Given that Sin (5$_x$ − 28)$^o$ = Cos(3$_x$ − 50)$^o$, o < x < 90$^o$ Find the value of x

Analyze the question: We have sin(5x − 28)° = cos(3x − 50)° with 0 < x < 90°. We need x.

Concept: Sine and cosine of complementary angles are equal: sin θ = cos(90° − θ). So if sin A = cos B, then A + B = 90°.

Solution:

Set A = 5x − 28 and B = 3x − 50.

A + B = 90 (5x − 28) + (3x − 50) = 90 8x − 78 = 90 8x = 168 x = 21°

Check:

• 5(21) − 28 = 105 − 28 = 77°, so sin 77°
• 3(21) − 50 = 63 − 50 = 13°, so cos 13°
• cos 13° = sin(90° − 13°) = sin 77° ✓

Both sides are equal, and 21° lies between 0° and 90°.

Answer: B. 21°

Exam tip: Whenever you see sin = cos (or the reverse) in the same equation, immediately add the two angles and set the sum equal to 90°. Then solve the simple linear equation, and verify by substitution if you have time.