Question

an airship flies in a direction of 20 degrees north of west at 420 mph. it encounters a 25 mph wind that is heading 17 degrees east of south. its ground speed of 405.4 mph can be calculated. equation a below is used to find the drift angle of the airship. a. \\( \frac{\sin a}{25}=\frac{\sin 53}{405.4} \\) b. \\( 420^{2}=25^{2}+405.4^{2}-2(25)(405.4) \cos (a) \\) c. \\( c^{2}=420^{2}+405.4^{2} \\) d. \\( \tan ^{-1}\left(\frac{420}{405.4}\
ight) \\) the value of the drift angle is

Explanation:

Step1: Analyze the problem

This is a problem related to vectors (airship velocity and wind velocity) and using the Law of Sines.

Step2: Recall the Law of Sines

The Law of Sines states that in a triangle, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$.

Step3: Identify the sides and angles in the vector - triangle

Let the side opposite angle $A$ (drift angle) be the wind speed ($25$ mph), and the side opposite the known angle ($53^{\circ}$) be the ground - speed ($405.4$ mph).
So, by the Law of Sines, we have $\frac{\sin A}{25}=\frac{\sin 53}{405.4}$.

For the value of the drift angle:
From $\frac{\sin A}{25}=\frac{\sin 53}{405.4}$, we can solve for $A$.

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Answer:

A. $\frac{\sin A}{25}=\frac{\sin 53}{405.4}$