Question

For the acute angles in a right triangle, sin(2x)=cos(4x+12°)
. What is the measure of the larger angle?(1 point)
°
Remaining Attempts : 3

Answer 1

To solve the equation \( \sin(2x) = \cos(4x + 12^\circ) \) where \( x \) is an acute angle in a right triangle, we'll use the identity \( \cos A = \sin(90^\circ - A) \).

First, let's rewrite the equation using the sine function:

\[
\sin(2x) = \sin(90^\circ - (4x + 12^\circ))
\]

This simplifies to:

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

From the property of the sine function, we know that if \( \sin A = \sin B \), then:

1. \( A = B + n \times 180^\circ \)
2. \( A = 180^\circ - B + n \times 180^\circ \)

where \( n \) is an integer.

For our case:

1. \( 2x = 78^\circ - 4x + n \times 180^\circ \) (Usually we use \( n = 0 \) since we are looking for acute angles)
2. \( 2x = 180^\circ - (78^\circ - 4x) + n \times 180^\circ \)

We'll solve the first equation:

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

Now, using the second equation:

\[
2x = 180^\circ - 78^\circ + 4x
\]
\[
2x = 102^\circ + 4x
\]
\[
2x - 4x = 102^\circ
\]
\[
-2x = 102^\circ
\]
\[
x = -51^\circ \quad (\text{not an acute angle})
\]

Thus, the only valid solution for \( x \) from our calculations is \( x = 13^\circ \).

In a right triangle, the other acute angle can be found as follows:

\[
\text{Other angle} = 90^\circ - x = 90^\circ - 13^\circ = 77^\circ
\]

Finally, the larger acute angle is:

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