First, let's calculate the horizontal and vertical components of the wind speed (W) and the airplane speed (A), knowing that south is a bearing of 270° and northeast is a bearing of 45°:


Now, let's add the components of the same direction:

To find the resultant bearing (theta), we can use the formula below:

The angle -86° is equivalent to -86 + 360 = 274°.
Therefore the correct option is b.
<h3>Answer:</h3>
Equation of the ellipse = 3x² + 5y² = 32
<h3>Step-by-step explanation:</h3>
<h2>Given:</h2>
- The centre of the ellipse is at the origin and the X axis is the major axis
- It passes through the points (-3, 1) and (2, -2)
<h2>To Find:</h2>
- The equation of the ellipse
<h2>Solution:</h2>
The equation of an ellipse is given by,

Given that the ellipse passes through the point (-3, 1)
Hence,

Cross multiplying we get,
- 9b² + a² = 1 ²× a²b²
- a²b² = 9b² + a²
Multiply by 4 on both sides,
- 4a²b² = 36b² + 4a²------(1)
Also by given the ellipse passes through the point (2, -2)
Substituting this,

Cross multiply,
- 4b² + 4a² = 1 × a²b²
- a²b² = 4b² + 4a²-------(2)
Subtracting equations 2 and 1,
- 3a²b² = 32b²
- 3a² = 32
- a² = 32/3----(3)
Substituting in 2,
- 32/3 × b² = 4b² + 4 × 32/3
- 32/3 b² = 4b² + 128/3
- 32/3 b² = (12b² + 128)/3
- 32b² = 12b² + 128
- 20b² = 128
- b² = 128/20 = 32/5
Substituting the values in the equation for ellipse,


Multiplying whole equation by 32 we get,
3x² + 5y² = 32
<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
The answer is A) 12
Hope this helped
Pls give me brainlest
Total area: pi*9^2 = 81pi
Shaded area = Total area / 3, because 120/360 = 1/3
So Shaded area = 81pi/3 = 27pi