91Ó°ÊÓ

From the given data we get the terms below

$$\begin{gathered} a+b\sin \frac{2\mathrm{τ}\mathrm{τ}×32}{366}+c\frac{2\mathrm{τ}\mathrm{τ}×32}{366}........\left(1\right) \\ a+b\sin \frac{2\mathrm{τ}\mathrm{τ}×77}{366}+c\frac{2\mathrm{τ}\mathrm{τ}×77}{366}........\left(2\right) \\ a+b\sin \frac{2\mathrm{τ}\mathrm{τ}×121}{366}+c\frac{2\mathrm{τ}\mathrm{τ}×121}{366}........\left(3\right) \\ a+b\sin \frac{2\mathrm{τ}\mathrm{τ}×152}{366}+c\frac{2\mathrm{τ}\mathrm{τ}×152}{366}........\left(4\right) \end{gathered}$$

Therefore, consider the system,

$\underset{A}{\underset{ÁåŸ}{\left[\begin{array}{ccc}1& \sin \frac{64\mathrm{Ï„}\mathrm{Ï„}}{366}& \cos \frac{154\mathrm{Ï„}\mathrm{Ï„}}{366}\\ 1& \sin \frac{154\mathrm{Ï„}\mathrm{Ï„}}{366}& \cos \frac{64\mathrm{Ï„}\mathrm{Ï„}}{366}\\ 1& \sin \frac{242\mathrm{Ï„}\mathrm{Ï„}}{366}& \cos \frac{242\mathrm{Ï„}\mathrm{Ï„}}{366}\\ 1& \sin \frac{304\mathrm{Ï„}\mathrm{Ï„}}{366}& \cos \frac{304\mathrm{Ï„}\mathrm{Ï„}}{366}\end{array}\right]}\underset{\stackrel{→}{x}}{\underset{ÁåŸ}{\left[\begin{array}{c}a\\ b\\ c\end{array}\right]}}}=\left[\begin{array}{c}10\\ 12\\ 14\\ 15\end{array}\right]$

If ker(A)=$\left\{\stackrel{âŠ\yen }{0}\right\}$then, $\stackrel{r}{x}*={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{r}{b}*$.

Since, the kernel of the matrix A is {$\stackrel{âŠ\yen }{0}$}.

So, it is written as,

$$\begin{gathered} A^{T}A=\left[\begin{array}{ccc}4.0000& 2.8732& -0.2475\\ 2.8732& 2.2341& -0.1773\\ 3183& & \end{array}\right] \\ {\left(A^{T}A\right)}^{-1}=\left[\begin{array}{ccc}3.2823& -4.2183& -0.0365\\ -4.2183& 5.8725& -0.0016\\ 0.0365& -0.0016& 0.5712\end{array}\right] \\ {\left(A^{T}A\right)}^{-1}{A}^T=\left[\begin{array}{cccc}1.1109& -0.7970& -0.4242& 1.1104\\ -1.1534& 1.4728& 0.9718& -1.2372\\ 0.5229& 0.1757& -0.2420& -0.4566\end{array}\right] \\ {\left(A^{T}A\right)}^{-1}{A}^T\stackrel{r}{b}*=\left[\begin{array}{c}12.2610\\ 0.4310\\ -2.8993\end{array}\right] \end{gathered}$$

Thus, it is written as,

role="math" localid="1659702486449" $\stackrel{r}{x}*=\left[\begin{array}{c}12.2610\\ 0.4310\\ -2.8993\end{array}\right]$

And the trigonometry function that best fits the data points is

$f*\left(t\right)=12.610+0.4310\sin \left(\frac{2\mathrm{Ï„}\mathrm{Ï„}}{366}t\right)-2.89993\cos \left(\frac{2\mathrm{Ï„}\mathrm{Ï„}}{366}t\right)$

Hence,$\stackrel{r}{x}*=\left[\begin{array}{c}12.2610\\ 0.4310\\ -2.8993\end{array}\right]$ and the required hours will be$f*\left(173\right)≈15.1913$ hrs.