Given that sin(a°) = cos(b°) and a and b are expressed in terms of k, find the value of k.

We are given:

- Two acute angles: $ a^\circ $ and $ b^\circ $
- $ \sin(a^\circ) = \cos(b^\circ) $
- $ a = 4k - 22 $
- $ b = 6k - 13 $
- We are to find the value of $ k $

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Step 1: Use Trigonometric Identity

We know that for acute angles:
$$
\sin(\theta) = \cos(90^\circ - \theta)
$$

So,
$$
\sin(a^\circ) = \cos(b^\circ) \Rightarrow \cos(90^\circ - a^\circ) = \cos(b^\circ)
$$

Since cosine is one-to-one for acute angles (i.e., in $ 0^\circ $ to $ 90^\circ $), we can equate the angles:
$$
90^\circ - a = b
$$

So:
$$
a + b = 90^\circ
$$

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Step 2: Substitute expressions for $ a $ and $ b $

Given:
- $ a = 4k - 22 $
- $ b = 6k - 13 $

Then:
$$
(4k - 22) + (6k - 13) = 90
$$

Simplify:
$$
10k - 35 = 90
$$

Add 35 to both sides:
$$
10k = 125
$$

Divide by 10:
$$
k = 12.5
$$

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Step 3: Check if angles are acute

Let’s verify that $ a $ and $ b $ are acute (less than 90°):

- $ a = 4(12.5) - 22 = 50 - 22 = 28^\circ $ → acute ✔
- $ b = 6(12.5) - 13 = 75 - 13 = 62^\circ $ → acute ✔

Also, check $ \sin(28^\circ) \approx \cos(62^\circ) $?
Yes, because $ \cos(62^\circ) = \sin(90^\circ - 62^\circ) = \sin(28^\circ) $ → matches.

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Final Answer:

$$
\boxed{C) \ 12.5}
$$