In a certain triangle, the measures of `angle A and angle B` are `(6k-8)^(@) and (7k-45)^(@),` respectively. If `(sin angle A)/(cos angle B)=1,` what is the value of k ?

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Set up the equation
We are given that:
- \( \angle A = 6k - 8 \)
- \( \angle B = 7k - 45 \)

The condition provided is:
\[ \frac{\sin \angle A}{\cos \angle B} = 1 \]
This implies:
\[ \sin \angle A = \cos \angle B \]

### Step 2: Use the trigonometric identity
From the identity \( \sin x = \cos y \), we know that:
\[ \sin x = \cos y \quad \text{if and only if} \quad x = 90^\circ - y \quad \text{or} \quad y = 90^\circ - x \]
Thus, we can write:
\[ 6k - 8 = 90 - (7k - 45) \]

### Step 3: Simplify the equation
Now, simplify the right side:
\[ 6k - 8 = 90 - 7k + 45 \]
Combine like terms:
\[ 6k - 8 = 135 - 7k \]

### Step 4: Rearrange the equation
Now, we will bring all terms involving \( k \) to one side and constant terms to the other side:
\[ 6k + 7k = 135 + 8 \]
This simplifies to:
\[ 13k = 143 \]

### Step 5: Solve for \( k \)
Now, divide both sides by 13:
\[ k = \frac{143}{13} \]
Calculating this gives:
\[ k = 11 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{11} \]