If the distance angles of triangle ABC as shown in the figure is 45° and 71°, compute the value of $\Delta A^2 + \Delta A \Delta B + \Delta B^2$.
- 6.48
- 7.42
- 4.64
- 3.43
\[
10 – \log \sin 45^{\circ} = 9.849485002
\]
\[
10 – \log \sin 45^{\circ}00’01” = 9.849487108
\]
\[
\Delta A = 9.849487108 – 9.849485002
\]
\[
\Delta A = 2.10550886
\]
\[
10 – \log \sin 71^{\circ} = 9.975670065
\]
\[
10 – \log \sin 71^{\circ}00’01” = 9.97567079
\]
\[
\Delta B = 9.97567079 – 9.975670065
\]
\[
\Delta B = 0.725
\]
\[
\Delta A^2 + \Delta A \Delta B + \Delta B^2 = 2.10550886^2 + (2.10550886)(0.0725) + 0.725^2
\]
\[
\Delta A^2 + \Delta A \Delta B + \Delta B^2 = 2.10550886^2 + (2.10550886)(0.0725) + 0.725^2
\]
\[
\Delta A^2 + \Delta A \Delta B + \Delta B^2 = 6.485286483
\]
\[
\Delta A^2 + \Delta A \Delta B + \Delta B^2 \approx 6.48 \leftarrow \text{answer}
\]
