The given function is,
[tex]y=A\cos (Bx)+C[/tex]The curve passes through the point (0,8), (60,78) and (120,8).
Substituting the point (0,8) in the given equation,
[tex]\begin{gathered} 8=A\cos (B\times0)+C \\ 8=A+C\ldots\ldots\mathrm{}(1) \end{gathered}[/tex]Substituting the value (60, 78) in the given equation,
[tex]78=A\cos (60B)+C\ldots\ldots\ldots\text{.}(2)[/tex]Substituting the value (120,8) in the given equation,
[tex]8=A\cos (120B)+C\ldots\ldots\ldots\text{.}(3)[/tex]Substracting equation (3) from equation (1),
[tex]\begin{gathered} A+C-(A\cos 120B+C)=0 \\ A(1-\cos 120B)=0 \\ A=0\text{ or cos120B=0} \\ A=0\text{ or 120B=}\frac{\pi}{2} \\ A=0\text{ or B=}\frac{\pi}{240} \end{gathered}[/tex]The value of A cannot be zero as the curvature won't be of the cosine.
Thus, the value of B will be pi/240.
Substituting the value of B in the equation (2),
[tex]\begin{gathered} 78=A\cos (60\times\frac{\pi}{240})+C \\ \\ 78=0.707A+C\ldots\ldots(4) \end{gathered}[/tex]Substracting equation (4) from (1),
[tex]\begin{gathered} 0.29A=70 \\ A=238.99 \end{gathered}[/tex]Substituting the value of A in the equation (1),
[tex]\begin{gathered} 238.99+C=8 \\ C=230.99 \end{gathered}[/tex]Thus, the required value of A is 238.99, the value of B is pi/240 and the value of C is 230.99.