To find the measure of the larger angle in the given right triangle, we start with the equation:
\[ \sin(2x) = \cos(4x + 12^\circ) \]
Using the co-function identity, we recognize that:
\[ \sin(2x) = \cos(90^\circ - 2x) \]
This gives us the equation:
\[ \cos(90^\circ - 2x) = \cos(4x + 12^\circ) \]
For the cosine function, the two angles are equal or differ by \(360^\circ\) or are supplementary, giving us two cases to consider:
Case 1:
\[ 90^\circ - 2x = 4x + 12^\circ \]
Rearranging this equation gives:
\[ 90^\circ - 12^\circ = 4x + 2x \]
\[ 78^\circ = 6x \]
\[ x = \frac{78^\circ}{6} = 13^\circ \]
Case 2:
\[ 90^\circ - 2x = -(4x + 12^\circ) + 360^\circ \]
Simplifying this, we have:
\[ 90^\circ - 2x = -4x - 12^\circ + 360^\circ \]
\[ 90^\circ - 2x = 348^\circ - 4x \]
Rearranging gives:
\[ 2x - 4x = 348^\circ - 90^\circ \]
\[ -2x = 258^\circ \]
\[ x = -129^\circ \]
Since \(x\) must be between \(0^\circ\) and \(90^\circ\), this case is invalid.
Thus, the only solution is from Case 1:
\[ x = 13^\circ \]
Now, in a right triangle, the other acute angle is:
\[ 90^\circ - x = 90^\circ - 13^\circ = 77^\circ \]
Now, we have the two acute angles \(13^\circ\) and \(77^\circ\). The larger angle in this triangle is:
\[ \boxed{77^\circ} \]
1 answer