In `Delta ABC` a =5 ,b=3 ,c=7 .Then value of `3 cos C +7 cos B` is equal :-

Options

A. 5
B. 10
C. 7
D. 3

Answer

To solve the problem, we need to find the value of \(3 \cos C + 7 \cos B\) given the sides of triangle \(ABC\) as \(a = 5\), \(b = 3\), and \(c = 7\).

### Step-by-Step Solution:

1. **Calculate \(\cos C\)**:
We use the cosine rule to find \(\cos C\):
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Substituting the values:
\[ a = 5, \quad b = 3, \quad c = 7 \]
\[ \cos C = \frac{5^2 + 3^2 - 7^2}{2 \cdot 5 \cdot 3} \]
\[ = \frac{25 + 9 - 49}{30} \]
\[ = \frac{34 - 49}{30} \]
\[ = \frac{-15}{30} = -\frac{1}{2} \]

2. **Calculate \(\cos B\)**:
Again, we use the cosine rule to find \(\cos B\):
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Substituting the values:
\[ \cos B = \frac{5^2 + 7^2 - 3^2}{2 \cdot 5 \cdot 7} \]
\[ = \frac{25 + 49 - 9}{70} \]
\[ = \frac{65}{70} = \frac{13}{14} \]

3. **Calculate \(3 \cos C + 7 \cos B\)**:
Now we substitute the values of \(\cos C\) and \(\cos B\) into the expression:
\[ 3 \cos C + 7 \cos B = 3 \left(-\frac{1}{2}\right) + 7 \left(\frac{13}{14}\right) \]
\[ = -\frac{3}{2} + \frac{91}{14} \]
To add these fractions, we need a common denominator. The least common multiple of 2 and 14 is 14:
\[ -\frac{3}{2} = -\frac{3 \cdot 7}{2 \cdot 7} = -\frac{21}{14} \]
Now we can add:
\[ -\frac{21}{14} + \frac{91}{14} = \frac{91 - 21}{14} = \frac{70}{14} = 5 \]

### Final Answer:
Thus, the value of \(3 \cos C + 7 \cos B\) is \(5\).