91Ó°ÊÓ

Assuming there was no wind, we must find out the velocity vector the pilot was aiming for. The given angle is \(10.0^{\circ}\) west of north, and airspeed is \(200 \mathrm{km} / \mathrm{h}\). We can find the x and y components of the velocity vector using the sine and cosine functions. \(v_x = \sin(10.0^{\circ}) \times 200 \mathrm{km} / \mathrm{h}\) \(v_y = \cos(10.0^{\circ}) \times 200 \mathrm{km} / \mathrm{h}\) Calculate \(v_x\) and \(v_y\).

After flying for \(1 \mathrm{h}\), the pilot is \(15 \mathrm{km}\) off course to the west of where he expected to be. The final position of the pilot can be described as follows: \(x' = \int_0^1 (v_x - w_x)dt = v_x - w_x\) \(y' = \int_0^1 (v_y - w_y)dt = v_y - w_y\) The pilot is \(15 \mathrm{km}\) west off course, so: \(x' + 15 = v_x - w_x\) Thus, \(w_x = v_x - (x' + 15)\). To find the y-component of the wind velocity (\(w_y\)), notice that in one hour of flying, the pilot would have traveled \(y'\) northward. \(y' = v_y - w_y\) Therefore, \(w_y = v_y - y'\)

Now that we have the x and y components of the wind velocity, we can calculate the magnitude of the wind velocity using the Pythagorean theorem: \(w = \sqrt{w_x^2 + w_y^2}\) To find the angle (direction) of the wind, we can use the arctangent function: \(\theta = \arctan\left(\frac{w_y}{w_x}\right)\) Calculate \(w\) and \(\theta\) to obtain the wind velocity's magnitude and direction.

Now we wish to find the heading that would have allowed the pilot to fly directly to Oklahoma City without drifting off course. To achieve this, we need to determine the heading where the x and y components of the airspeed vector (\(v'_x\) and \(v'_y\)) combined with the wind vector (\(w_x\) and \(w_y\)) result in the desired angle of \(10.0^{\circ}\) west of north. \(v'_x = \sin(10.0^{\circ}) \times 330/1\) \(v'_y = \cos(10.0^{\circ}) \times 330/1\) The relation between the desired components (\(v'_x\) and \(v'_y\)) and the wind components (\(w_x\) and \(w_y\)) can be described as: \((v'_x - w_x) = 200 \cdot \sin(\phi)\) \((v'_y - w_y) = 200 \cdot \cos(\phi)\) Where \(\phi\) is the new heading angle we wish to find.

We can now solve the system of equations for the new heading angle \(\phi\). It's easier to first use the \(v'_y\) equation: \(\phi = \arccos\left(\frac{v'_y - w_y}{200}\right)\) Now, calculate \(\phi\) to find the direction the pilot should have headed to fly directly to Oklahoma City without being blown off course.