if the sides of a triangle are in the ratio `2:sqrt6 : sqrt3 + 1, `then the largest angle of the trangle will be (1) 60 (3) 72 (2) 75 (4) 90

Options

A. `60^(@)`
B. `72^(@)`
C. `75^(@)`
D. `90^(@)`

Answer

To solve the problem step by step, we need to find the largest angle of a triangle whose sides are in the ratio \(2 : \sqrt{6} : \sqrt{3} + 1\).

### Step 1: Assign Variables to the Sides
Let the sides of the triangle be:
- \( a = 2k \)
- \( b = \sqrt{6}k \)
- \( c = (\sqrt{3} + 1)k \)

Here, \( k \) is a positive constant.

### Step 2: Identify the Largest Side
To find the largest angle, we need to identify the largest side. We will compare the values of \( a \), \( b \), and \( c \):
- \( a = 2k \)
- \( b = \sqrt{6}k \approx 2.45k \)
- \( c = (\sqrt{3} + 1)k \approx (1.73 + 1)k = 2.73k \)

From this, we see that \( c \) is the largest side.

### Step 3: Use the Cosine Rule
The angle opposite to the largest side \( c \) will be the largest angle, which we will denote as \( C \). We will use the cosine rule to find \( C \):
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

### Step 4: Calculate \( a^2 \), \( b^2 \), and \( c^2 \)
Calculating each square:
- \( a^2 = (2k)^2 = 4k^2 \)
- \( b^2 = (\sqrt{6}k)^2 = 6k^2 \)
- \( c^2 = ((\sqrt{3} + 1)k)^2 = (3 + 2\sqrt{3} + 1)k^2 = (4 + 2\sqrt{3})k^2 \)

### Step 5: Substitute into the Cosine Rule
Substituting these values into the cosine rule:
\[ \cos C = \frac{4k^2 + 6k^2 - (4 + 2\sqrt{3})k^2}{2(2k)(\sqrt{6}k)} \]
This simplifies to:
\[ \cos C = \frac{(10 - 4 - 2\sqrt{3})k^2}{4\sqrt{6}k^2} = \frac{6 - 2\sqrt{3}}{4\sqrt{6}} \]
\[ \cos C = \frac{3 - \sqrt{3}}{2\sqrt{6}} \]

### Step 6: Recognize the Value of \( \cos C \)
We need to recognize that \( \frac{3 - \sqrt{3}}{2\sqrt{6}} \) corresponds to a known angle. It can be shown that:
\[ \cos 75^\circ = \frac{\sqrt{6} - \sqrt{2}}{4} \]
Through trigonometric identities, we can establish that:
\[ \cos 75^\circ = \frac{3 - \sqrt{3}}{2\sqrt{6}} \]

### Step 7: Conclusion
Thus, the angle \( C \) is:
\[ C = 75^\circ \]

### Final Answer
The largest angle of the triangle is \( 75^\circ \).

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