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(III) An airplanc, whose air speed is 580 $\mathrm{km} / \mathrm{h}$ , is supposedto fly in a straight path $38.0^{\circ} \mathrm{N}$ of E. But a steady 72 $\mathrm{km} / \mathrm{h}$wind is blowing from the north. In what direction should theplane head?

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$43.6^{\circ}$ north of east

Physics 101 Mechanics

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

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So it's draw our triangle. And so this would be the velocity of the air relative to the ground. Ah, this angle, he would be 38 degrees. We can say that this factor here would be the velocity of the plane relative to the ground. Uh, this would be our angle, Fada. And then this sector here are high pop news. Essentially would be the vector, the velocity vector of the plane relative to the air. And so we can say that we're gonna call the East Positive and the North positive. Uh, and so we can say that we can apply the law of signs here in order to solve this. We know that this angle here also is 128 degrees. That's how well be able to use the law of signs. And so we can sit at the velocity of the plane relative to the air, divided by sign of 128 degrees would be equal to the velocity of the air relative to the ground, divided by sign. If Ada and so sign of data rather recon, simply say the data is gonna be the arc sine of velocity of the air relative to the ground, divided by be velocity of the plane relative to the air and then multiplied by signed brother multiplied by sine of 128 degrees closed parentheses. And so we can simply so solve this. So Fada is gonna equal arc sine of 72 divided by 580 times. Sign of 128 degrees. Uh, this is gonna give us 43.6 degrees north of east. Given that we have positive X and positive right components, that is the end of the solution. Thank you for

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