If cos x = sin(20 + x)° and 0° < x < 90°, then the value of x is 

Solution:

Given that, cos x = sin(20 + x)° and 0° < x < 90°

Clearly, it is given that x lies in 1st Quadrant

We know that sine and cosine are co-functions, which means cos x = sin (90^° - x) if x lies in Q1

So, we can replace cos x with sin(90 - x) in the given equation

⇒ cos x = sin(20 + x)°

⇒ sin(90 - x) = sin(20 + x)

Apply sin^-1 on both sides, we get sin^-1 and sin get cancelled

90° - x = 20° + x

90° - 20° = x + x

70° = 2x

x = 35°

Therefore, the value of x is 35°.

If cos x = sin(20 + x)° and 0° < x < 90°, then the value of x is 

Summary:

If cos x = sin(20 + x)° and 0° < x < 90°, then the value of x is 35°.