Respuesta :

stephaniep7nsa2

Answer:

BC=[tex]\sqrt{95}[/tex]

Angle A (or BAC)=54.31466...

Angle C (or ACB)=35.68533...

Step-by-step explanation:

[tex]\frac{7}{\sin \left(\angle \:ACB\right)}=\frac{12}{\sin \left(90\right)}[/tex]

[tex]BC=\sqrt{12^2-7^2}[/tex]

[tex]\sqrt{12^2-7^2}=\sqrt{95}[/tex]

[tex]\frac{\sqrt{95}}{\sin \left(\angle \:BAC\right)}=\frac{12}{\sin \left(90\right)}\quad :\quad \angle \:BAC=54.31466[/tex]

[tex]\frac{7}{\sin \left(\angle \:ACB\right)}=\frac{12}{\sin \left(90\right)}\quad :\quad \angle \:ACB=35.68533[/tex]

jaimemenendezalvarez

Using the pythagorean theorem

[tex]AC^2=AB^2+BC^2\\BC=\sqrt{AC^2-AB^2}\\BC=\sqrt{12^2-7^2}\\BC=\sqrt{144-49}\\BC=\sqrt{95}[/tex]

[tex]sin A=\frac{BC}{AC}=\frac{\sqrt{95} }{12}\\A = sin^{-1}(\frac{\sqrt{95} }{12})\\A = 54.31[/tex]

[tex]A + C = 90\\C = 90-A\\C= 90- 54.31\\C = 35.69[/tex]

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