Question

For the acute angles in a right triangle, sin(3x+10°)=cos(x+24°)
sin
⁡
(
3
𝑥
+
10
°
)
=
cos
⁡
(
𝑥
+
24
°
)
. What is the value of x
𝑥
?(1 point)
Responses

28°
28
°
28 degrees

16°
16
°
16 degrees

31°
31
°
31 degrees

14°

Answer 1

To solve the equation \( \sin(3x + 10^\circ) = \cos(x + 24^\circ) \), we can use the co-function identity, which states that \( \sin(\theta) = \cos(90^\circ - \theta) \).

Thus, we can rewrite the equation as:

\[
\sin(3x + 10^\circ) = \sin(90^\circ - (x + 24^\circ))
\]

This simplifies to:

\[
\sin(3x + 10^\circ) = \sin(66^\circ - x)
\]

Now, we can set up two equations from this identity:

1. \( 3x + 10^\circ = 66^\circ - x + 360^\circ n \) where \( n \) is any integer.
2. \( 3x + 10^\circ = 180^\circ - (66^\circ - x) + 360^\circ n \) where \( n \) is any integer.

We'll solve the first equation first:

1. \( 3x + 10 = 66 - x \)

Adding \( x \) to both sides and subtracting \( 10 \):

\[
4x = 56
\]

Dividing by \( 4 \):

\[
x = 14^\circ
\]

Now let's check the second equation, but usually the first solution is sufficient for acute angles. However, for completeness:

2. \( 3x + 10 = 114 + x \)

Rearranging gives:

\[
3x - x = 114 - 10
\]
\[
2x = 104
\]
\[
x = 52^\circ
\]

Since \( 52^\circ \) is not an acute angle, we discard it.

Therefore, the only solution for \( x \) that is an acute angle is:

\[
\boxed{14^\circ}
\]