If G,D and `theta` be the number of grades, degrees and radians respectively in any angle,prove that:
(i) `D/90=G/100=(2theta)/pi` (ii) `G-D=(20theta)/pi`

Answer

To prove the given statements, we will start by establishing the relationships between grades (G), degrees (D), and radians (θ).

### Step 1: Establish Relationships
1. **Degrees to Radians**: We know that \( 180^\circ \) is equal to \( \pi \) radians. Therefore, we can express degrees in terms of radians:
\[ D \text{ degrees} = \frac{D \cdot \pi}{180} \text{ radians} \]

2. **Grades to Radians**: Similarly, \( 100 \) grades correspond to \( 90^\circ \) or \( \frac{\pi}{2} \) radians. Thus:
\[ G \text{ grades} = \frac{G \cdot \pi}{200} \text{ radians} \]

### Step 2: Prove Part (i)
We need to show that:
\[ \frac{D}{90} = \frac{G}{100} = \frac{2\theta}{\pi} \]

1. **Express θ in terms of D**:
From the relationship established in Step 1, we can write:
\[ \theta = \frac{D \cdot \pi}{180} \]
Therefore, substituting this into \( \frac{2\theta}{\pi} \):
\[ \frac{2\theta}{\pi} = \frac{2 \cdot \frac{D \cdot \pi}{180}}{\pi} = \frac{2D}{180} = \frac{D}{90} \]

2. **Express θ in terms of G**:
Similarly, from the grades relationship:
\[ \theta = \frac{G \cdot \pi}{200} \]
Substituting this into \( \frac{2\theta}{\pi} \):
\[ \frac{2\theta}{\pi} = \frac{2 \cdot \frac{G \cdot \pi}{200}}{\pi} = \frac{2G}{200} = \frac{G}{100} \]

3. **Conclude Part (i)**:
Since we have shown:
\[ \frac{D}{90} = \frac{2\theta}{\pi} \quad \text{and} \quad \frac{G}{100} = \frac{2\theta}{\pi} \]
It follows that:
\[ \frac{D}{90} = \frac{G}{100} = \frac{2\theta}{\pi} \]

### Step 3: Prove Part (ii)
We need to show that:
\[ G - D = \frac{20\theta}{\pi} \]

1. **Substituting D and G**:
From our earlier results:
\[ D = \frac{90 \cdot 2\theta}{\pi} \quad \text{and} \quad G = \frac{100 \cdot 2\theta}{\pi} \]

2. **Calculate G - D**:
\[ G - D = \left(\frac{100 \cdot 2\theta}{\pi}\right) - \left(\frac{90 \cdot 2\theta}{\pi}\right) \]
Simplifying this:
\[ G - D = \frac{(100 - 90) \cdot 2\theta}{\pi} = \frac{10 \cdot 2\theta}{\pi} = \frac{20\theta}{\pi} \]

### Conclusion
Thus, we have shown that:
\[ G - D = \frac{20\theta}{\pi} \]
Both parts of the problem are proved.