Question

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

Answer 1

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:

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