If (48°+ k) is an acute angle and sin(48°+ k) = cos13°, what is the value of k(in °)?

Correct — D. Use the co-function identity sin θ = cos(90° − θ).

sin(48° + k) = cos(90° − 48° − k) = cos(42° − k)
So cos(42° − k) = cos13°
Both angles are acute, so 42° − k = 13°
k = 42° − 13° = 29 → option (d)

Check: 48° + 29° = 77°, still acute, and sin77° = cos13°.

Why the others are wrong

• (a)
  23 — k = 23 makes the angle 71°, and sin71° = cos19°. The complement needed is 13°, not 19°, so the equation does not hold.
• (b)
  17 — k = 17 makes the angle 65°, whose sine equals cos25°. That is 12° away from the cos13° the stem gives.
• (c)
  37 — k = 37 makes the angle 85°, and sin85° = cos5°. The two angles must sum to 90°, and 85° + 13° = 98°.

Sine and cosine are co-functions: what sine does to an angle, cosine does to its complement. So sin θ = cos(90° − θ) for every θ.

When a question equates a sine and a cosine, it is asking you to make the two angles add to 90°. Here 48° + k and 13° must be complementary, which pins k at a single value.

The acute condition on 48° + k matters: without it, sine and cosine can be equated at other angles, and more than one k would work.

Key facts

• sin θ = cos(90° − θ) and cos θ = sin(90° − θ) hold for every angle θ.
• For acute A and B, sin A = cos B exactly when A + B = 90°.
• The same pairing runs through the other co-functions: tan θ = cot(90° − θ) and sec θ = cosec(90° − θ).
• sin77° = cos13°, the identity this question is built on.

Common traps

• Equating the angles directly, 48° + k = 13°, which has no acute solution.
• Adding rather than subtracting: 48° + 13° = 61° is not k.
• Ignoring the acute condition, which is the clause that fixes one value of k.

No directly related past PYQ was found.