This is a good exercise to apply the law of cosines and the law of sines:
First, we are going to use the law of cosines to find any angle, and then we can use the law of sines to find the other one; last but not least we are going to use the internal angles definition (the sum of the interior angles of a triangle is 180°) to fin our final angle and compare them:
Using the law of cosines: we know that [tex]COSA= \frac{b^{2}+ c^{2}- a^{2} }{2bc} [/tex], but we need the vale of the angle A, so we are going to use the inverse cosine function, ARCCOS, to solve for A:
[tex]A=ARCCOS(\frac{b^{2}+ c^{2}- a^{2} }{2bc})[/tex]
We also know that a=121, b=140, and c=158, so the only thing we have to do is replace the values into our equation:
[tex]A=ARCCOS(\frac{140^{2}+ 158^{2}- 121^{2} }{2(140)(158)})[/tex]
[tex]A=ARCCOS( \frac{29923}{44240}) [/tex]
[tex]A=47.4[/tex]°
Now that we know the value of the angle A, we can use the law of sines to find the angle B:
[tex] \frac{SinA}{a} = \frac{SinB}{b} [/tex]
[tex]SinB= \frac{bSinA}{a} [/tex]
And just like before we'll use the inverse sine function, ARCSIN, to solve for B:
[tex]B:ARCSin( \frac{bSinA}{a} )[/tex]
We know that b=140, a=121, and A=47.4, so we can replace the vales to get:
[tex]B=ARCSin( \frac{140Sin(47.4)}{121} )[/tex]
[tex]B=58.4[/tex]°
Now that we have two angles, we can use the interior angle definition. If the sum of the interiors angle of a triangle is 180°, then:
[tex]C=180-(A+B)[/tex]
[tex]C=180-(47.4+58.4)[/tex]
[tex]C=108-105.8[/tex]
[tex]C=74.2[/tex]
We can conclude that the camera that has to cover the greatest angle is camera 3, and it has to cover an angle of 74.2°.
