​ Trig Word Problem

This is what i got so far.....

From Nov 13 200 to Oct 1 2002 is 687 days

687-365.25=321.75 days [1 cycle]

365.25-321.75=43.5 days [full cycle - the remaining days = days in between?]

1. Days between A and B

\(\begin{array}{|rcll|} \hline d &=& 2\cdot 365,25\ \text{days} - 687\ \text{days} \\ d &=& 43.5\ \text{days} \\ \hline \end{array}\)

2. Angle \(\alpha\) = ACB:

\( \begin{array}{|rcll|} \hline \alpha &=& \frac{ 43.5\ \text{days} } {365.25\ \text{days} } \cdot 360^{\circ}\\ \alpha &=& 42.8747433265^{\circ}\\ \hline \end{array}\)

3. Angle \(\beta \) = ABC = BAC ( isosceles triangle )

\( \begin{array}{|rcll|} \hline 2\cdot \beta + \alpha &=& 180^{\circ} \\ 2\cdot \beta &=& 180^{\circ} - \alpha \\ \beta &=& 90^{\circ} - \frac{ \alpha } {2} \\ && \beta = 68.5626283368^{\circ}\\ \hline \end{array} \)

4. \(b = \overline{BA}:\)

\( \begin{array}{|rcll|} \hline \frac{ \sin(\alpha) } {b} &=& \frac{ \sin(\beta) } {1\ \text{AU}} \\ b &=& \frac{ \sin(\alpha) } { \sin(\beta) } \cdot 1\ \text{AU} \qquad & | \qquad \beta = 90^{\circ} - \frac{ \alpha } {2} \\ b &=& \frac{ \sin(\alpha) } { \sin(90^{\circ} - \frac{ \alpha } {2}) } \cdot 1\ \text{AU} \\ b &=& \frac{ \sin(\alpha) } { \cos(\frac{ \alpha } {2}) } \cdot 1\ \text{AU} \\ b &=& \frac{ \sin(\alpha) } { \cos(\frac{ \alpha } {2}) } \qquad & | \qquad b \text{ in AU}\\ && b = 0.70212731514\ \text{AU}\\ \hline \end{array}\)

5. Angle \(\delta\) = BDA:
 \( \begin{array}{|rcll|} \hline \delta + (\beta -17.1^{\circ}) + (\beta+42.8^{\circ}) &=& 180^{\circ} \\ \delta + 2\cdot \beta -17.1^{\circ} + 42.8^{\circ} &=& 180^{\circ} \\ \delta + 2\cdot \beta + 25.7^{\circ} &=& 180^{\circ} \\ \delta + 2\cdot \beta &=& 180^{\circ} - 25.7^{\circ}\\ \delta &=& 180^{\circ} - 25.7^{\circ} - 2\cdot \beta \qquad & | \qquad \beta = 90^{\circ} - \frac{ \alpha } {2} \\ \delta &=& 180^{\circ} - 25.7^{\circ} - 2\cdot ( 90^{\circ} - \frac{ \alpha } {2} ) \\ \delta &=& 180^{\circ} - 25.7^{\circ} - 180^{\circ} + \alpha \\ \delta &=& \alpha - 25.7^{\circ} \\ && \delta = 17.1747433265^{\circ} \\ \hline \end{array} \)

6. \( a = \overline{AD}:\)

\(\begin{array}{|rcll|} \hline \frac{ \sin(\delta) } {b} &=& \frac{ \sin(\beta+42.8^{\circ} ) } {a} \\ a &=& \frac{ \sin(\beta+42.8^{\circ} ) } { \sin(\delta) } \cdot b \qquad & | \qquad \beta = 90^{\circ} - \frac{ \alpha } {2} \qquad \delta = \alpha - 25.7^{\circ} \\ a &=& \frac{ \sin(90^{\circ} - \frac{ \alpha } {2}+42.8^{\circ} ) } { \sin(\alpha - 25.7^{\circ}) } \cdot b \\ a &=& \frac{ \sin(90^{\circ} - [~\frac{ \alpha } {2}-42.8^{\circ} ~]) } { \sin(\alpha - 25.7^{\circ}) } \cdot b \\ a &=& \frac{ \cos(\frac{ \alpha } {2}-42.8^{\circ} ) } { \sin(\alpha - 25.7^{\circ}) } \cdot b \qquad & | \qquad b = \frac{ \sin(\alpha) } { \cos(\frac{ \alpha } {2}) }\\ a &=& \frac{ \cos(\frac{ \alpha } {2}-42.8^{\circ} ) } { \sin(\alpha - 25.7^{\circ}) } \cdot \frac{ \sin(\alpha) } { \cos(\frac{ \alpha } {2}) }\\ && a = 2.30537070178\ \text{AU}\\ \hline \end{array} \)

7. \(x = \overline{CD}\)

\(\begin{array}{|rcll|} \hline x^2 &=& 1^2 + a^2 - 2\cdot 1 \cdot a \cdot \cos(17.1^{\circ}) \qquad & | \qquad x \text{ in AU}\\ x^2 &=& 1 + a^2 - 2 \cdot a \cdot \cos(17.1^{\circ}) \qquad & | \qquad a = 2.30537070178\ \text{AU}\\ x^2 &=& 1 + 2.30537070178^2 - 2 \cdot 2.30537070178 \cdot \cos(17.1^{\circ}) \\ x^2 &=& 1.90781964606\\ x &=& 1.38123844649\ \text{AU} \\ && x = 1.38123844649\ \text{AU} \cdot 93000000 \frac{ \text{miles} } {\text{AU}} \\ && x = 128455175.524\ \text{miles} \\ \hline \end{array} \)

The distance of mars from the sun is 1.38123844649 AU or 128,455,175.524 miles

But according to astronomy books, Mars distance from the Sun is:

Average=1.52 AU

Perihelion=1.41 AU

Aphelion =1.64 AU