If in a triangle ABC, `(cosA)/a=(cosB)/b=(cosC)/c`,then the triangle is

Options

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

Answer

To solve the problem, we need to analyze the given condition in the triangle ABC, which states that:

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

### Step 1: Understanding the given condition
We know from the law of sines that:

\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R} \]

where \( R \) is the circumradius of the triangle. The given condition relates the cosines of the angles to the sides of the triangle.

### Step 2: Relating cosines and sines
From the relationship of sine and cosine, we can express the sides in terms of the angles and the circumradius:

\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

### Step 3: Substituting into the given condition
Substituting these expressions into the given condition, we have:

\[ \frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C} \]

This simplifies to:

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

### Step 4: Using the tangent function
The ratio \(\frac{\cos A}{\sin A}\) can be rewritten as:

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

### Step 5: Concluding the triangle type
If \(\cot A = \cot B = \cot C\), it implies that the angles \(A\), \(B\), and \(C\) are all equal. Therefore, we can conclude that:

\[ A = B = C = 60^\circ \]

This means that triangle ABC is equilateral.

### Final Answer:
Thus, the triangle is **equilateral**.

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