Question:medium

In \(\triangle ABC\), \(\sin A, \sin B, \sin C\) are in A.P. and \(C>90^\circ\). Then \(\cos A\) is:

Show Hint

Always convert sine A.P. into side relation using sine rule, then apply cosine rule.
Updated On: Apr 17, 2026
  • \( \frac{3c - 4b}{2b} \)
  • \( \frac{3c - 4b}{2c} \)
  • \( \frac{4c - 3b}{2b} \)
  • \( \frac{4c - 3b}{2c} \)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
According to the Sine Rule, \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\). Thus, if the sines are in Arithmetic Progression (AP), the side lengths \(a, b, c\) are also in AP.
Step 2: Key Formula or Approach:
1. For AP: \(2b = a + c \implies a = 2b - c\).
2. Cosine Rule: \(\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}\).
: Detailed Explanation:
Substitute \(a = 2b - c\) into the cosine rule for \(\cos A\):
\[ \cos A = \frac{b^{2} + c^{2} - (2b - c)^{2}}{2bc} \]
Expand the squared term:
\[ \cos A = \frac{b^{2} + c^{2} - (4b^{2} + c^{2} - 4bc)}{2bc} \]
\[ \cos A = \frac{b^{2} + c^{2} - 4b^{2} - c^{2} + 4bc}{2bc} \]
\[ \cos A = \frac{4bc - 3b^{2}}{2bc} \]
Factor out \(b\) from the numerator:
\[ \cos A = \frac{b(4c - 3b)}{2bc} = \frac{4c - 3b}{2c} \]
Step 3: Final Answer:
\(\cos A = \frac{4c - 3b}{2c}\).
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