The value of `alpha` such that `sin^(-1)2/(sqrt(5)),sin^(-1)3/(sqrt(10)),sin^(-1)alpha` are the angles of a triangle is `(-1)/(sqrt(2))` (b) `1/2` (c) `1/(sqrt(3))` (d) `1/(sqrt(2))`

Options

A. `(-1)/(sqrt2)`
B. `(1)/(2)`
C. `(1)/(sqrt3)`
D. `(1)/(sqrt2)`

Explanation

To find the value of `alpha` such that `sin^(-1)(2/sqrt(5))`, `sin^(-1)(3/sqrt(10))`, and `sin^(-1)(alpha)` are the angles of a triangle, we can follow these steps:

### Step 1: Understand the Triangle Angle Sum Property
The sum of the angles in a triangle is equal to \( \pi \) radians (or 180 degrees). Therefore, we can write:
\[ \sin^{-1}\left(\frac{2}{\sqrt{5}}\right) + \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) + \sin^{-1}(\alpha) = \pi \]

### Step 2: Use the Identity for Sine Inverse
We can use the identity for the sine inverse function:
\[ \sin^{-1}(x) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \]
This means we can convert each sine inverse into a tangent inverse.

### Step 3: Convert Each Term
For \( \sin^{-1}\left(\frac{2}{\sqrt{5}}\right) \):
\[ \sin^{-1}\left(\frac{2}{\sqrt{5}}\right) = \tan^{-1}\left(\frac{\frac{2}{\sqrt{5}}}{\sqrt{1 - \left(\frac{2}{\sqrt{5}}\right)^2}}\right) \]
Calculating \( \sqrt{1 - \left(\frac{2}{\sqrt{5}}\right)^2} \):
\[ \sqrt{1 - \frac{4}{5}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]
Thus,
\[ \sin^{-1}\left(\frac{2}{\sqrt{5}}\right) = \tan^{-1}\left(\frac{2/\sqrt{5}}{1/\sqrt{5}}\right) = \tan^{-1}(2) \]

For \( \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \):
\[ \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) = \tan^{-1}\left(\frac{\frac{3}{\sqrt{10}}}{\sqrt{1 - \left(\frac{3}{\sqrt{10}}\right)^2}}\right) \]
Calculating \( \sqrt{1 - \left(\frac{3}{\sqrt{10}}\right)^2} \):
\[ \sqrt{1 - \frac{9}{10}} = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}} \]
Thus,
\[ \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) = \tan^{-1}\left(\frac{3/\sqrt{10}}{1/\sqrt{10}}\right) = \tan^{-1}(3) \]

### Step 4: Substitute Back into the Angle Sum Equation
Now substituting back into the equation:
\[ \tan^{-1}(2) + \tan^{-1}(3) + \sin^{-1}(\alpha) = \pi \]

### Step 5: Use the Tangent Addition Formula
Using the formula for the tangent of the sum:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
where \( a = 2 \) and \( b = 3 \):
\[ \tan^{-1}(2) + \tan^{-1}(3) = \tan^{-1}\left(\frac{2 + 3}{1 - 2 \cdot 3}\right) = \tan^{-1}\left(\frac{5}{1 - 6}\right) = \tan^{-1}\left(-\frac{5}{5}\right) = \tan^{-1}(-1) = -\frac{\pi}{4} \]

### Step 6: Solve for \( \sin^{-1}(\alpha) \)
Now, substituting back:
\[ -\frac{\pi}{4} + \sin^{-1}(\alpha) = \pi \]
Rearranging gives:
\[ \sin^{-1}(\alpha) = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \]
This is not possible since \( \sin^{-1}(\alpha) \) must be between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). Thus we need to consider the properties of angles in a triangle.

### Step 7: Find \( \alpha \)
Using the sine addition formula:
\[ \sin^{-1}(\alpha) = \pi - \left(\tan^{-1}(2) + \tan^{-1}(3)\right) \]
This means:
\[ \alpha = \sin\left(\pi - \left(\tan^{-1}(2) + \tan^{-1}(3)\right)\right) = \sin\left(\tan^{-1}(2) + \tan^{-1}(3)\right) \]
Using the sine addition formula:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
Where \( A = \tan^{-1}(2) \) and \( B = \tan^{-1}(3) \):
\[ \sin A = \frac{2}{\sqrt{5}}, \quad \cos A = \frac{1}{\sqrt{5}}, \quad \sin B = \frac{3}{\sqrt{10}}, \quad \cos B = \frac{1}{\sqrt{10}} \]
Thus:
\[ \alpha = \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}} + \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{5}} = \frac{2 + 3}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = \frac{1}{\sqrt{2}} \]