Finding the Ratio of Side Squares from Angle Ratio in a Triangle

The problem asks for the ratio of the squares of the sides (\(a^2 : b^2 : c^2\)) of a triangle ABC, given that its angles A, B, and C are in the ratio \(3 : 5 : 4\). To solve this, we first need to find the actual measures of the angles and then use the Sine Rule to determine the ratio of the sides.

Step 1: Calculate the Angles of the Triangle

The sum of the angles in any triangle is \(180^\circ\). Let the angles be \(A = 3x\), \(B = 5x\), and \(C = 4x\), where \(x\) is a constant. Their sum is:

\(A + B + C = 3x + 5x + 4x = 12x\)

Setting the sum equal to \(180^\circ\):

\(12x = 180^\circ\)

Solving for \(x\):

\(x = \frac{180^\circ}{12} = 15^\circ\)

Now we can find the measure of each angle:

• Angle \(A = 3x = 3 \times 15^\circ = 45^\circ\)
• Angle \(B = 5x = 5 \times 15^\circ = 75^\circ\)
• Angle \(C = 4x = 4 \times 15^\circ = 60^\circ\)

We can verify that \(45^\circ + 75^\circ + 60^\circ = 180^\circ\).

Step 2: Use the Sine Rule to Find the Ratio of Sides

The Sine Rule states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. That is:

\(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)

This implies that the ratio of the sides \(a : b : c\) is equal to the ratio of the sines of their opposite angles:

\(a : b : c = \sin A : \sin B : \sin C\)

Substitute the calculated angle values:

\(a : b : c = \sin 45^\circ : \sin 75^\circ : \sin 60^\circ\)

We need the values of \(\sin 45^\circ\), \(\sin 60^\circ\), and \(\sin 75^\circ\):

• \(\sin 45^\circ = \frac{\sqrt{2}}{2}\)
• \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
• To find \(\sin 75^\circ\), we can use the sum identity \(\sin(X+Y) = \sin X \cos Y + \cos X \sin Y\). Let \(X=45^\circ\) and \(Y=30^\circ\):
  \(\sin 75^\circ = \sin (45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ\)
  \(\sin 75^\circ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)\)
  \(\sin 75^\circ = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

So, the ratio of sides is:

\(a : b : c = \frac{\sqrt{2}}{2} : \frac{\sqrt{6} + \sqrt{2}}{4} : \frac{\sqrt{3}}{2}\)

To simplify this ratio, we can multiply all terms by the least common multiple of the denominators, which is 4:

\(a : b : c = 4 \times \frac{\sqrt{2}}{2} : 4 \times \frac{\sqrt{6} + \sqrt{2}}{4} : 4 \times \frac{\sqrt{3}}{2}\)

\(a : b : c = 2\sqrt{2} : \sqrt{6} + \sqrt{2} : 2\sqrt{3}\)

Step 3: Calculate the Ratio of the Squares of the Sides

Now we need to find the ratio \(a^2 : b^2 : c^2\). We square each term in the ratio \(a : b : c\):

\(a^2 : b^2 : c^2 = (2\sqrt{2})^2 : (\sqrt{6} + \sqrt{2})^2 : (2\sqrt{3})^2\)

Calculate each square:

• \((2\sqrt{2})^2 = 2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8\)
• \((2\sqrt{3})^2 = 2^2 \times (\sqrt{3})^2 = 4 \times 3 = 12\)
• \((\sqrt{6} + \sqrt{2})^2 = (\sqrt{6})^2 + (\sqrt{2})^2 + 2(\sqrt{6})(\sqrt{2})\)
  \(= 6 + 2 + 2\sqrt{12}\)
  \(= 8 + 2\sqrt{4 \times 3}\)
  \(= 8 + 2 \times 2\sqrt{3}\)
  \(= 8 + 4\sqrt{3}\)

So the ratio \(a^2 : b^2 : c^2\) is:

\(a^2 : b^2 : c^2 = 8 : (8 + 4\sqrt{3}) : 12\)

Step 4: Simplify the Ratio and Compare with Options

To simplify the ratio \(8 : (8 + 4\sqrt{3}) : 12\), we can divide all terms by the greatest common divisor of the coefficients, which is 4:

\(\frac{8}{4} : \frac{8 + 4\sqrt{3}}{4} : \frac{12}{4}\)

\(2 : (2 + \sqrt{3}) : 3\)

This simplified ratio is \(2 : 2 + \sqrt{3} : 3\).

Comparing this result with the given options:

• Option 1: \(2 : 2 + \sqrt{3} : 3\)
• Option 2: \(2 : 2 - \sqrt{3} : 2\)
• Option 3: \(2 : 2 + \sqrt{3} : 2\)
• Option 4: \(2 : 2 - \sqrt{3} : 3\)

The calculated ratio matches Option 1.

Thus, the ratio \(a^2 : b^2 : c^2\) is \(2 : 2 + \sqrt{3} : 3\).