frac{\frac{\frac}\frac{\frac{\frac}{\frac}{\frac\frac{\frac{\frac\frac{}\frac{\frac\frac{\frac}{\frac}{\frac{\frac\frac{\frac}{\frac\frac\frac{\frac{\frac\frac{\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac\frac{\frac\frac{\frac{\frac\frac{\frac\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac{\frac\frac\frac{\frac{\frac\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac\frac{\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac{\frac\frac{\frac{\frac\frac{\frac\frac\frac\frac{\frac{\frac{\frac\frac\frac\frac{{\frac\frac\frac{\frac\frac\frac{\frac\frac{{\frac{\frac\frac{\frac\frac\frac{\frac{\frac{\frac\frac\frac\frac{\frac{\frac\frac\frac{\frac{\frac\frac{\frac\frac\frac{\frac{\frac{\frac\frac\frac\frac{\frac\frac{\frac\frac{\frac\frac{\frac\frac\frac{\frac\frac\frac{\frac\frac{\frac\frac\frac\frac{\frac\frac\frac\frac\frac{\frac{\frac\frac\frac\frac{\frac\frac\frac\frac\frac\frac\frac{\frac\frac\frac{\frac\frac\frac\frac\frac\frac{\frac\frac\frac\frac

Luego, cotan 45^{\circ}=1

g)

\sec45^{\circ}=\frac{Hipotenusa}{Cateto adjacente a45^{\circ}}=\frac{\sqrt{2}}{1}=\sqrt{2}=1.4142...

Luego, sec 45^{\circ}=\sqrt{2}=1.4142

h)

h)\csc45^{\circ}=\frac{Hipotenusa}{Cateto opuesto a45^{\circ}}=\frac{\sqrt{2}}{1}=\sqrt{2}=1.4142...

Luego, csc 45^{\circ}=\sqrt{2}=1.4142

Aplicación

Los ejercicios 1, 2 y 3, se resuelven con la información y gráfica siguientes:

Dado el triángulo equilátero de lado 1 unidad Calcular:

1. El valor de la altura sobre uno de sus lados.

2.Hallar

a) Sen 60°b) Cos 60°c) Tan 60^{\circ}

d)Cotan 60°e) Sec60°f)\mathsf{Csc}60^{\circ}

a) Sen 30°d)Cotan 30°b)Cos 30°e)Sec30c)Tan 30°f)Csc30^{\circ}

Tema 2 // Razones trigonométricas




95

Capitulo 1. Razones geométricas y trigonométricas