Answer :
The fighter aircraft is located 50 km away at an angle of 36.9 degrees west of south from the bomber aircraft. So, option(a) is correct from given options.
The fighter aircraft travels 30 km due south and the bomber travels 40 km due east. These two paths form a right angle triangle. By using the Pythagorean theorem (a^2 + b^2 = c^2), we can calculate the hypotenuse c, which represents the straight-line distance between the fighter and the bomber:
c = sqrt(30km^2 + 40km^2 ) = 50 km
Next, we find the angle. In a right-angle triangle, the tangent of an angle θ, is the ratio of the opposite side to the adjacent side. The angle between the direction south (where the fighter is) and the line joining the fighter and the bomber is θ. Therefore, tan θ = 40km (opposite) / 30km (adjacent)). Using the inverse tan function, we find θ = arctan(40/30) = 53.1 degrees.
However, when describing the position of one point with respect to another, the reference angle (Φ) is 0 degrees along the reference line (south in this case) and increases clockwise. As such, Φ = 90 degrees - θ = 36.9 degrees.
Then we say the fighter is 50 km away, 36.9 degrees west of south from the bomber.
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