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 \( \sin(2x) = \cos(4x + 12^\circ) \), we can use the identity \( \cos(\theta) = \sin(90^\circ - \theta) \):

\[
\cos(4x + 12^\circ) = \sin(90^\circ - (4x + 12^\circ)) = \sin(78^\circ - 4x)
\]

Now we can rewrite our equation:

\[
\sin(2x) = \sin(78^\circ - 4x)
\]

This leads to two possibilities due to the properties of sine:

1. \( 2x = 78^\circ - 4x + k \cdot 360^\circ \) for some integer \( k \)
2. \( 2x = 180^\circ - (78^\circ - 4x) + k \cdot 360^\circ \) for some integer \( k \)

### Case 1:

Starting with the first equation:

\[
2x + 4x = 78^\circ
\]
\[
6x = 78^\circ
\]
\[
x = 13^\circ
\]

### Case 2:

Now for the second equation:

\[
2x = 180^\circ - 78^\circ + 4x
\]
\[
2x = 102^\circ + 4x
\]
\[
2x - 4x = 102^\circ
\]
\[
-2x = 102^\circ
\]
\[
x = -51^\circ \quad (\text{not valid since we want acute angles})
\]

So the only valid solution is \( x = 13^\circ \).

In a right triangle with angles \( x \), \( 90^\circ \), and \( 90^\circ - x \), the larger acute angle is:

\[
90^\circ - x = 90^\circ - 13^\circ = 77^\circ
\]

Thus, the measure of the larger angle is:

\[
\boxed{77^\circ}
\]