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

Options

A. `(pi)/(4)`
B. `(pi)/(3)`
C. `cos^(-1)((1)/(3sqrt(7)))`
D. `cos^(-1)((1)/(sqrt(7)))`

Answer

To solve the problem, we need to find the angle \( \angle ACB \) 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 that the weighted sum of the altitudes is zero. This means that the altitudes can be represented as vectors, and their respective coefficients (9, 4, and 7) suggest a relationship between the sides of the triangle.

**Hint**: Recognize that the coefficients indicate a relationship between the sides opposite to the angles in triangle \( ABC \).

2. **Setting Up the Triangle**:
Let's denote the lengths of the sides opposite to vertices \( A \), \( B \), and \( C \) as \( a \), \( b \), and \( c \) respectively. The altitudes \( AP \), \( BQ \), and \( CR \) can be expressed in terms of the area \( K \) of triangle \( ABC \):
\[ AP = \frac{2K}{a}, \quad BQ = \frac{2K}{b}, \quad CR = \frac{2K}{c} \]

**Hint**: Use the formula for the area of a triangle in terms of its sides and altitudes.

3. **Substituting the Altitudes**:
Substitute the expressions for the altitudes into the equation:
\[ 9 \left(\frac{2K}{a}\right) + 4 \left(\frac{2K}{b}\right) + 7 \left(\frac{2K}{c}\right) = 0 \]
This simplifies to:
\[ 2K \left( \frac{9}{a} + \frac{4}{b} + \frac{7}{c} \right) = 0 \]

**Hint**: Since \( K \) (the area) cannot be zero for a non-degenerate triangle, the expression in parentheses must equal zero.

4. **Setting Up the Relationship**:
We can set up the equation:
\[ \frac{9}{a} + \frac{4}{b} + \frac{7}{c} = 0 \]
Rearranging gives:
\[ 9bc + 4ac + 7ab = 0 \]

**Hint**: This equation relates the sides of the triangle and can help us find the angles using the law of cosines.

5. **Using the Law of Cosines**:
We can express \( \cos C \) using the law of cosines:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

**Hint**: Substitute the values of \( a \), \( b \), and \( c \) in terms of a common ratio based on the coefficients from the altitude equation.

6. **Finding the Ratios**:
From the coefficients, we can assume:
\[ a = 3k, \quad b = 2k, \quad c = \sqrt{7}k \]
Substitute these into the cosine formula:
\[ \cos C = \frac{(3k)^2 + (2k)^2 - (\sqrt{7}k)^2}{2(3k)(2k)} \]
Simplifying gives:
\[ \cos C = \frac{9k^2 + 4k^2 - 7k^2}{12k^2} = \frac{6k^2}{12k^2} = \frac{1}{2} \]

**Hint**: Recognize that \( \cos C = \frac{1}{2} \) corresponds to a specific angle.

7. **Finding the Angle**:
The angle \( C \) for which \( \cos C = \frac{1}{2} \) is:
\[ C = \frac{\pi}{3} \text{ or } 60^\circ \]

**Final Result**: Thus, the angle \( \angle ACB \) is \( \frac{\pi}{3} \).