Question

If \( \cos B = \frac{\sin A}{2 \sin C} \), then the triangle is

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

Hint:

When solving problems involving triangles, it's often useful to express the given conditions in terms of the side lengths using the sine and cosine rules. This allows for algebraic manipulation to find relationships between the sides or angles.

Answer 1

The Correct Option is B

Step 1: Apply 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: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \), where \( k \) is a constant. From this, we have \( \sin A = ak \) and \( \sin C = ck \). Step 2: Substitute the expressions from the Sine Rule into the given equation.
The given equation is \( \cos B = \frac{\sin A}{2 \sin C} \). Substituting the expressions for \( \sin A \) and \( \sin C \):
$$\cos B = \frac{ak}{2(ck)} = \frac{a}{2c}$$ Step 3: Apply the Cosine Rule for angle B.
The cosine rule states that \( b^2 = a^2 + c^2 - 2ac \cos B \), which can be rearranged to express \( \cos B \) as:
$$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$$ Step 4: Equate the two expressions for \( \cos B \). From Step 2 and Step 3, we have:
$$\frac{a}{2c} = \frac{a^2 + c^2 - b^2}{2ac}$$ Step 5: Simplify the equation.
Multiply both sides by \( 2ac \):
$$a^2 = a^2 + c^2 - b^2$$ Subtract \( a^2 \) from both sides:
$$0 = c^2 - b^2$$ Rearrange the terms:
$$b^2 = c^2$$ Step 6: Deduce the relationship between the sides.
Taking the square root of both sides, and knowing that side lengths must be positive, we get: $$b = c$$ Step 7: Identify the type of triangle.
Since two sides of the triangle (\( b \) and \( c \)) are equal in length, the triangle is an isosceles triangle.