A 305 kg block is pulled by two forces. The first force is 124 N at a 29.6° angle, and the second force is 187 N at a 79.4° angle. What is the direction of the acceleration of the block?

Let's find the direction of the acceleration of the block when pulled by two forces at specific angles.

### Step-by-Step Solution:

1. Understand the Forces and Angles:
- We have two forces acting on the block:
- Force 1: [tex]$124 \, \text{N}$[/tex] at an angle of [tex]$29.6^\circ$[/tex]
- Force 2: [tex]$187 \, \text{N}$[/tex] at an angle of [tex]$79.4^\circ$[/tex]

2. Break Down Each Force into Components:
- For each force, we'll calculate the components in the horizontal (x-axis) and vertical (y-axis) directions.

3. Calculate Components for Force 1:
- Horizontal component ([tex]$F_{1x}$[/tex]):
[tex]\[
F_{1x} = 124 \, \text{N} \times \cos(29.6^\circ)
\][/tex]
- Vertical component ([tex]$F_{1y}$[/tex]):
[tex]\[
F_{1y} = 124 \, \text{N} \times \sin(29.6^\circ)
\][/tex]

4. Calculate Components for Force 2:
- Horizontal component ([tex]$F_{2x}$[/tex]):
[tex]\[
F_{2x} = 187 \, \text{N} \times \cos(79.4^\circ)
\][/tex]
- Vertical component ([tex]$F_{2y}$[/tex]):
[tex]\[
F_{2y} = 187 \, \text{N} \times \sin(79.4^\circ)
\][/tex]

5. Find Net Force Components:
- Net force in the x direction ([tex]$F_{\text{net}_x}$[/tex]):
[tex]\[
F_{\text{net}_x} = F_{1x} + F_{2x} = 142.22 \, \text{N}
\][/tex]
- Net force in the y direction ([tex]$F_{\text{net}_y}$[/tex]):
[tex]\[
F_{\text{net}_y} = F_{1y} + F_{2y} = 245.06 \, \text{N}
\][/tex]

6. Calculate the Direction of the Net Force:
- The direction (angle [tex]$\theta$[/tex]) of the net force with the horizontal can be calculated using the arctangent function:
[tex]\[
\theta = \tan^{-1}\left(\frac{F_{\text{net}_y}}{F_{\text{net}_x}}\right)
\][/tex]
- This gives us the angle:
[tex]\[
\theta \approx 59.87^\circ
\][/tex]

Thus, the direction of the acceleration of the block is approximately [tex]$59.87^\circ$[/tex] above the horizontal.