Answer :
Sure, let's thoroughly address each of these questions one by one:
### 1. Convert [tex]\(43.1025^\circ\)[/tex] to degrees, minutes, and seconds.
To convert a decimal degree measurement into degrees, minutes, and seconds, we follow these steps:
- The integer part of the decimal degree is the degree.
- Multiply the fractional part by 60 to get minutes.
- Take the fractional part of the minutes obtained and multiply it by 60 to get seconds.
Given:
[tex]\[ 43.1025^\circ \][/tex]
1. The integer part is 43 degrees.
2. The fractional part is 0.1025. Converting it to minutes:
[tex]\[ 0.1025 \times 60 = 6.15 \text{ minutes} \][/tex]
3. The integer part of the minutes is 6 minutes.
4. The fractional part is 0.15. Converting it to seconds:
[tex]\[ 0.15 \times 60 = 9 \text{ seconds} \][/tex]
Thus, [tex]\(43.1025^\circ\)[/tex] converts to:
[tex]\[ 43^\circ\, 6'\, 9'' \][/tex]
### 2. Find the positive co-terminal angle of [tex]\(30^\circ\)[/tex].
A positive co-terminal angle is an angle that lies within the range of [tex]\(0^\circ\)[/tex] to [tex]\(360^\circ\)[/tex] and has a difference of [tex]\(360^\circ\)[/tex] from the given angle.
Given:
[tex]\[ 30^\circ \][/tex]
Adding [tex]\(360^\circ\)[/tex] to the given angle:
[tex]\[ 30^\circ + 360^\circ = 390^\circ \][/tex]
So, the positive co-terminal angle of [tex]\(30^\circ\)[/tex] is:
[tex]\[ 390^\circ \][/tex]
### 3. Find the slope of a line if its angle of inclination is [tex]\(135^\circ\)[/tex].
The slope [tex]\(m\)[/tex] of a line is given by the tangent of its angle of inclination [tex]\(\theta\)[/tex]:
[tex]\[ m = \tan(\theta) \][/tex]
Given:
[tex]\[ \theta = 135^\circ \][/tex]
We find the tangent of [tex]\(135^\circ\)[/tex]:
[tex]\[ \tan(135^\circ) = -1 \][/tex]
Thus, the slope of the line is:
[tex]\[ -1 \][/tex]
### 4. Evaluate [tex]\(\sec\left(\frac{\pi}{2}\right) + \cos\left(\frac{2\pi}{3}\right)\)[/tex].
To solve this, we need the values of both [tex]\(\sec\left(\frac{\pi}{2}\right)\)[/tex] and [tex]\(\cos\left(\frac{2\pi}{3}\right)\)[/tex].
1. First, evaluate [tex]\(\sec\left(\frac{\pi}{2}\right)\)[/tex]:
[tex]\[ \sec\left(\frac{\pi}{2}\right) = \frac{1}{\cos\left(\frac{\pi}{2}\right)} \][/tex]
[tex]\[ \cos\left(\frac{\pi}{2}\right) = 0 \][/tex]
Since division by zero is undefined, this is considered to be:
[tex]\[ \sec\left(\frac{\pi}{2}\right) \to \infty \][/tex]
2. Then evaluate [tex]\(\cos\left(\frac{2\pi}{3}\right)\)[/tex]:
[tex]\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \][/tex]
So, the expression becomes:
[tex]\[ \sec\left(\frac{\pi}{2}\right) + \cos\left(\frac{2\pi}{3}\right) = \infty + \left(-\frac{1}{2}\right) = \infty \][/tex]
While [tex]\(\sec\left(\frac{\pi}{2}\right)\)[/tex] is technically undefined, for practical purposes, this expression evaluates to a very large number.
### 5. An angle inscribed in an arc less than a semicircle is an acute angle.
By definition, an angle inscribed in an arc that is smaller than a semicircle must be acute. This is because:
- An acute angle measures less than [tex]\(90^\circ\)[/tex].
- In a circle, an angle inscribed in an arc that is less than a semicircle is less than a right angle, making it acute.
Thus, it is true that an angle inscribed in an arc less than a semicircle is an acute angle.
### Summary:
The results of our computations are:
1. [tex]\( 43.1025^\circ = 43^\circ\, 6'\, 9'' \)[/tex]
2. The positive co-terminal angle of [tex]\(30^\circ\)[/tex] is [tex]\(390^\circ\)[/tex].
3. The slope of a line with an angle of inclination of [tex]\(135^\circ\)[/tex] is [tex]\(-1\)[/tex].
4. [tex]\(\sec\left(\frac{\pi}{2}\right) + \cos\left(\frac{2\pi}{3}\right)\)[/tex] approximates to a very large figure (technically undefined).
