We start with the equation \(\sin(x + 12^\circ) = \cos(18^\circ + 2x)\). Using the co-function identity, \(\cos(\theta) = \sin(90^\circ - \theta)\), we can rewrite it as:
\[ \sin(x + 12^\circ) = \sin(72^\circ - 2x) \]
This leads to two possible equations to solve:
- \(x + 12^\circ = 72^\circ - 2x + k \cdot 180^\circ\), or
- \(x + 12^\circ = 180^\circ - (72^\circ - 2x) + k \cdot 180^\circ\) (for some integer \(k\)).
For the first case where \(k = 0\):
\[ x + 2x = 72^\circ - 12^\circ \] \[ 3x = 60^\circ \Rightarrow x = 20^\circ \]
For the second case, simplifying gives us:
\[ x + 12^\circ = 108^\circ + 2x \] \[ 12^\circ - 108^\circ = 2x - x \Rightarrow -96^\circ = x \] (not valid for an angle x)
Thus, the only valid solution is \(x = 20^\circ\).
The smaller angle in the triangle is \(x\), which is \(20^\circ\).
I arrived at my answer by applying the co-function identity and simplifying the resulting equations to find the angle \(x\).