The range of values of `alpha` in the interval `(0,pi)` such that the points (3,5) and `(sinalpha, cos alpha)` lie on the same side of the line `x+y-1=0` is

Options

A. `(0,pi//4)`
B. `(pi//4,pi//2)`
C. `(0,pi//2)`
D. None

Answer

To find the range of values of \(\alpha\) in the interval \((0, \pi)\) such that the points \(A(3, 5)\) and \(B(\sin \alpha, \cos \alpha)\) lie on the same side of the line \(x + y - 1 = 0\), we will follow these steps:

### Step 1: Determine the line equation
The line equation is given by:
\[ x + y - 1 = 0 \]
We can rewrite it in the form \(y = -x + 1\).

### Step 2: Evaluate the position of point A
Substituting the coordinates of point \(A(3, 5)\) into the line equation:
\[ L_A = 3 + 5 - 1 = 7 \]
Since \(L_A > 0\), point \(A\) is above the line.

### Step 3: Evaluate the position of point B
Now we need to evaluate the position of point \(B(\sin \alpha, \cos \alpha)\) with respect to the line:
\[ L_B = \sin \alpha + \cos \alpha - 1 \]
For points \(A\) and \(B\) to be on the same side of the line, we require:
\[ L_B > 0 \]
This simplifies to:
\[ \sin \alpha + \cos \alpha - 1 > 0 \]
or
\[ \sin \alpha + \cos \alpha > 1 \]

### Step 4: Use the identity for sine and cosine
We can use the identity:
\[ \sin \alpha + \cos \alpha = \sqrt{2} \sin\left(\alpha + \frac{\pi}{4}\right) \]
Thus, we need:
\[ \sqrt{2} \sin\left(\alpha + \frac{\pi}{4}\right) > 1 \]
Dividing both sides by \(\sqrt{2}\):
\[ \sin\left(\alpha + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}} \]

### Step 5: Determine the range of \(\alpha\)
The inequality \(\sin\left(\alpha + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}}\) implies:
\[ \alpha + \frac{\pi}{4} \in \left(\frac{\pi}{4}, \frac{3\pi}{4}\right) \]
This leads to:
\[ \frac{\pi}{4} < \alpha + \frac{\pi}{4} < \frac{3\pi}{4} \]
Subtracting \(\frac{\pi}{4}\) from all parts:
\[ 0 < \alpha < \frac{\pi}{2} \]

### Step 6: Final range of \(\alpha\)
Thus, the range of values of \(\alpha\) such that both points lie on the same side of the line is:
\[ \alpha \in \left(0, \frac{\pi}{2}\right) \]