5. True, an angle inscribed in an arc less than a semicircle is an acute angle.
### 1. Convert [tex]\(43.1025^\circ\)[/tex] to degrees, minutes, and seconds.
To convert a decimal degree measurement into degrees, minutes, and seconds, we follow these steps:
- The integer part of the decimal degree is the degree.
- Multiply the fractional part by 60 to get minutes.
- Take the fractional part of the minutes obtained and multiply it by 60 to get seconds.
Given:
[tex]\[ 43.1025^\circ \][/tex]
1. The integer part is 43 degrees.
2. The fractional part is 0.1025. Converting it to minutes:
[tex]\[ 0.1025 \times 60 = 6.15 \text{ minutes} \][/tex]
3. The integer part of the minutes is 6 minutes.
4. The fractional part is 0.15. Converting it to seconds:
[tex]\[ 0.15 \times 60 = 9 \text{ seconds} \][/tex]
Thus, [tex]\(43.1025^\circ\)[/tex] converts to:
[tex]\[ 43^\circ\, 6'\, 9'' \][/tex]
### 2. Find the positive co-terminal angle of [tex]\(30^\circ\)[/tex].
A positive co-terminal angle is an angle that lies within the range of [tex]\(0^\circ\)[/tex] to [tex]\(360^\circ\)[/tex] and has a difference of [tex]\(360^\circ\)[/tex] from the given angle.
Given:
[tex]\[ 30^\circ \][/tex]
Adding [tex]\(360^\circ\)[/tex] to the given angle:
[tex]\[ 30^\circ + 360^\circ = 390^\circ \][/tex]
So, the positive co-terminal angle of [tex]\(30^\circ\)[/tex] is:
[tex]\[ 390^\circ \][/tex]
### 3. Find the slope of a line if its angle of inclination is [tex]\(135^\circ\)[/tex].
The slope [tex]\(m\)[/tex] of a line is given by the tangent of its angle of inclination [tex]\(\theta\)[/tex]:
[tex]\[ m = \tan(\theta) \][/tex]
Given:
[tex]\[ \theta = 135^\circ \][/tex]
We find the tangent of [tex]\(135^\circ\)[/tex]:
[tex]\[ \tan(135^\circ) = -1 \][/tex]
Thus, the slope of the line is:
[tex]\[ -1 \][/tex]
### 4. Evaluate [tex]\(\sec\left(\frac{\pi}{2}\right) + \cos\left(\frac{2\pi}{3}\right)\)[/tex].
To solve this, we need the values of both [tex]\(\sec\left(\frac{\pi}{2}\right)\)[/tex] and [tex]\(\cos\left(\frac{2\pi}{3}\right)\)[/tex].
1. First, evaluate [tex]\(\sec\left(\frac{\pi}{2}\right)\)[/tex]:
[tex]\[ \sec\left(\frac{\pi}{2}\right) = \frac{1}{\cos\left(\frac{\pi}{2}\right)} \][/tex]
[tex]\[ \cos\left(\frac{\pi}{2}\right) = 0 \][/tex]
Since division by zero is undefined, this is considered to be:
[tex]\[ \sec\left(\frac{\pi}{2}\right) \to \infty \][/tex]
2. Then evaluate [tex]\(\cos\left(\frac{2\pi}{3}\right)\)[/tex]:
[tex]\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \][/tex]
So, the expression becomes:
[tex]\[ \sec\left(\frac{\pi}{2}\right) + \cos\left(\frac{2\pi}{3}\right) = \infty + \left(-\frac{1}{2}\right) = \infty \][/tex]
While [tex]\(\sec\left(\frac{\pi}{2}\right)\)[/tex] is technically undefined, for practical purposes, this expression evaluates to a very large number.
### 5. An angle inscribed in an arc less than a semicircle is an acute angle.
By definition, an angle inscribed in an arc that is smaller than a semicircle must be acute. This is because:
- An acute angle measures less than [tex]\(90^\circ\)[/tex].
- In a circle, an angle inscribed in an arc that is less than a semicircle is less than a right angle, making it acute.
Thus, it is true that an angle inscribed in an arc less than a semicircle is an acute angle.
### Summary:
The results of our computations are:
1. [tex]\( 43.1025^\circ = 43^\circ\, 6'\, 9'' \)[/tex]
2. The positive co-terminal angle of [tex]\(30^\circ\)[/tex] is [tex]\(390^\circ\)[/tex].
3. The slope of a line with an angle of inclination of [tex]\(135^\circ\)[/tex] is [tex]\(-1\)[/tex].
4. [tex]\(\sec\left(\frac{\pi}{2}\right) + \cos\left(\frac{2\pi}{3}\right)\)[/tex] approximates to a very large figure (technically undefined).
5. True, an angle inscribed in an arc less than a semicircle is an acute angle.
