Question

Ayuda Porfavor, tiene que ser con teorema del coseno

Answer 1

Problema 1:

Consideraciones iniciales:

Según el teorema del coseno

$a^2=b^2+c^2-2ab(cos(\alpha ))$

Solución:  

Cálculo de α

a = 10

b = 13

c = 15

α = x

Reemplazando en la ecuación del coseno

$a^2=b^2+c^2-2bc(cos(\alpha ))\\10^2=13^2+15^2-2(13)(15)(cos(\alpha))\\100=169+225-390(cos(\alpha))\\390(cos(\alpha))=294\\cos(\alpha) = 294/390\\cos(\alpha) = \frac{49}{65} \\\alpha =41.08$

Cálculo de β

a = 15

b = 13

c = 10

α = x

$a^2=b^2+c^2-2bc(cos(\beta ))\\15^2=13^2+10^2-2(13)(10)(cos(\beta )\\225=169+100-260(cos(\beta ))\\260(cos(\beta ))=44\\cos(\beta )= \frac{44}{260} \\cos(\beta ) = \frac{11}{65} \\ \beta =80.26$

Cálculo de γ

a = 13

b = 10

c = 15

α = x

$a^2=b^2+c^2-2bc(cos(y)\\13^2=10^2+15^2-2(10)(15)(cos(y))\\169=100+225-300(cos(y)\\300(cos(\alpha))=156\\cos(y) = 156/300\\cos(y)= \frac{13}{25} \\\ y =58.67$

Respuesta:  

α = 41.08 °

β = 80.26 °

γ = 58.67 °

Problema 2:

Consideraciones iniciales:

a = 12

b = x

c = 10

α = x

Solución:  

Cálculo de b

$a^2=b^2+c^2-2bc(cos(50)\\12^2=x^2+10^2-2(x)(10)(cos(50)\\144=x^2+100-12.86x\\x^2-12.86x-44=0\\x=15.7$

Cálculo de β

$a^2=b^2+c^2-2bc(cos(\beta ))\\15.7^2=12^2+10^2-2(12)(10)(cos(\beta )\\246.49=144+100-240(cos(\beta ))\\240(cos(\beta ))=-2.49\\cos(\beta )= \frac{-2.49}{240} \\ \beta =90.59$

Cálculo de γ

$a^2=b^2+c^2-2bc(cos(y)\\10^2=12^2+15.7^2-2(12)(15.7)(cos(y))\\100=144+246.49-376.8(cos(y)\\376.8(cos(\alpha))=290.5\\cos(y) = 290.5/376.8\\ y =39.50$

Respuesta:  

BC = 15.7

β = 90.56 °

γ = 39.50 °