Question

me podrían ayudar con este ejercicio por favor es para hoy antes de las cinco ​

Answer 1

¡Hola!

$\bold{Sustimos \ los \ datos \ del \ triangulo \ rectangulo \ en \ las \ funciones \ trigonometrica.}$Si los datos son:

Lado a: 8 cm

Lado b: 10 cm

Hipotenusa: $2\sqrt{41}$

$\bold{Procedemos \ a \ resolver.}$

Cos (A).

$\cos(A)=\dfrac{L. A}{HIP} \\\\\\ \cos(A)=\dfrac{b}{c} \rightarrow \dfrac{10}{2\sqrt{41}} \\\\\\ \textrm{Vamos a racionalizar.} \\\\\\ \dfrac{10}{2\sqrt{41}} \rightarrow \dfrac{5}{\sqrt{41}} \cdot \dfrac{\sqrt{41}}{\sqrt{41}}= \dfrac{5 \cdot \sqrt{41}}{\sqrt{41} \cdot \sqrt{41}}= \dfrac{5\sqrt{41}}{41} \\\\\\ \Rightarrow \boxed{\boxed{\cos(A)=\dfrac{5\sqrt{41}}{41}}}$

Tan(A).

$\tan(a)=\dfrac{L.O}{L.A} \\\\\\ \tan(a)=\dfrac{a}{b} \rightarrow \dfrac{8}{10} \\\\\\ \Rightarrow \boxed{\boxed{\tan(a)=\dfrac{4}{5}}}$

Sen(B)

$\sin(B)=\dfrac{L.O}{HIP} \\\\\\ \sin(B)=\dfrac{b}{c} \rightarrow \dfrac{10}{2\sqrt{41}} \\\\\\ \textrm{Vamos a racionalizar.} \\\\\\ \dfrac{10}{2\sqrt{41}} \rightarrow \dfrac{5}{\sqrt{41}} \cdot \dfrac{\sqrt{41}}{\sqrt{41}}= \dfrac{5 \cdot \sqrt{41}}{\sqrt{41} \cdot \sqrt{41}}= \dfrac{5\sqrt{41}}{41} \\\\\\ \Rightarrow \boxed{\boxed{\sin(B)=\dfrac{5\sqrt{41}}{41}}}$

Cos(B).

$\cos(B)=\dfrac{L.A}{HIP} \\\\\\ \cos(B)=\dfrac{a}{c} \rightarrow \dfrac{8}{2\sqrt{41}} \\\\\\ \textrm{Vamos a racionalizar.} \\\\\\ \dfrac{8}{2\sqrt{41}} \rightarrow \dfrac{4}{\sqrt{41}} \cdot \dfrac{\sqrt{41}}{\sqrt{41}}= \dfrac{4 \cdot \sqrt{41}}{\sqrt{41} \cdot \sqrt{41}}= \dfrac{4\sqrt{41}}{41} \\\\\\ \Rightarrow \boxed{\boxed{\cos(B)=\dfrac{4\sqrt{41}}{41}}}$

Tan(B).

$\tan(B)=\dfrac{L.O}{L.A} \\\\\\ \tan(B)=\dfrac{b}{a} \rightarrow \dfrac{10}{8} \\\\\\ \Rightarrow \boxed{\boxed{\tan(B)=\dfrac{5}{4}}}}$

Espero que sirva y saludos.