`SinA +cosB=1`, `A` =`30^(@)` and `B` is an acute angle ,then find the value of `B`.

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `0^(@)`

Answer

To solve the equation \( \sin A + \cos B = 1 \) given that \( A = 30^\circ \) and \( B \) is an acute angle, we can follow these steps:

### Step 1: Substitute the value of A
We know that \( A = 30^\circ \). Therefore, we can substitute this value into the equation:
\[ \sin 30^\circ + \cos B = 1 \]

### Step 2: Calculate \( \sin 30^\circ \)
The value of \( \sin 30^\circ \) is \( \frac{1}{2} \). So, we can rewrite the equation:
\[ \frac{1}{2} + \cos B = 1 \]

### Step 3: Isolate \( \cos B \)
To find \( \cos B \), we need to isolate it on one side of the equation. We can do this by subtracting \( \frac{1}{2} \) from both sides:
\[ \cos B = 1 - \frac{1}{2} \]
\[ \cos B = \frac{1}{2} \]

### Step 4: Find the value of B
We know that \( \cos B = \frac{1}{2} \). The angle \( B \) that satisfies this equation in the range of acute angles (0° to 90°) is:
\[ B = 60^\circ \]

### Final Answer
Thus, the value of \( B \) is:
\[ B = 60^\circ \]

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