If sin 3A = cos(A − 26°),then A = ________, where 3A is an acute angle.

Correct — C. The stem is sin 3A = cos(A − 26°), a sine set equal to a cosine, so turn one into the other with the complement rule.

sin 3A = cos(90° − 3A)
So cos(90° − 3A) = cos(A − 26°)
Equate the angles: 90° − 3A = A − 26°

116° = 4A
A = 29° → option (c)

Check the stem's condition: 3A = 87°, which is acute.

Why the others are wrong

• (a)
  78° is 3 × 26°, the value 3A would take if A were 26°. The complement rule fixes 3A + (A − 26°) = 90°, which gives A = 29°, not 3A = 78°.
• (b)
  26° is the constant copied from the stem. Put it back: sin 78° ≈ 0.98 while cos 0° = 1, so the equation fails.
• (d)
  52° is 2 × 26°. It makes 3A = 156°, which is not acute, so the stem's own condition rules it out.

Sine and cosine are co-functions: sin θ = cos(90° − θ) for every θ. So an equation with a sine on one side and a cosine on the other is solved by rewriting one side and then equating angles.

Here that gives 3A + (A − 26°) = 90°, a single linear equation. The two angles inside the ratios must add to 90°, which is the whole content of the question.

The acute condition on 3A is not decoration. It picks out the one solution in the first quadrant and lets you reject anything the general solution would otherwise allow.

Key facts

• sin θ = cos(90° − θ) and cos θ = sin(90° − θ).
• sin 3A = cos(A − 26°) therefore forces 3A + (A − 26°) = 90°.
• 4A = 116° gives A = 29°, and 3A = 87° satisfies the acute condition.
• 52° is 2 × 26° and 78° is 3 × 26°, both built from the constant in the stem.

Common traps

• Equating the angles directly as 3A = A − 26°, which ignores the co-function step.
• Marking the constant 26° straight off the stem without solving.
• Solving for 3A and reporting 87° when the question asks for A.

No directly related past PYQ was found.