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 know that \( 2\sin^2 \theta = 1 - \cos(2\theta) \). Therefore, we can rewrite \( 2\sin^2 55^\circ \) as:
\[ 2\sin^2 55^\circ = 1 - \cos(110^\circ) \]
So, substituting this into the equation gives:
\[ \cos 20^\circ + 1 - \cos 110^\circ \]

### Step 2: Combine the terms
Now we have:
\[ \cos 20^\circ + 1 - \cos 110^\circ \]
We can rearrange this to:
\[ 1 + \cos 20^\circ - \cos 110^\circ \]

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

### Step 4: Substitute back into the equation
Now substituting back into our equation:
\[ 1 + \frac{2 \sin(65^\circ)}{\sqrt{2}} = 1 + \sqrt{2} \sin K^\circ \]

### Step 5: Compare both sides
From this, we can compare the coefficients of \( \sin K^\circ \):
\[ \frac{2 \sin(65^\circ)}{\sqrt{2}} = \sqrt{2} \sin K^\circ \]
Dividing both sides by \( \sqrt{2} \):
\[ \frac{2 \sin(65^\circ)}{2} = \sin K^\circ \]
Thus:
\[ \sin K^\circ = \sin(65^\circ) \]

### Step 6: Solve for \( K \)
Since \( K \) is in the range \( (0, 90) \), we have:
\[ K = 65^\circ \]

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