Determining the Type of Triangle Using Trigonometric Relation

We are given a triangle ABC with the relation \( \cos A = \dfrac{\sin B}{2 \sin C} \). We need to determine the type of this triangle.

To solve this problem and find the type of triangle, we can use the fundamental laws of trigonometry related to triangles: the Sine Rule and the Cosine Rule.

Applying the Sine Rule to the Triangle

The Sine Rule for a triangle states that the ratio of the length of a side to the sine of its opposite angle is constant for all sides and angles in a given triangle. Mathematically, it is expressed as:

\( \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R \)

where \(a, b, c\) are the lengths of the sides opposite to angles A, B, C respectively, and \(R\) is the circumradius of the triangle.

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

• \( \sin A = \dfrac{a}{2R} \)
• \( \sin B = \dfrac{b}{2R} \)
• \( \sin C = \dfrac{c}{2R} \)

Substituting into the Given Equation

Now, we substitute the expressions for \( \sin B \) and \( \sin C \) from the Sine Rule into the given equation \( \cos A = \dfrac{\sin B}{2 \sin C} \):

\( \cos A = \dfrac{\dfrac{b}{2R}}{2 \cdot \dfrac{c}{2R}} \)

Simplify the right side of the equation:

\( \cos A = \dfrac{\dfrac{b}{2R}}{\dfrac{2c}{2R}} \)

\( \cos A = \dfrac{\dfrac{b}{2R}}{\dfrac{c}{R}} \)

To simplify further, multiply the numerator by the reciprocal of the denominator:

\( \cos A = \dfrac{b}{2R} \times \dfrac{R}{c} \)

\( \cos A = \dfrac{b \cdot R}{2R \cdot c} \)

Cancel out \(R\) from the numerator and denominator:

\( \cos A = \dfrac{b}{2c} \)

So, the given relation simplifies to \( \cos A = \dfrac{b}{2c} \).

Applying the Cosine Rule

Next, we use the Cosine Rule for angle A, which relates the cosine of an angle to the lengths of the sides of the triangle. The Cosine Rule states:

\( \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \)

Equating Cosine Expressions and Solving

Now we have two expressions for \( \cos A \). We can equate them:

\( \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{b}{2c} \)

To eliminate the denominators, we can multiply both sides of the equation by \( 2bc \):

\( 2bc \cdot \left( \dfrac{b^2 + c^2 - a^2}{2bc} \right) = 2bc \cdot \left( \dfrac{b}{2c} \right) \)

\( b^2 + c^2 - a^2 = b \cdot b \)

\( b^2 + c^2 - a^2 = b^2 \)

Subtract \( b^2 \) from both sides of the equation:

\( c^2 - a^2 = 0 \)

\( c^2 = a^2 \)

Since \(a\) and \(c\) represent lengths of the sides of a triangle, they must be positive values. Therefore, taking the square root of both sides gives:

\( c = a \)

Conclusion: Type of Triangle

The result \( c = a \) means that the side length opposite to angle C is equal to the side length opposite to angle A. A triangle with two sides of equal length is defined as an isosceles triangle.

Thus, based on the given relation \( \cos A = \dfrac{\sin B}{2 \sin C} \), the triangle ABC must be an isosceles triangle.

Summary of the Steps

1. Start with the given trigonometric relation in the triangle: \( \cos A = \dfrac{\sin B}{2 \sin C} \).
2. Apply the Sine Rule to express \( \sin B \) and \( \sin C \) in terms of side lengths \(b\) and \(c\) and circumradius \(R\).
3. Substitute these expressions into the given equation and simplify to get \( \cos A = \dfrac{b}{2c} \).
4. Apply the Cosine Rule to express \( \cos A \) in terms of side lengths \(a, b, c\), i.e., \( \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \).
5. Equate the two expressions for \( \cos A \) and solve the resulting algebraic equation.
6. The solution leads to \( c = a \), indicating that two sides of the triangle are equal.
7. Conclude that the triangle is an isosceles triangle.

This step-by-step process shows clearly how the given condition defines the type of triangle.