Certainly! Let's prove the given relations step by step.
Given:
[tex]\[ \frac{G}{100} = \frac{D}{90} = \frac{2R}{\pi} \][/tex]
Where:
- [tex]\( G \)[/tex] is the angle in grades,
- [tex]\( D \)[/tex] is the angle in degrees,
- [tex]\( R \)[/tex] is the angle in radians.
To prove this, let's consider two relations and then combine them to establish the final equation.
### Step 1: Proving the Relation [tex]\(\frac{G}{100} = \frac{D}{90}\)[/tex]
1. By definition, the total measurement around a circle can be expressed in different units:
- A full circle is [tex]\(400\)[/tex] grades,
- A full circle is [tex]\(360\)[/tex] degrees.
2. Therefore, the proportion of grades to degrees for any angle would be:
[tex]\[ \frac{G \text{ (grades)}}{400} = \frac{D \text{ (degrees)}}{360} \][/tex]
3. Simplifying this equation:
[tex]\[ \frac{G}{400} = \frac{D}{360} \][/tex]
4. Multiplying both sides of the equation by [tex]\( \frac{9}{10} \)[/tex]:
[tex]\[ \frac{G}{400} \cdot \frac{9}{10} = \frac{D}{360} \cdot \frac{9}{10} \][/tex]
[tex]\[ \frac{G}{400} \cdot \frac{9}{10} = \frac{D}{40} \cdot \frac{1}{10} \][/tex]
[tex]\[ \frac{G}{100} = \frac{D}{90} \][/tex]
Thus, we have proved that:
[tex]\[ \frac{G}{100} = \frac{D}{90} \][/tex]
### Step 2: Proving the Relation [tex]\(\frac{D}{90} = \frac{2R}{\pi}\)[/tex]
1. By definition, the total measurement around a circle can also be expressed in radians:
- A full circle is [tex]\(2\pi\)[/tex] radians,
- A full circle is [tex]\(360\)[/tex] degrees.
2. Therefore, the proportion of degrees to radians for any angle would be:
[tex]\[ \frac{D \text{ (degrees)}}{360} = \frac{R \text{ (radians)}}{2\pi} \][/tex]
3. Simplifying this equation:
[tex]\[ \frac{D}{360} = \frac{R}{2\pi} \][/tex]
4. Multiplying both sides of the equation by [tex]\(4\)[/tex]:
[tex]\[ 4 \cdot \frac{D}{360} = 4 \cdot \frac{R}{2\pi} \][/tex]
[tex]\[ \frac{D}{90} = \frac{2R}{\pi} \][/tex]
Thus, we have proved that:
[tex]\[ \frac{D}{90} = \frac{2R}{\pi} \][/tex]
### Step 3: Combining Both Relations
Since we have established the following equalities:
[tex]\[ \frac{G}{100} = \frac{D}{90} \][/tex]
[tex]\[ \frac{D}{90} = \frac{2R}{\pi} \][/tex]
We can combine them to get:
[tex]\[ \frac{G}{100} = \frac{D}{90} = \frac{2R}{\pi} \][/tex]
Therefore, we have successfully proven that:
[tex]\[ \frac{G}{100} = \frac{D}{90} = \frac{2R}{\pi} \][/tex]
