If AP, BQ and CR are the altitudes of acute `triangleABC and 9AP+4BQ+7CR=0` `angleABC` is equal to

Options

A. a. `cos^(-1)(2)/(sqrt(7))`
B. b. `(pi)/(2)`
C. c. `"cos"^(-1)((sqrt(7))/(3))`
D. d. `(pi)/(3)`

Answer

To solve the problem, we need to find the angle \( \angle ABC \) given the equation \( 9AP + 4BQ + 7CR = 0 \), where \( AP, BQ, \) and \( CR \) are the altitudes of triangle \( ABC \).

### Step-by-Step Solution:

1. **Understanding the Equation**:
The equation \( 9AP + 4BQ + 7CR = 0 \) implies a relationship between the altitudes of triangle \( ABC \). Since the altitudes are directed towards the opposite vertices, we can interpret this as a vector equation.

2. **Setting Up the Triangle**:
Let \( AP = h_a \), \( BQ = h_b \), and \( CR = h_c \). The equation can be rewritten as:
\[ 9h_a + 4h_b + 7h_c = 0 \]
This suggests that the altitudes are proportional to the sides of the triangle.

3. **Using the Area of Triangle**:
The area \( A \) of triangle \( ABC \) can be expressed using any of the altitudes:
\[ A = \frac{1}{2} \times BC \times h_a = \frac{1}{2} \times AC \times h_b = \frac{1}{2} \times AB \times h_c \]
From this, we can derive relationships between the sides and the altitudes.

4. **Setting Ratios**:
From the equation \( 9h_a + 4h_b + 7h_c = 0 \), we can express the ratios of the sides:
\[ \frac{h_a}{9} = \frac{h_b}{4} = \frac{h_c}{7} \]
Let \( k \) be a constant such that:
\[ h_a = 9k, \quad h_b = 4k, \quad h_c = 7k \]

5. **Finding the Sides**:
Using the area relationships:
\[ A = \frac{1}{2} \times BC \times h_a = \frac{1}{2} \times AC \times h_b = \frac{1}{2} \times AB \times h_c \]
We can express the sides \( a, b, c \) as:
\[ a = 7k, \quad b = 9k, \quad c = 4k \]

6. **Using the Cosine Rule**:
To find \( \angle ABC \), we can use the cosine rule:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Substituting the values:
\[ a = 7k, \quad b = 9k, \quad c = 4k \]
\[ \cos B = \frac{(7k)^2 + (4k)^2 - (9k)^2}{2 \cdot (7k)(4k)} \]
Simplifying:
\[ \cos B = \frac{49k^2 + 16k^2 - 81k^2}{56k^2} = \frac{-16k^2}{56k^2} = -\frac{2}{7} \]

7. **Finding the Angle**:
Now, we find \( B \):
\[ B = \cos^{-1}\left(-\frac{2}{7}\right) \]

### Final Answer:
Thus, the angle \( \angle ABC \) is given by:
\[ \angle ABC = \cos^{-1}\left(-\frac{2}{7}\right) \]