If a `Delta PQR " if" 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P =1 , " then" angleR` is equal to

Options

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

Answer

To solve the problem, we are given two equations involving the angles of triangle PQR:

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

We need to find the value of angle \( R \).

### Step 1: Square both equations

First, we will square both equations to eliminate the sine and cosine terms.

From Equation 1:
\[ (3 \sin P + 4 \cos Q)^2 = 6^2 \]
This expands to:
\[ 9 \sin^2 P + 24 \sin P \cos Q + 16 \cos^2 Q = 36 \quad \text{(Equation 3)} \]

From Equation 2:
\[ (4 \sin Q + 3 \cos P)^2 = 1^2 \]
This expands to:
\[ 16 \sin^2 Q + 24 \sin Q \cos P + 9 \cos^2 P = 1 \quad \text{(Equation 4)} \]

### Step 2: Add the two squared equations

Now, we will add Equation 3 and Equation 4.

\[ (9 \sin^2 P + 16 \cos^2 Q + 24 \sin P \cos Q) + (16 \sin^2 Q + 9 \cos^2 P + 24 \sin Q \cos P) = 36 + 1 \]

This simplifies to:
\[ 9 (\sin^2 P + \cos^2 P) + 16 (\sin^2 Q + \cos^2 Q) + 24 (\sin P \cos Q + \sin Q \cos P) = 37 \]

### Step 3: Use Pythagorean identity

Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 9(1) + 16(1) + 24(\sin P \cos Q + \sin Q \cos P) = 37 \]
This simplifies to:
\[ 25 + 24(\sin P \cos Q + \sin Q \cos P) = 37 \]

### Step 4: Solve for sine terms

Subtract 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 5: Use 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 6: Determine angles

The angles for which \( \sin(P + Q) = \frac{1}{2} \) are:
\[ P + Q = 30^\circ \quad \text{or} \quad P + Q = 150^\circ \]

### Step 7: Find angle R

Since \( P + Q + R = 180^\circ \):
1. If \( P + Q = 30^\circ \), then \( R = 180^\circ - 30^\circ = 150^\circ \) (not possible since angles in a triangle must be less than 180°).
2. If \( P + Q = 150^\circ \), then \( R = 180^\circ - 150^\circ = 30^\circ \).

### Step 8: Convert to radians

Convert \( R \) to radians:
\[ R = 30^\circ = \frac{\pi}{6} \]

### Final Answer

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