Question

an airship flies in a direction of 20 degrees north of west at 420 mph. it encounters a 2 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: Use the Law of Sines

The Law of Sines states that in a triangle, \(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\). Here, we assume a triangle - like situation for the airship's motion (with sides representing speeds and angles representing directions). If we consider the side lengths as \(a = 25\) (speed of the wind component), and \(b=405.4\) (ground - speed of the airship), and an angle \(B = 53^{\circ}\) (derived from the direction angles of the airship and wind).

Step2: Substitute into the Law of Sines formula

By substituting \(a = 25\), \(b = 405.4\) and \(B = 53^{\circ}\) into the Law of Sines formula \(\frac{\sin A}{a}=\frac{\sin B}{b}\), we get \(\frac{\sin A}{25}=\frac{\sin 53}{405.4}\).

For option B: The Law of Cosines \(a^{2}=b^{2}+c^{2}-2bc\cos A\) is used when we know two sides and the included angle or all three sides of a triangle. But in the context of finding the drift angle (which is more about the ratio of sides and angles, not about finding a side when two sides and an included angle are known), this formula is not appropriate.

For option C: \(c^{2}=a^{2}+b^{2}\) is the Pythagorean theorem which is applicable only for right - angled triangles. There is no indication that the triangle formed by the airship's motion (with respect to wind) is a right - angled triangle.

For option D: \(\tan^{-1}(\frac{420}{405.4})\) is the formula for finding an angle in a right - angled triangle (if we assume a right - angled triangle with opposite side \(420\) and adjacent side \(405.4\)). But again, there is no evidence that the relevant triangle is right - angled.

Answer:

A. \(\frac{\sin A}{25}=\frac{\sin 53}{405.4}\)