Two forces of magnitudes 7N and 3N act at right angles to each other. The angle \(\theta\) between the resultant and the 7N force is given by
cos\(\theta\) = \(\frac{3}{7}\)
sin\(\theta\) = \(\frac{3}{7}\)
tan\(\theta\) = \(\frac{3}{7}\)
cot\(\theta\) = \(\frac{3}{7}\)
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7N force acts horizontally, and the 3N force acts vertically.
Because they’re at right angles, we can apply Pythagoras to find the resultant.
R = √(7² + 3²) = √(49 + 9) = √58
But wait... we don’t actually need the magnitude of R to solve this.
We’re finding the angle θ between the 7N force and the resultant.
Adjacent = 7N
Opposite = 3N
Resultant = hypotenuse
So, the angle θ is between the adjacent (7N) and hypotenuse (resultant).
To find angle θ, we use this identity:
cosθ = adjacent / hypotenuse
But we don’t have the hypotenuse... so let’s try another identity. 
Let’s use: tanθ = opposite / adjacent
So: tanθ = 3 / 7 
And boom!
That’s the answer! (C) 

The selected answer is wrong:
Given that the 2 forces are at right angle to each other, the angle to be calculated is between the force 7 and the resultant. Resolving it, you'll get tantita=3/7

Find by means of vector diagrams the resultant of two force 7n and 3n perpendicular to each other


