If : `cos^(2) A + cos^(2) B + cos^(2) C = 1, "then" : Delta ABC ` is A)scalane B)equilateral C)isosceles D)right angled

Options

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

Answer

To solve the problem, we start with the given equation:

\[ \cos^2 A + \cos^2 B + \cos^2 C = 1 \]

### Step 1: Use the identity for the sum of squares of cosines in a triangle.

In any triangle ABC, there is a known identity:

\[ \cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C \]

### Step 2: Substitute the given equation into the identity.

Given that:

\[ \cos^2 A + \cos^2 B + \cos^2 C = 1 \]

We can substitute this into the identity:

\[ 1 = 1 - 2 \cos A \cos B \cos C \]

### Step 3: Simplify the equation.

Subtract 1 from both sides:

\[ 0 = -2 \cos A \cos B \cos C \]

### Step 4: Rearrange the equation.

This implies:

\[ 2 \cos A \cos B \cos C = 0 \]

### Step 5: Analyze the product.

Since the product of the cosines is zero, at least one of the cosines must be zero:

\[ \cos A = 0 \quad \text{or} \quad \cos B = 0 \quad \text{or} \quad \cos C = 0 \]

### Step 6: Determine the implications of cosine being zero.

If \(\cos A = 0\), then angle \(A\) is \(90^\circ\). Similarly, if \(\cos B = 0\) or \(\cos C = 0\), then angle \(B\) or angle \(C\) is \(90^\circ\) respectively. This means that triangle ABC has a right angle.

### Conclusion:

Thus, triangle ABC is a right-angled triangle. Therefore, the correct option is:

**D) right angled**

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