A

Step 1: Identify the forces acting on the block.
The weight of the block (W) is given by:
$$ W = m imes g = 10 ext{kg} imes 10 ext{m/s}^2 = 100 ext{N} $$
There is also an external force (F) of 100N applied at an angle of 37° to the horizontal.

Step 2: Resolve the applied force (F) into its horizontal (F_x) and vertical (F_y) components:
$$ F_x = F imes ext{cos}( heta) = 100 ext{N} imes ext{cos}(37^ ext{o}) $$
$$ F_y = F imes ext{sin}( heta) = 100 ext{N} imes ext{sin}(37^ ext{o}) $$
Using the known values of sin(37°) and cos(37°) (approximately 0.6 and 0.8, respectively), we calculate:
$$ F_x = 100 imes 0.8 = 80 ext{N} $$
$$ F_y = 100 imes 0.6 = 60 ext{N} $$

Step 3: Calculate the normal force (N):
The normal force is the weight of the block plus the vertical component of the applied force:
$$ N = W - F_y = 100 ext{N} - 60 ext{N} = 40 ext{N} $$

Step 4: Frictional force (F_friction) is given by:
$$ F_{friction} = ext{Coefficient of Friction} imes N $$

Step 5: Apply the condition of impending motion:
At the point of sliding, the frictional force equals the horizontal component of the applied force:
$$ F_{friction} = F_x $$
Hence, $$ ext{Coefficient of Friction} imes 40 = 80 $$
Therefore, the coefficient of friction is:
$$ ext{Coefficient of Friction} = rac{80}{40} = 2 $$

Step 6: Find the angle of friction (φ):
The angle of friction is related to the coefficient of friction by:
$$ an( ext{φ}) = ext{Coefficient of Friction} $$
Therefore:
$$ ext{φ} = an^{-1}(2) $$
This value can be approximated to find the angle of friction. Using a calculator, we find:
$$ ext{φ} ext{ is approximately } 63.4^ ext{o} $$

Therefore, the angle of friction is approximately 63.4°.
Hence, the final answer is: A.