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Question

If $\sin(3x - 20)^\circ = \cos(20 - 3y)^\circ$, then value of x - y will be:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$30^\circ$

The problem asks for the value of $x - y$ given the trigonometric equation:

$ \sin(3x - 20)^\circ = \cos(20 - 3y)^\circ $

Using Trigonometric Identities to Solve for x - y

We can use the co-function identity $\sin(\theta) = \cos(90^\circ - \theta)$. Apply this to the left side of the equation:

  • $ \cos(90^\circ - (3x - 20^\circ)) = \cos(20 - 3y)^\circ $
  • $ \cos(90^\circ - 3x + 20^\circ) = \cos(20 - 3y)^\circ $
  • $ \cos(110^\circ - 3x) = \cos(20 - 3y)^\circ $

Since the cosine values are equal, the angles must be related. Assuming the simplest case where the angles are equal:

  • $ 110^\circ - 3x = 20 - 3y $

Rearrange the equation to solve for $x - y$:

  • $ 110^\circ - 20^\circ = 3x - 3y $
  • $ 90^\circ = 3(x - y) $
  • $ x - y = \frac{90^\circ}{3} $
  • $ x - y = 30^\circ $

Therefore, the value of $x - y$ is $30^\circ$.

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