Calculating Triangle Perimeter from Side Lengths and Sine Relation

We are given a triangle ABC with the following information:

• Side a (opposite angle A), which is BC = $\sqrt{3}$
• Side b (opposite angle B), which is CA = 1
• The perimeter of triangle ABC is 6 times the arithmetic mean (AM) of the sines of its angles.

Let the angles of the triangle be A, B, and C, and the sides opposite to these angles be a, b, and c respectively. The perimeter of the triangle is $P = a + b + c$.

The arithmetic mean of the sines of the angles is $\frac{\sin A + \sin B + \sin C}{3}$.

According to the given condition:

$\text{Perimeter} = 6 \times \left( \frac{\sin A + \sin B + \sin C}{3} \right)$

$a + b + c = 2 (\sin A + \sin B + \sin C)$

Applying the Sine Rule to Find Circumradius

The Sine Rule states that for any triangle ABC with circumradius R, the ratio of a side to the sine of its opposite angle is constant and equal to twice the circumradius:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$

From the Sine Rule, we can express the sines of the angles in terms of sides and the circumradius R:

• $\sin A = \frac{a}{2R}$
• $\sin B = \frac{b}{2R}$
• $\sin C = \frac{c}{2R}$

Substitute these expressions for $\sin A$, $\sin B$, and $\sin C$ into the perimeter equation:

$a + b + c = 2 \left( \frac{a}{2R} + \frac{b}{2R} + \frac{c}{2R} \right)$

$a + b + c = 2 \left( \frac{a + b + c}{2R} \right)$

$a + b + c = \frac{a + b + c}{R}$

Since $a+b+c$ represents the perimeter of a valid triangle, it must be a non-zero value. We can safely divide both sides of the equation by $a+b+c$:

$1 = \frac{1}{R}$

This equation implies that the circumradius of the triangle is $R=1$.

Finding the Third Side Using Cosine and Sine Rules

We know sides $a = \sqrt{3}$ and $b = 1$, and the circumradius $R = 1$. We need to find the length of the third side, $c$ (AB).

We can use the Cosine Rule, which relates the sides and angles of a triangle:

$c^2 = a^2 + b^2 - 2ab \cos C$

Substitute the known values $a = \sqrt{3}$ and $b = 1$ into the Cosine Rule formula:

$c^2 = (\sqrt{3})^2 + (1)^2 - 2(\sqrt{3})(1) \cos C$

$c^2 = 3 + 1 - 2\sqrt{3} \cos C$

$c^2 = 4 - 2\sqrt{3} \cos C$

We can also use the Sine Rule with the derived circumradius $R=1$ to relate side $c$ and angle $C$:

$\frac{c}{\sin C} = 2R \implies \frac{c}{\sin C} = 2(1) \implies c = 2 \sin C$

Now, we have two expressions involving $c$ and $\cos C$ or $\sin C$. Substitute the expression for $c$ from the Sine Rule into the equation derived from the Cosine Rule:

$(2 \sin C)^2 = 4 - 2\sqrt{3} \cos C$

$4 \sin^2 C = 4 - 2\sqrt{3} \cos C$

Using the trigonometric identity $\sin^2 C = 1 - \cos^2 C$ to express everything in terms of $\cos C$:

$4 (1 - \cos^2 C) = 4 - 2\sqrt{3} \cos C$

$4 - 4 \cos^2 C = 4 - 2\sqrt{3} \cos C$

Rearrange the terms to form a quadratic equation in terms of $\cos C$:

$4 \cos^2 C - 2\sqrt{3} \cos C = 0$

Factor out the common term $2 \cos C$:

$2 \cos C (2 \cos C - \sqrt{3}) = 0$

This equation yields two possible solutions for $\cos C$:

• Possibility 1: $2 \cos C = 0 \implies \cos C = 0$
• Possibility 2: $2 \cos C - \sqrt{3} = 0 \implies \cos C = \frac{\sqrt{3}}{2}$

Evaluating the Possible Triangle Perimeters

Since C is an angle within a triangle, its value must be between 0° and 180° ($0 < C < \pi$).

Possibility 1: $\cos C = 0$

If $\cos C = 0$ and $0 < C < \pi$, the angle $C = \frac{\pi}{2}$ (90°). This means the triangle is right-angled at C.

Using the relationship $c = 2 \sin C$, we find the length of side $c$:

$c = 2 \sin \left( \frac{\pi}{2} \right) = 2(1) = 2$

In this case, the side lengths of the triangle are $a = \sqrt{3}$, $b = 1$, and $c = 2$. The perimeter is calculated as:

$P = a + b + c = \sqrt{3} + 1 + 2 = \sqrt{3} + 3$.

Possibility 2: $\cos C = \frac{\sqrt{3}}{2}$

If $\cos C = \frac{\sqrt{3}}{2}$ and $0 < C < \pi$, the angle $C = \frac{\pi}{6}$ (30°).

Using the relationship $c = 2 \sin C$, we find the length of side $c$:

$c = 2 \sin \left( \frac{\pi}{6} \right) = 2 \left( \frac{1}{2} \right) = 1$

In this case, the side lengths of the triangle are $a = \sqrt{3}$, $b = 1$, and $c = 1$. The perimeter is calculated as:

$P = a + b + c = \sqrt{3} + 1 + 1 = \sqrt{3} + 2$.

Final Perimeter Result

Our analysis shows that there are two possible triangles that satisfy the initial condition and have sides of length $\sqrt{3}$ and 1, leading to perimeters of $\sqrt{3} + 3$ and $\sqrt{3} + 2$. Comparing these results with the given options, the value $\sqrt{3} + 3$ is listed.

The perimeter of the triangle is $\sqrt{3} + 3$.

Revision Table: Essential Triangle Formulas

Additional Information on Triangle Geometry

The relationship between the perimeter and the sum of sines of angles involves the circumradius (R), which is a fundamental property of a triangle. The circumradius is the radius of the circle that circumscribes the triangle, passing through all three vertices.

A condition like "Perimeter is k times the AM of sine of angles" implies a specific value for the circumradius R, because $\sum \sin A = \frac{a+b+c}{2R}$. Substituting this into the condition allows us to solve for R.

In this problem, we found R=1. This means the unique circle passing through vertices A, B, and C has a radius of 1 unit. Combining this information with given side lengths a and b restricts the possible shape(s) of the triangle, which can then be solved using standard trigonometric laws like the Sine Rule and Cosine Rule.