If `theta` and `phi` are angles in the first quadrant such that `tan theta = 1//7` and `sin phi =1 // sqrt( 10)` , then

Options

A. `theta + 2 phi = 90^(@)`
B. `theta = 2phi = 30^(@)`
C. ` theta + 2 phi = 75^(@)`
D. `theta + 2phi = 45^(@)`

Answer

To solve the problem, we need to find a relationship between the angles \( \theta \) and \( \phi \) given that \( \tan \theta = \frac{1}{7} \) and \( \sin \phi = \frac{1}{\sqrt{10}} \).

### Step 1: Find \( \sin \theta \) and \( \cos \theta \)

Given:
\[ \tan \theta = \frac{1}{7} \]
This means:
\[ \frac{\sin \theta}{\cos \theta} = \frac{1}{7} \]
We can represent \( \sin \theta \) and \( \cos \theta \) in terms of a right triangle where the opposite side (perpendicular) is 1 and the adjacent side (base) is 7.

Using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{1^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \]

Now we can find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{5\sqrt{2}}, \quad \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{7}{5\sqrt{2}} \]

### Step 2: Find \( \cos \phi \) and \( \tan \phi \)

Given:
\[ \sin \phi = \frac{1}{\sqrt{10}} \]
Using the Pythagorean identity \( \sin^2 \phi + \cos^2 \phi = 1 \):
\[ \cos^2 \phi = 1 - \sin^2 \phi = 1 - \left(\frac{1}{\sqrt{10}}\right)^2 = 1 - \frac{1}{10} = \frac{9}{10} \]
Thus:
\[ \cos \phi = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}} \]
Now we can find \( \tan \phi \):
\[ \tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{\frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}} = \frac{1}{3} \]

### Step 3: Find \( \sin 2\phi \)

Using the double angle formula:
\[ \sin 2\phi = 2 \sin \phi \cos \phi \]
Substituting the values:
\[ \sin 2\phi = 2 \cdot \frac{1}{\sqrt{10}} \cdot \frac{3}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5} \]

### Step 4: Find \( \cos 2\phi \)

Using the identity \( \cos 2\phi = 1 - 2\sin^2 \phi \):
\[ \cos 2\phi = 1 - 2\left(\frac{1}{\sqrt{10}}\right)^2 = 1 - 2 \cdot \frac{1}{10} = 1 - \frac{2}{10} = \frac{4}{5} \]

### Step 5: Use the sine addition formula

We want to find \( \sin(\theta + 2\phi) \):
\[ \sin(\theta + 2\phi) = \sin \theta \cos 2\phi + \cos \theta \sin 2\phi \]
Substituting the values:
\[ \sin(\theta + 2\phi) = \left(\frac{1}{5\sqrt{2}}\right) \left(\frac{4}{5}\right) + \left(\frac{7}{5\sqrt{2}}\right) \left(\frac{3}{5}\right) \]
Calculating each term:
\[ = \frac{4}{25\sqrt{2}} + \frac{21}{25\sqrt{2}} = \frac{25}{25\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 6: Relate \( \sin(\theta + 2\phi) \) to an angle

We know that:
\[ \sin(\theta + 2\phi) = \frac{1}{\sqrt{2}} \implies \theta + 2\phi = 45^\circ \]

### Final Result

Thus, we have established the relationship:
\[ \theta + 2\phi = 45^\circ \]