91Ó°ÊÓ

Adam is flying home from Oklahoma city to St. Louis.

One option is to take the direct flight to St. Louis which is $461.1$miles long.

The other option is to take a connecting flight to Kansas City and then to St. Louis.

Bearing from Oklahoma City to Kansas City $N29.6^{∘}E$Bearing from Oklahoma City to St. Louis:$N57.7^{∘}E$Bearing from St. Louis to Kansas City $N79.4^{∘}W$.Bearing from St. Louis to Oklahoma City$S57.7^{∘}W$

The sketch is shown as

The measure of the angle $\mathrm{Î\pm }$that completes a straight line with angles$79.4°$and role="math" localid="1649919204922" $57.7^{∘}$:

$$\begin{gathered} \mathrm{Î\pm }+79.4+57.7=180 \\ \mathrm{Î\pm }=180-79.4-57.7 \\ \mathrm{Î\pm }=42.9^{∘} \end{gathered}$$

The angle $\mathrm{β}$at the vertex Oklahoma city

$$\begin{gathered} \mathrm{β}+29.6=57.7 \\ \mathrm{β}=28.1^{∘} \end{gathered}$$

The last angle inside the triangle $\mathrm{γ}$

$$\begin{gathered} 42.9+28.1+\mathrm{γ}=180 \\ \mathrm{γ}={109}^{∘} \end{gathered}$$

$$\begin{gathered} \frac{\sin \left(42.9^{∘}\right)}{d_1}=\frac{\sin \left({109}^{∘}\right)}{461.1} \\ {d}_1=\sin \left(42.9^{∘}\right)\frac{\sin \left({109}^{∘}\right)}{461.1} \\ {d}_1=331.97miles \end{gathered}$$

$$\begin{gathered} \frac{\sin \left(28.1^{∘}\right)}{d_2}=\frac{\sin \left({109}^{∘}\right)}{461.1} \\ {d}_2=\sin \left(28.1^{∘}\right)\frac{\sin \left({109}^{∘}\right)}{461.1} \\ {d}_2=229.7miles \end{gathered}$$

The length of the connecting flight is then equal to:

$d_1+d_2=331.97+229.7=561.67miles$

which means that the connecting flight is:$561.67−461.1=100.57$miles longer.