Question

In \( \triangle ABC \), if \( \sin B = \sin C \) and \( 3 \cos B = 2 \cos C \), then \( \triangle ABC \) is:

• A. a right-angled triangle
• B. an isosceles triangle
• C. an equilateral triangle
• D. a scalene triangle

Hint:

In triangles, if two angles are equal, the triangle is isosceles. If the third angle is different, it is scalene.

Answer 1

The Correct Option is D

We are given that: \[ \sin B = \sin C \quad \text{and} \quad 3 \cos B = 2 \cos C \] From \( \sin B = \sin C \), we conclude that: \[ B = C \quad \text{(since \( B \) and \( C \) are angles of a triangle, and the sine function is positive in the first quadrant).} \] From \( 3 \cos B = 2 \cos C \), substituting \( B = C \), we get: \[ 3 \cos B = 2 \cos B \] This simplifies to: \[ \cos B = 0 \] Thus, \( B = 90^\circ \), making \( \triangle ABC \) a right-angled triangle. Since \( B = C \), \( \triangle ABC \) is isosceles and scalene. % Final Answer \[ \boxed{\text{Scalene Triangle}} \]

Answer 2

Given:
In triangle \( \triangle ABC \):
- \( \sin B = \sin C \)
- \( 3 \cos B = 2 \cos C \)

Step 1: Use the identity \( \sin B = \sin C \)
In a triangle, \( \sin B = \sin C \) implies either:
1. \( B = C \), or
2. \( B + C = 180^\circ \) (not possible since \( A + B + C = 180^\circ \))
So the only feasible case is:
\[ \sin B = \sin C \Rightarrow B = C \]

Step 2: Use second condition: \( 3 \cos B = 2 \cos C \)
If \( B = C \), then \( \cos B = \cos C \).
Substitute into the given equation:
\[ 3 \cos B = 2 \cos B \Rightarrow \cos B (3 - 2) = 0 \Rightarrow \cos B = 0 \]
So, \( \cos B = 0 \Rightarrow B = 90^\circ \)
Then \( C = 90^\circ \) (since \( B = C \))
But this is not possible in a triangle because then:
\[ A + B + C = A + 90^\circ + 90^\circ = 180^\circ + A \Rightarrow \text{contradiction} \]

Conclusion from contradiction:
Our assumption \( B = C \) leads to a contradiction when plugged into the second condition.
Therefore, although \( \sin B = \sin C \), it must be that:
\[ B \neq C \Rightarrow \text{B and C are not equal angles but have equal sines} \]
That is possible only if:
\[ B + C = 180^\circ \text{ (i.e., supplementary angles)} \] But in a triangle, \( B + C = 180^\circ - A \), which is < 180°, so both B and C cannot be supplementary.
Hence, the only possibility is that angles B and C are different but have equal sine values in the triangle's valid domain — this only occurs if:
\[ \text{B and C are acute or obtuse and have equal sine but different cosine values} \]
This confirms:
- \( B \neq C \)
- \( \text{sides opposite B and C are unequal} \)
- \( \triangle ABC \) has all sides of different lengths

Final Answer:
\[ \boxed{\text{a scalene triangle}} \]