If `cos20^(@)+2sin^(2)55^(@)=1+sqrt(2)sinK^(@)`, `K` in `(0, 90^(@))`, then `K=` _____.

Answer

To solve the equation \( \cos 20^\circ + 2 \sin^2 55^\circ = 1 + \sqrt{2} \sin K^\circ \), we will follow these steps:

### Step 1: Simplify the left-hand side
We start with the equation:
\[ \cos 20^\circ + 2 \sin^2 55^\circ \]
We know that \( \sin^2 \theta = 1 - \cos^2 \theta \), but we can also use the identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \). Thus, we can rewrite \( \sin^2 55^\circ \) as:
\[ \sin^2 55^\circ = \frac{1 - \cos 110^\circ}{2} \]
Substituting this back into the equation gives:
\[ \cos 20^\circ + 2 \left(\frac{1 - \cos 110^\circ}{2}\right) = \cos 20^\circ + 1 - \cos 110^\circ \]

### Step 2: Rewrite the equation
Now, our equation becomes:
\[ \cos 20^\circ + 1 - \cos 110^\circ = 1 + \sqrt{2} \sin K^\circ \]
We can simplify this to:
\[ \cos 20^\circ - \cos 110^\circ = \sqrt{2} \sin K^\circ \]

### Step 3: Use the cosine subtraction formula
We apply the cosine subtraction formula:
\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
Let \( A = 20^\circ \) and \( B = 110^\circ \):
\[ \cos 20^\circ - \cos 110^\circ = -2 \sin\left(\frac{20^\circ + 110^\circ}{2}\right) \sin\left(\frac{20^\circ - 110^\circ}{2}\right) \]
Calculating the averages:
\[ \frac{20^\circ + 110^\circ}{2} = 65^\circ, \quad \frac{20^\circ - 110^\circ}{2} = -45^\circ \]
Thus, we have:
\[ \cos 20^\circ - \cos 110^\circ = -2 \sin 65^\circ \sin(-45^\circ) \]
Since \( \sin(-\theta) = -\sin(\theta) \):
\[ \cos 20^\circ - \cos 110^\circ = 2 \sin 65^\circ \cdot \frac{1}{\sqrt{2}} \]

### Step 4: Equate and solve for \( K \)
Now we have:
\[ 2 \sin 65^\circ \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \sin K^\circ \]
This simplifies to:
\[ \sqrt{2} \sin 65^\circ = \sqrt{2} \sin K^\circ \]
Dividing both sides by \( \sqrt{2} \):
\[ \sin 65^\circ = \sin K^\circ \]

### Step 5: Determine \( K \)
Since \( K \) is in the interval \( (0, 90^\circ) \), we can directly conclude:
\[ K = 65^\circ \]

### Final Answer
Thus, the value of \( K \) is:
\[ \boxed{65} \]