Question

For the acute angles in a right triangle, sin(2x)=cos(4x+12°)
. What is the measure of the larger angle?(1 point)

Answer 1

To solve the equation \( \sin(2x) = \cos(4x + 12°) \), we can use the identity that relates sine to cosine:

\[
\sin(2x) = \cos(90° - 2x)
\]

This implies that:

\[
\cos(4x + 12°) = \cos(90° - 2x)
\]

From the properties of cosine, we know that:

\[
4x + 12° = 90° - 2x + k \cdot 360° \quad \text{or} \quad 4x + 12° = - (90° - 2x) + k \cdot 360°
\]

for some integer \( k \).

### First equation:
\[
4x + 12° = 90° - 2x
\]

Rearranging gives:

\[
4x + 2x = 90° - 12°
\]
\[
6x = 78°
\]
\[
x = 13°
\]

### Second equation:
\[
4x + 12° = -90° + 2x + k \cdot 360°
\]

Rearranging this gives:

\[
4x - 2x = -90° - 12° + k \cdot 360°
\]
\[
2x = -102° + k \cdot 360°
\]
\[
x = -51° + k \cdot 180°
\]

This solution does not yield an acute angle since \( x \) must be positive.

### Conclusion:
The valid solution is \( x = 13° \).

The angles in the right triangle are \( 90° \), \( x \), and \( 90° - x \). Thus, the larger acute angle is:

\[
90° - x = 90° - 13° = 77°
\]

Therefore, the measure of the larger acute angle is:

\[
\boxed{77°}
\]