Given that [tex]\(\sin 61^\circ = \sqrt{p}\)[/tex], we want to determine:
1. [tex]\(\sin 241^\circ\)[/tex]
2. [tex]\(\sin 29^\circ\)[/tex]
### Step-by-Step Solution:
#### Finding [tex]\(\sin 241^\circ\)[/tex]:
To find [tex]\(\sin 241^\circ\)[/tex], we use the fact that 241° can be expressed as [tex]\(180^\circ + 61^\circ\)[/tex]:
[tex]\[ 241^\circ = 180^\circ + 61^\circ \][/tex]
Using the sine addition formula, we know that:
[tex]\[ \sin(180^\circ + x) = -\sin(x) \][/tex]
Therefore,
[tex]\[ \sin(241^\circ) = \sin(180^\circ + 61^\circ) = -\sin(61^\circ) \][/tex]
Given that [tex]\(\sin 61^\circ = \sqrt{p}\)[/tex], we substitute:
[tex]\[ \sin(241^\circ) = -\sqrt{p} \][/tex]
So, the correct answer for [tex]\(\sin 241^\circ\)[/tex] is:
a. [tex]\(-\sqrt{p}\)[/tex]
#### Finding [tex]\(\sin 29^\circ\)[/tex]:
Next, we need to find [tex]\(\sin 29^\circ\)[/tex].
Using the complementary angle identity:
[tex]\[ 29^\circ = 90^\circ - 61^\circ \][/tex]
We know that:
[tex]\[ \sin(90^\circ - x) = \cos(x) \][/tex]
Thus,
[tex]\[ \sin(29^\circ) = \cos(61^\circ) \][/tex]
Now, we use the Pythagorean identity:
[tex]\[ \cos^2(x) + \sin^2(x) = 1 \][/tex]
Substituting [tex]\(\sin 61^\circ = \sqrt{p}\)[/tex], we get:
[tex]\[ \cos^2(61^\circ) + (\sqrt{p})^2 = 1 \][/tex]
[tex]\[ \cos^2(61^\circ) + p = 1 \][/tex]
[tex]\[ \cos^2(61^\circ) = 1 - p \][/tex]
Therefore,
[tex]\[ \cos(61^\circ) = \sqrt{1 - p} \text{ (since cosine is positive in the first quadrant)} \][/tex]
So,
[tex]\[ \sin(29^\circ) = \sqrt{1 - p} \][/tex]
The correct answer for [tex]\(\sin 29^\circ\)[/tex] is:
c. [tex]\(\sqrt{1 - p}\)[/tex]
### Final Answer:
1. [tex]\(\sin 241^\circ\)[/tex]:
a. [tex]\(-\sqrt{p}\)[/tex]
2. [tex]\(\sin 29^\circ\)[/tex]:
c. [tex]\(\sqrt{1 - p}\)[/tex]
