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 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 } ( \frac { 420 } { 405.4 } ) \\) the value of the drift angle is

Explanation:

Step1: Analyze the problem

This is a problem about vector addition (airship velocity and wind velocity) and using the Law of Sines.

Step2: Recall the Law of Sines formula

The Law of Sines states that in a triangle with sides \(a,b,c\) and opposite angles \(A,B,C\) respectively, \(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)

Step3: Determine the angles

The airship's direction is \(20^{\circ}\) North of West and the wind is \(17^{\circ}\) East of South. The angle between the airship's velocity vector and the wind's velocity vector is \(180-(20 + 17)=143^{\circ}\). Let the drift - related angle be \(A\) and the other non - drift related angle in the triangle formed by the vectors be \(180 - 143=37^{\circ}\). But if we consider the Law of Sines application for the vectors (airship velocity \(v_{a} = 420\), wind velocity \(v_{w}=25\), ground - speed \(v_{g}=405.4\)), we can find that the angle related to the Law of Sines proportion for the drift angle \(A\) and the known - angle (by geometric vector - addition and angle - calculation) which is \(53^{\circ}\) (since \(90-(37)=53\)) gives the proportion \(\frac{\sin A}{25}=\frac{\sin 53}{405.4}\) (using Law of Sines \(\frac{\sin A}{v_{w}}=\frac{\sin\theta}{v_{g}}\) where \(\theta = 53^{\circ}\))

Step4: Solve for \(A\)

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Answer:

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