In a `DeltaPQR`. if `3 sinP +4 cosQ=6` and `4sinQ+3cosP=1`, then the angle `R` is equal to:

Options

A. `( 5pi)/(6)`
B. `(pi)/(6)`
C. `(pi)/(4)`
D. `(3pi)/(4)`

Explanation

To solve the problem, we start with the given equations:

1. \( 3 \sin P + 4 \cos Q = 6 \) (Equation 1)
2. \( 4 \sin Q + 3 \cos P = 1 \) (Equation 2)

### Step 1: Square both equations and add them.

Squaring Equation 1:
\[ (3 \sin P + 4 \cos Q)^2 = 6^2 \]
Expanding this:
\[ 9 \sin^2 P + 24 \sin P \cos Q + 16 \cos^2 Q = 36 \quad (1) \]

Squaring Equation 2:
\[ (4 \sin Q + 3 \cos P)^2 = 1^2 \]
Expanding this:
\[ 16 \sin^2 Q + 24 \sin Q \cos P + 9 \cos^2 P = 1 \quad (2) \]

### Step 2: Add the two squared equations.

Adding (1) and (2):
\[ 9 \sin^2 P + 16 \cos^2 Q + 16 \sin^2 Q + 9 \cos^2 P + 24 (\sin P \cos Q + \sin Q \cos P) = 36 + 1 \]
This simplifies to:
\[ 9 (\sin^2 P + \cos^2 P) + 16 (\sin^2 Q + \cos^2 Q) + 24 (\sin P \cos Q + \sin Q \cos P) = 37 \]
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\[ 9(1) + 16(1) + 24 (\sin P \cos Q + \sin Q \cos P) = 37 \]
This simplifies to:
\[ 25 + 24 (\sin P \cos Q + \sin Q \cos P) = 37 \]
Subtracting 25 from both sides:
\[ 24 (\sin P \cos Q + \sin Q \cos P) = 12 \]
Dividing by 24:
\[ \sin P \cos Q + \sin Q \cos P = \frac{1}{2} \]

### Step 3: Use the sine addition formula.

Using the sine addition formula:
\[ \sin(P + Q) = \sin P \cos Q + \cos P \sin Q \]
We have:
\[ \sin(P + Q) = \frac{1}{2} \]

### Step 4: Determine \( P + Q \).

The equation \( \sin(P + Q) = \frac{1}{2} \) implies:
\[ P + Q = \frac{\pi}{6} \quad \text{or} \quad P + Q = \frac{5\pi}{6} \]

### Step 5: Determine angle \( R \).

Since the angles in a triangle sum to \( \pi \):
\[ P + Q + R = \pi \]
Thus:
\[ R = \pi - (P + Q) \]

#### Case 1: If \( P + Q = \frac{\pi}{6} \)
\[ R = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

#### Case 2: If \( P + Q = \frac{5\pi}{6} \)
\[ R = \pi - \frac{5\pi}{6} = \frac{\pi}{6} \]

### Step 6: Check the validity of the cases.

1. For \( R = \frac{5\pi}{6} \):
- \( P + Q = \frac{\pi}{6} \) leads to contradictions in the original equations.

2. For \( R = \frac{\pi}{6} \):
- \( P + Q = \frac{5\pi}{6} \) is valid.

### Conclusion:

Thus, the angle \( R \) is:
\[ \boxed{\frac{\pi}{6}} \]