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This is Giancoli Answers with Mr. Dychko. So the intention is to direct the plane such that it goes straight down with respect to the ground. So when you combine the velocity of the plane with respect to the air which has to be directed along some angle that we are gonna find here and then add to that the velocity of the air with respect to the ground the resultant, which is this, should be straight down. So in order for that to happen the x-component of the velocity of the plane with respect to the air has to be equal in magnitude but in the opposite direction so the negative of, in other words, of the velocity of the air with respect to the ground x-component. And so the x-component of the velocity of the Earth with respect to the ground is v AG times cos 45. And so that's 90 times cos 45 and so we now know the x-component of the velocity of the plane with respect to the air— negative 63.64 kilometers an hour. And so our velocity triangle for the plane with respect to the air vector looks like this; it's going along this direction Θ that's to the west of south and has this x-component and has this hypotenuse and we'll use inverse sign to figure out the angle Θ. So the inverse sign of the opposite divided by the hypotenuse is 63.64 kilometers per hour divided by 688 which gives 5.31 degrees towards the west of south.