In `Delta ABC`, if `(cos A)/(a) =(cos B)/(b) =(cos C)/(c )`, then the triangle is

Options

A. right angled
B. obtuse angled
C. equilateral
D. isosceles

Answer

To solve the problem, we start with the given condition in triangle \( ABC \):

\[ \frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c} \]

Let's denote this common ratio as \( k \). Therefore, we can express the sides in terms of the cosines of the angles:

\[ \cos A = k \cdot a, \quad \cos B = k \cdot b, \quad \cos C = k \cdot c \]

### Step 1: Use the Law of Sines
According to the Law of Sines, we have:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = K \]

### Step 2: Relate Cosines and Sines
From the above, we can express \( a, b, c \) in terms of \( K \):

\[ a = K \sin A, \quad b = K \sin B, \quad c = K \sin C \]

### Step 3: Substitute into the Cosine Ratios
Now substituting these into our original equation gives:

\[ \cos A = k \cdot K \sin A, \quad \cos B = k \cdot K \sin B, \quad \cos C = k \cdot K \sin C \]

### Step 4: Rearranging the Equations
Rearranging these equations, we have:

\[ \frac{\cos A}{\sin A} = kK, \quad \frac{\cos B}{\sin B} = kK, \quad \frac{\cos C}{\sin C} = kK \]

### Step 5: Equal Ratios
Since all three ratios are equal, we can set them equal to each other:

\[ \frac{\cos A}{\sin A} = \frac{\cos B}{\sin B} = \frac{\cos C}{\sin C} \]

### Step 6: Use the Cotangent Function
This can be rewritten using the cotangent function:

\[ \cot A = \cot B = \cot C \]

### Step 7: Conclusion on Angles
If \( \cot A = \cot B = \cot C \), it implies that:

\[ A = B = C \]

### Step 8: Type of Triangle
Since all angles are equal, triangle \( ABC \) is an equilateral triangle.

Thus, the final answer is that triangle \( ABC \) is an **equilateral triangle**.

---