Question
For the acute angles in a right triangle, sin(2x)=cos(4x+12°)
. What is the measure of the larger angle?(1 point)
°
. What is the measure of the larger angle?(1 point)
°
Answers
GPT-4o mini
To solve the equation \( \sin(2x) = \cos(4x + 12^\circ) \), we can use the co-function identity for sine and cosine:
\[
\sin(2x) = \cos(90^\circ - 2x)
\]
This implies:
\[
\cos(4x + 12^\circ) = \cos(90^\circ - 2x)
\]
Since the cosine function has the property that \( \cos A = \cos B \) implies \( A = B + k \cdot 360^\circ \) or \( A = -B + k \cdot 360^\circ \) for any integer \( k \), we can set up the following cases:
### Case 1:
\[
4x + 12^\circ = 90^\circ - 2x + k \cdot 360^\circ
\]
### Case 2:
\[
4x + 12^\circ = - (90^\circ - 2x) + k \cdot 360^\circ
\]
Let's solve Case 1 first.
1. Rearranging the equation gives:
\[
4x + 2x = 90^\circ - 12^\circ + k \cdot 360^\circ
\]
\[
6x = 78^\circ + k \cdot 360^\circ
\]
\[
x = 13^\circ + k \cdot 60^\circ
\]
#### For \( k = 0 \):
\[
x = 13^\circ
\]
#### For \( k = 1 \):
\[
x = 13^\circ + 60^\circ = 73^\circ
\]
#### For \( k = 2 \):
\[
x = 13^\circ + 120^\circ = 133^\circ \quad \text{(not acute, discard)}
\]
### Now check Case 2:
1. Rearranging gives:
\[
4x + 12^\circ = -90^\circ + 2x + k \cdot 360^\circ
\]
\[
4x - 2x = -90^\circ - 12^\circ + k \cdot 360^\circ
\]
\[
2x = -102^\circ + k \cdot 360^\circ
\]
\[
x = -51^\circ + k \cdot 180^\circ
\]
Only \( k = 1 \) gives a positive angle:
\[
x = -51^\circ + 180^\circ = 129^\circ \quad \text{(not acute, discard)}
\]
The valid acute angles we found are:
- \( x = 13^\circ \)
- \( x = 73^\circ \)
### Finding the larger angle in the triangle:
Since a right triangle has one angle of \(90^\circ\), the two acute angles \(x\) must sum to \(90^\circ\):
1. If \( x = 13^\circ\), then the other angle is \(77^\circ\).
2. If \( x = 73^\circ\), then the other angle is \(17^\circ\).
Given these sets of angles \( (13^\circ, 77^\circ) \) and \( (73^\circ, 17^\circ) \), the larger angle among the acute angles in the triangle is:
\[
\text{The larger acute angle is } 77^\circ.
\]
Thus, the measure of the larger angle is **77°**.
\[
\sin(2x) = \cos(90^\circ - 2x)
\]
This implies:
\[
\cos(4x + 12^\circ) = \cos(90^\circ - 2x)
\]
Since the cosine function has the property that \( \cos A = \cos B \) implies \( A = B + k \cdot 360^\circ \) or \( A = -B + k \cdot 360^\circ \) for any integer \( k \), we can set up the following cases:
### Case 1:
\[
4x + 12^\circ = 90^\circ - 2x + k \cdot 360^\circ
\]
### Case 2:
\[
4x + 12^\circ = - (90^\circ - 2x) + k \cdot 360^\circ
\]
Let's solve Case 1 first.
1. Rearranging the equation gives:
\[
4x + 2x = 90^\circ - 12^\circ + k \cdot 360^\circ
\]
\[
6x = 78^\circ + k \cdot 360^\circ
\]
\[
x = 13^\circ + k \cdot 60^\circ
\]
#### For \( k = 0 \):
\[
x = 13^\circ
\]
#### For \( k = 1 \):
\[
x = 13^\circ + 60^\circ = 73^\circ
\]
#### For \( k = 2 \):
\[
x = 13^\circ + 120^\circ = 133^\circ \quad \text{(not acute, discard)}
\]
### Now check Case 2:
1. Rearranging gives:
\[
4x + 12^\circ = -90^\circ + 2x + k \cdot 360^\circ
\]
\[
4x - 2x = -90^\circ - 12^\circ + k \cdot 360^\circ
\]
\[
2x = -102^\circ + k \cdot 360^\circ
\]
\[
x = -51^\circ + k \cdot 180^\circ
\]
Only \( k = 1 \) gives a positive angle:
\[
x = -51^\circ + 180^\circ = 129^\circ \quad \text{(not acute, discard)}
\]
The valid acute angles we found are:
- \( x = 13^\circ \)
- \( x = 73^\circ \)
### Finding the larger angle in the triangle:
Since a right triangle has one angle of \(90^\circ\), the two acute angles \(x\) must sum to \(90^\circ\):
1. If \( x = 13^\circ\), then the other angle is \(77^\circ\).
2. If \( x = 73^\circ\), then the other angle is \(17^\circ\).
Given these sets of angles \( (13^\circ, 77^\circ) \) and \( (73^\circ, 17^\circ) \), the larger angle among the acute angles in the triangle is:
\[
\text{The larger acute angle is } 77^\circ.
\]
Thus, the measure of the larger angle is **77°**.