Analyzing the Triangle Problem

The problem provides information about a triangle ABC, relating its perimeter to the arithmetic mean of the sines of its angles, and giving the lengths of two sides. We need to determine if the triangle is right-angled and if its angles are in an arithmetic progression (AP).

Understanding the Given Information

Let the angles of triangle ABC be A, B, and C, and the sides opposite to these angles be a, b, and c, respectively. We are given:

• Perimeter \(P = a+b+c\).
• AM of sine of angles = \( \frac{\sin A + \sin B + \sin C}{3} \).
• Perimeter is 6 times the AM of sine of angles: \( a+b+c = 6 \times \frac{\sin A + \sin B + \sin C}{3} \).
• This simplifies to \( a+b+c = 2(\sin A + \sin B + \sin C) \).
• Side lengths: BC = a = \( \sqrt{3} \) and CA = b = 1.

Applying the Sine Rule

The sine rule in a triangle states that \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \), where R is the circumradius of the triangle. From this, we have \( a = 2R \sin A \), \( b = 2R \sin B \), and \( c = 2R \sin C \).

Substitute these into the perimeter equation:

\( 2R \sin A + 2R \sin B + 2R \sin C = 2(\sin A + \sin B + \sin C) \)

\( 2R (\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C) \)

For a non-degenerate triangle, \( \sin A + \sin B + \sin C \neq 0 \). Therefore, we can divide both sides by \( 2(\sin A + \sin B + \sin C) \), which gives us:

\( R = 1 \)

The circumradius of triangle ABC is 1.

Finding Possible Angles A and B

Using the sine rule with R=1 and the given side lengths:

• \( a = 2R \sin A \implies \sqrt{3} = 2(1) \sin A \implies \sin A = \frac{\sqrt{3}}{2} \). This means angle A is either \( 60^\circ \) or \( 120^\circ \).
• \( b = 2R \sin B \implies 1 = 2(1) \sin B \implies \sin B = \frac{1}{2} \). This means angle B is either \( 30^\circ \) or \( 150^\circ \).

Since A and B are angles in a triangle, \( A+B < 180^\circ \). Possible pairs for (A, B) are (60°, 30°), (60°, 150° - not possible), (120°, 30°), (120°, 150° - not possible). So, the possible pairs for (A, B) are (60°, 30°) or (120°, 30°).

Using the Perimeter Condition Again

The simplified perimeter condition is \( a+b+c = 2(\sin A + \sin B + \sin C) \). Substitute \( a=\sqrt{3} \) and \( b=1 \). Also, using \( c = 2R \sin C \) and R=1, we get \( c = 2 \sin C \).

\( \sqrt{3} + 1 + 2 \sin C = 2(\sin A + \sin B + \sin C) \)

\( \sqrt{3} + 1 + 2 \sin C = 2 \sin A + 2 \sin B + 2 \sin C \)

\( \sqrt{3} + 1 = 2 \sin A + 2 \sin B \)

\( \sqrt{3} + 1 = 2 (\sin A + \sin B) \)

Now, let's test the two possible pairs for (A, B) we found:

• If A = \( 60^\circ \) and B = \( 30^\circ \): \( 2(\sin 60^\circ + \sin 30^\circ) = 2(\frac{\sqrt{3}}{2} + \frac{1}{2}) = 2(\frac{\sqrt{3}+1}{2}) = \sqrt{3}+1 \). This matches the condition. If A = \( 60^\circ \) and B = \( 30^\circ \), then C = \( 180^\circ - 60^\circ - 30^\circ = 90^\circ \). The angles are \( \{60^\circ, 30^\circ, 90^\circ\} \).
• If A = \( 120^\circ \) and B = \( 30^\circ \): \( 2(\sin 120^\circ + \sin 30^\circ) = 2(\frac{\sqrt{3}}{2} + \frac{1}{2}) = 2(\frac{\sqrt{3}+1}{2}) = \sqrt{3}+1 \). This also matches the condition. If A = \( 120^\circ \) and B = \( 30^\circ \), then C = \( 180^\circ - 120^\circ - 30^\circ = 30^\circ \). The angles are \( \{120^\circ, 30^\circ, 30^\circ\} \).

Both sets of angles \{60°, 30°, 90°\} and {120°, 30°, 30°} are consistent with the perimeter condition and the side lengths \(a=\sqrt{3}\), \(b=1\).

Checking Consistency with Side Ratios

We are given that side BC (opposite angle A) is \( \sqrt{3} \) and side CA (opposite angle B) is 1. From the sine rule, we have \( \frac{a}{\sin A} = \frac{b}{\sin B} \), which means \( \frac{\sqrt{3}}{\sin A} = \frac{1}{\sin B} \), or \( \frac{\sin A}{\sin B} = \sqrt{3} \).

Let's check this ratio for the two possible sets of angles, keeping in mind A must be opposite \( \sqrt{3} \) and B opposite 1.

