Answer:
Camera 2nd has to cover the maximum angle, i.e. [tex]78.70^\circ[/tex].
Step-by-step explanation:
Please have a look at the triangular park represented as a triangle [tex]\triangle ABC[/tex] with sides
a = 110 ft
b = 158 ft
c = 137 ft
1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.
We can use law of cosines here, to find out the angles [tex]\angle A, \angle B, \angle C[/tex]
As per Law of cosine:
[tex]cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}[/tex]
Putting the values of a,b and c to find out angles [tex]\angle A, \angle B, \angle C[/tex].
[tex]cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ[/tex]
[tex]cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ[/tex]
[tex]cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ[/tex]
Camera 2nd has to cover the maximum angle, i.e. [tex]78.70^\circ[/tex].
