Okay I am soo confused with these two questions:

The resultant of two forces acting on a body has a magnitude of 80 pounds. The angles between the resultant and the forces are 20° and 52°. Find the magnitude of the larger force.

Without the wind, a plane would fly due east at a rate of 150 mph. The wind is blowing southeast at a rate of 50 mph. The wind is blowing at a 45° angle from due east. How far off of the due east path does the wind blow the plane?

for this i got 10.8 degrees?

Mcos52+mcos20=80

Msin52+msin20=0
Solve for M.

second question

50cos45=windspeed south.
50sin45= windspeed east

plane vector: 150E+50sin45E +50cos45 S

so the angle is arc tan (35/(185))= 10.7 deg. check my math, I did most of it in my head.

To find the magnitude of the larger force in the first question, you can use the law of cosines. Here's how you can solve it step by step:

1. Label the forces as F₁ and F₂. Let's assume F₁ is the larger force.

2. Use the given angles to find the angles between the resultant and each force. The angle between the resultant and F₁ is 20°, and the angle between the resultant and F₂ is 52°.

3. Apply the law of cosines to find the magnitude of the resultant (R):
R² = F₁² + F₂² - 2F₁F₂cosθ,
where θ is the angle between the two forces.

4. Substitute the given values into the equation:
80² = F₁² + F₂² - 2F₁F₂cos20°.

5. Since we want to find the magnitude of the larger force (F₁), we need to express the equation in terms of F₁:
6400 = F₁² + F₂² - 2F₁F₂cos20°.

6. We can't directly solve this equation because we have two variables (F₁ and F₂). However, we can use the given information to eliminate one variable. If we rearrange the equation, we can express F₂ in terms of F₁:
F₂ = (F₁² + F₂² - 6400) / (2F₁cos20°).

7. Substitute this expression for F₂ into the equation:
6400 = F₁² + ((F₁² + F₂² - 6400) / (2F₁cos20°))² - 2F₁((F₁² + F₂² - 6400) / (2F₁cos20°))cos20°.

8. Simplify the equation:
6400 = F₁² + ((F₁² + ((F₁² + F₂² - 6400) / (2F₁cos20°)))² - 2F₁((F₁² + ((F₁² + F₂² - 6400) / (2F₁cos20°)))cos20°).

9. Solve this equation to find F₁.

For the second question, let's find the off-path distance caused by the wind step by step:

1. Visualize the scenario: The plane is flying due east at 150 mph, and the wind is blowing southeast at a rate of 50 mph.

2. The wind is blowing at a 45° angle from due east, forming a right-angled triangle with the plane's due east path.

3. The wind's speed is 50 mph. Since it's blowing at a 45° angle, we can split it into its components: one component is parallel to the plane's path, and the other is perpendicular to it. The component parallel to the path is 50 * cos(45°), and the perpendicular component is 50 * sin(45°).

4. The perpendicular component represents the off-path distance. Using the trigonometric relationship for right-angled triangles, we can calculate it as 50 * sin(45°).

5. Calculate the off-path distance:
off-path distance = 50 * sin(45°) = 50 * 0.7071 ≈ 35.36 miles.

So, the off-path distance caused by the wind is approximately 35.36 miles. It seems there might be an error in your calculation, as you mentioned 10.8 degrees, which does not align with the given information in the second question.