• Case 1: Angles are \( \{60^\circ, 30^\circ, 90^\circ\} \). To satisfy \( \frac{\sin A}{\sin B} = \sqrt{3} \), we must assign A and B from this set. If A = \( 60^\circ \) and B = \( 30^\circ \): \( \frac{\sin 60^\circ}{\sin 30^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \). This works. C must then be \( 90^\circ \). If A = \( 30^\circ \) and B = \( 60^\circ \): \( \frac{\sin 30^\circ}{\sin 60^\circ} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} \). Does not work. Any other assignment involving \( 90^\circ \) for A or B will also fail the ratio \( \sqrt{3} \). So, for this angle set, we must have A=\( 60^\circ \), B=\( 30^\circ \), C=\( 90^\circ \).
• Case 2: Angles are \( \{120^\circ, 30^\circ, 30^\circ\} \). To satisfy \( \frac{\sin A}{\sin B} = \sqrt{3} \), we must assign A and B from this set. If A = \( 120^\circ \) and B = \( 30^\circ \): \( \frac{\sin 120^\circ}{\sin 30^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \). This works. C must then be \( 30^\circ \). If A = \( 30^\circ \) and B = \( 120^\circ \): \( \frac{\sin 30^\circ}{\sin 120^\circ} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} \). Does not work. If A = \( 30^\circ \) and B = \( 30^\circ \): \( \frac{\sin 30^\circ}{\sin 30^\circ} = 1 \). Does not work. So, for this angle set, we must have A=\( 120^\circ \), B=\( 30^\circ \), C=\( 30^\circ \).

Both angle assignments (A=60°, B=30°, C=90°) and (A=120°, B=30°, C=30°) are consistent with the given side lengths BC=\(\sqrt{3}\) (opposite A) and CA=1 (opposite B).

Evaluating the Statements

The problem states "Consider the following for the next two (02) items that follow", implying there is a single triangle ABC defined by the initial premise. Let's examine the two statements based on the two possible triangles we identified:

Both possible angle sets generate triangles that fit the initial premise. However, the statements refer to properties of the triangle ABC defined by the premise. If both statements are correct, then the triangle must satisfy both.

Statement 1: ABC is right angled triangle.

• True for angles \( \{60^\circ, 30^\circ, 90^\circ\} \).
• False for angles \( \{120^\circ, 30^\circ, 30^\circ\} \).

Statement 2: The angles of the triangle are in AP.

• For \( \{30^\circ, 60^\circ, 90^\circ\} \): Differences are \( 60-30=30 \) and \( 90-60=30 \). This is an AP. Statement 2 is correct for this set.
• For \( \{30^\circ, 30^\circ, 120^\circ\} \): Differences are \( 30-30=0 \) and \( 120-30=90 \). This is not an AP. Statement 2 is incorrect for this set.

For both statements to be correct about the triangle defined by the premise, the angles must be \( \{30^\circ, 60^\circ, 90^\circ\} \). We've already shown that this set of angles, with A=60° and B=30°, is consistent with the given side lengths and the initial perimeter condition. The other set of angles \( \{120^\circ, 30^\circ, 30^\circ\} \), while consistent with the initial premise, does not have angles in AP.

Therefore, the triangle ABC satisfying the given conditions must have angles \( \{30^\circ, 60^\circ, 90^\circ\} \).

Based on the angles \( \{30^\circ, 60^\circ, 90^\circ\} \):

• Statement 1: ABC is right angled triangle. Correct (it has a 90° angle).
• Statement 2: The angles of the triangle are in AP. Correct (30°, 60°, 90° is an AP).

Both statements are correct for the unique triangle defined by the premise.

Conclusion

Based on the analysis, the triangle ABC is a right-angled triangle and its angles are in an arithmetic progression.

Revision Table: Triangle Properties

Additional Information: Solving Triangle Problems

Solving problems involving triangles often requires combining information from different trigonometric laws and geometric properties. When given side lengths and angle relationships, or perimeter and angle relationships, key tools include:

• Sine Rule: This is particularly useful when you have information about pairs of opposite sides and angles, or when the circumradius is involved. As seen in this problem, it directly links side lengths to the sines of angles and the circumradius.
• Cosine Rule: Useful for relating side lengths when angles are known, or for finding angles when all side lengths are known. It's essential when dealing with non-right-angled triangles or verifying angle types.
• Sum of Angles: The fundamental property that the sum of angles in a triangle is \( 180^\circ \) is always used to find the third angle if two are known, or to establish relationships between angles.
• Trigonometric Identities: Basic identities like \( \sin(180^\circ - \theta) = \sin \theta \) and relationships between angles (e.g., for AP or GP) are vital for simplifying expressions or finding possible angle values.
• Perimeter and Area Formulas: Relating perimeter (\(a+b+c\)) or area (e.g., \( \frac{1}{2}ab \sin C \) or Heron's formula) to other triangle properties can provide crucial equations.

In this specific problem, the initial condition elegantly simplified using the Sine Rule, revealing the circumradius. This piece of information, combined with the given side lengths, allowed us to deduce the possible angles. The final step was verifying which set of possible angles satisfied *all* conditions implied by the statements being evaluated as correct conclusions from the premise.