Solving Triangle Problems: Angles in Ratio

This problem involves a triangle where the angles are given in a specific ratio. We need to find the value of an expression involving the side lengths of this triangle. To solve this, we will first determine the actual measure of each angle and then use the Sine Rule to relate the sides to these angles.

Step 1: Calculate the Angles of the Triangle

The angles A, B, and C of triangle ABC are in the ratio 3 ∶ 5 ∶ 4. The sum of the angles in any triangle is 180 degrees.

Let the common ratio factor be $x$. Then the angles are $3x$, $5x$, and $4x$.

The sum of the angles is:

$\qquad 3x + 5x + 4x = 180^\circ$

$\qquad 12x = 180^\circ$

$\qquad x = \frac{180^\circ}{12} = 15^\circ$

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 = $4x = 4 \times 15^\circ = 60^\circ$

Let's verify the sum: $45^\circ + 75^\circ + 60^\circ = 180^\circ$. The angle calculations are correct.

Step 2: Using the Sine Rule

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, for triangle ABC with sides a, b, c opposite to angles A, B, C respectively:

$\qquad \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ (where k is a constant)

From this, we can express the sides in terms of the constant k and the sine of the angles:

• $a = k \sin A = k \sin 45^\circ$
• $b = k \sin B = k \sin 75^\circ$
• $c = k \sin C = k \sin 60^\circ$

Step 3: Calculate Sine Values for the Angles

We need the values of $\sin 45^\circ$, $\sin 60^\circ$, and $\sin 75^\circ$.

• $\sin 45^\circ = \frac{1}{\sqrt{2}}$
• $\sin 60^\circ = \frac{\sqrt{3}}{2}$
• $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$
• $\sin 75^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

So, the sides are:

• $a = k \cdot \frac{1}{\sqrt{2}} = \frac{k}{\sqrt{2}}$
• $b = k \cdot \frac{\sqrt{3}+1}{2\sqrt{2}}$
• $c = k \cdot \frac{\sqrt{3}}{2}$

Step 4: Evaluate the Expression $a + b + \sqrt{2} c$

Now we substitute the expressions for a, b, and c into the given expression $a + b + \sqrt{2} c$:

$\qquad a + b + \sqrt{2} c = \left(\frac{k}{\sqrt{2}}\right) + \left(k \cdot \frac{\sqrt{3}+1}{2\sqrt{2}}\right) + \sqrt{2} \left(k \cdot \frac{\sqrt{3}}{2}\right)$

Simplify the last term:

$\qquad \sqrt{2} \left(k \cdot \frac{\sqrt{3}}{2}\right) = k \cdot \frac{\sqrt{2}\sqrt{3}}{2} = k \cdot \frac{\sqrt{6}}{2}$

Now combine the terms:

$\qquad a + b + \sqrt{2} c = \frac{k}{\sqrt{2}} + \frac{k(\sqrt{3}+1)}{2\sqrt{2}} + \frac{k\sqrt{6}}{2}$

To add these terms, we can use a common denominator, $2\sqrt{2}$. Note that $\frac{k}{\sqrt{2}} = \frac{k \cdot 2}{\sqrt{2} \cdot 2} = \frac{2k}{2\sqrt{2}}$ and $\frac{k\sqrt{6}}{2} = \frac{k\sqrt{6} \cdot \sqrt{2}}{2 \cdot \sqrt{2}} = \frac{k\sqrt{12}}{2\sqrt{2}} = \frac{k \cdot 2\sqrt{3}}{2\sqrt{2}} = \frac{2k\sqrt{3}}{2\sqrt{2}}$.

So the expression becomes:

$\qquad a + b + \sqrt{2} c = \frac{2k}{2\sqrt{2}} + \frac{k(\sqrt{3}+1)}{2\sqrt{2}} + \frac{2k\sqrt{3}}{2\sqrt{2}}$

$\qquad a + b + \sqrt{2} c = \frac{2k + k(\sqrt{3}+1) + 2k\sqrt{3}}{2\sqrt{2}}$

$\qquad a + b + \sqrt{2} c = \frac{2k + k\sqrt{3} + k + 2k\sqrt{3}}{2\sqrt{2}}$

Combine like terms (terms with k and terms with $k\sqrt{3}$):

$\qquad a + b + \sqrt{2} c = \frac{(2k + k) + (k\sqrt{3} + 2k\sqrt{3})}{2\sqrt{2}}$

$\qquad a + b + \sqrt{2} c = \frac{3k + 3k\sqrt{3}}{2\sqrt{2}}$

Factor out 3k from the numerator:

$\qquad a + b + \sqrt{2} c = \frac{3k(1 + \sqrt{3})}{2\sqrt{2}}$

Recall the expression for b:

$\qquad b = k \cdot \frac{\sqrt{3}+1}{2\sqrt{2}} = k \cdot \frac{1+\sqrt{3}}{2\sqrt{2}}$

Comparing the expression for $a + b + \sqrt{2} c$ with the expression for $b$, we see that:

$\qquad a + b + \sqrt{2} c = 3 \cdot \left( \frac{k(1 + \sqrt{3})}{2\sqrt{2}} \right) = 3 \cdot b$

Therefore, the value of $a + b + \sqrt{2} c$ is equal to $3b$.

Revision Table: Triangle Angles and Side Lengths

Additional Information: Sine Rule Applications

The Sine Rule is a fundamental tool in solving triangles. It is particularly useful when you know:

• Two angles and one side (AAS or ASA).
• Two sides and an angle opposite to one of those sides (SSA - the ambiguous case).

In this problem, although we weren't given side lengths directly, the Sine Rule allowed us to express the relative lengths of the sides based on the calculated angles. This relationship is key to evaluating expressions involving the sides.

The value of $\sin 75^\circ$ is often needed in trigonometry problems. Remember it can be derived using $\sin(45^\circ+30^\circ)$ or $\sin(90^\circ-15^\circ)$. The value $\frac{\sqrt{6}+\sqrt{2}}{4}$ is equivalent to $\frac{\sqrt{3}+1}{2\sqrt{2}}$ after rationalizing the denominator: $\frac{\sqrt{3}+1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{4}$.