= Solution
Set $x_i=\sin(2\theta_i)$. The <maximum likelihood estimation> formulas for <simple linear regression> give
$$
\boxed{\widehat\beta=\frac{\sum_i(x_i-\bar x)(Y_i-\bar Y)}{\sum_i(x_i-\bar x)^2}=\frac{S_{xy}}{S_{xx}},\qquad\widehat\alpha=\bar Y-\widehat\beta\bar x.}
$$
Using the supplied rounded summaries gives $\widehat\beta\approx21082.7$ metres and $\widehat\alpha\approx312.0$ metres. Recomputing directly from the printed table gives approximately $21082.4$ and $312.0$ metres, respectively, so the small summary rounding differences do not affect the conclusion.
I would not accept the model merely because the <Student t-test> fails to reject a zero <regression intercept>. That <hypothesis test> tests one parameter restriction within the assumed model, not the adequacy of its error assumptions or its functional form. In particular, complementary elevations $\theta$ and $90^\circ-\theta$ have exactly the same $x_i$ and hence the same fitted mean, but their observed low-angle minus high-angle ranges are
$$
864,\quad1871,\quad1913,\quad1173\ \text{metres}
$$
for the four corresponding pairs. All four are positive, suggesting an elevation-dependent effect omitted by the mean model, such as air resistance, rather than random scatter determined only by $\sin(2\theta)$. This is a <regression diagnostics> concern, not a claim that these finite data logically rule out the normal-error model. The correct conclusion is \b[nonrejection of the intercept restriction does not validate the regression model]; further checks of the residual pattern and error assumptions are needed.
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