In`DeltaPQR` measure of angle Q is `90^(@)`.If `sinP=(12)/(13),andPQ=1` cm ,then what is the length (in cm. of side QR?

Options

A. 2.6
B. 3
C. 2.4
D. 4

Answer

To solve the problem, we will use the properties of right triangles and trigonometric ratios. Here’s a step-by-step solution:

### Step 1: Understand the triangle and given information
In triangle PQR, angle Q is 90 degrees. We know:
- \( \sin P = \frac{12}{13} \)
- \( PQ = 1 \) cm

### Step 2: Identify the sides
In a right triangle, the sine of an angle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. Here:
- \( \sin P = \frac{\text{Opposite (QR)}}{\text{Hypotenuse (PR)}} \)

### Step 3: Set up the equation using sine
From the sine definition:
\[ \sin P = \frac{QR}{PR} \]
Substituting the known value:
\[ \frac{12}{13} = \frac{QR}{PR} \]

### Step 4: Express PR in terms of QR
From the equation, we can express PR as:
\[ PR = \frac{13}{12} \cdot QR \]

### Step 5: Apply the Pythagorean theorem
In triangle PQR, we can apply the Pythagorean theorem:
\[ PQ^2 + QR^2 = PR^2 \]
Substituting \( PQ = 1 \) cm:
\[ 1^2 + QR^2 = PR^2 \]
This simplifies to:
\[ 1 + QR^2 = PR^2 \]

### Step 6: Substitute PR in the Pythagorean theorem
Now substitute \( PR \) from Step 4 into the equation:
\[ 1 + QR^2 = \left(\frac{13}{12} \cdot QR\right)^2 \]
Expanding the right side:
\[ 1 + QR^2 = \frac{169}{144} QR^2 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 1 = \frac{169}{144} QR^2 - QR^2 \]
Factoring out \( QR^2 \):
\[ 1 = QR^2 \left(\frac{169}{144} - 1\right) \]
Calculating \( \frac{169}{144} - 1 \):
\[ 1 = QR^2 \left(\frac{169 - 144}{144}\right) = QR^2 \left(\frac{25}{144}\right) \]

### Step 8: Solve for QR^2
Now, multiply both sides by \( \frac{144}{25} \):
\[ QR^2 = \frac{144}{25} \]

### Step 9: Take the square root to find QR
Taking the square root gives:
\[ QR = \sqrt{\frac{144}{25}} = \frac{12}{5} = 2.4 \text{ cm} \]

### Final Answer
The length of side QR is \( 2.4 \) cm.

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