Question

Can anybody help me on this please I will be giving 15 points..

Answer 1

Answer:

BC=$\sqrt{95}$

Angle A (or BAC)=54.31466...

Angle C (or ACB)=35.68533...

Step-by-step explanation:

$\frac{7}{\sin \left(\angle \:ACB\right)}=\frac{12}{\sin \left(90\right)}$

$BC=\sqrt{12^2-7^2}$

$\sqrt{12^2-7^2}=\sqrt{95}$

$\frac{\sqrt{95}}{\sin \left(\angle \:BAC\right)}=\frac{12}{\sin \left(90\right)}\quad :\quad \angle \:BAC=54.31466$

$\frac{7}{\sin \left(\angle \:ACB\right)}=\frac{12}{\sin \left(90\right)}\quad :\quad \angle \:ACB=35.68533$

Answer 2

Using the pythagorean theorem

$AC^2=AB^2+BC^2\\BC=\sqrt{AC^2-AB^2}\\BC=\sqrt{12^2-7^2}\\BC=\sqrt{144-49}\\BC=\sqrt{95}$

$sin A=\frac{BC}{AC}=\frac{\sqrt{95} }{12}\\A = sin^{-1}(\frac{\sqrt{95} }{12})\\A = 54.31$

$A + C = 90\\C = 90-A\\C= 90- 54.31\\C = 35.69$