Question

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

• A. \( \frac{3c - 4b}{2b} \)
• B. \( \frac{3c - 4b}{2c} \)
• C. \( \frac{4c - 3b}{2b} \)
• D. \( \frac{4c - 3b}{2c} \)

Hint:

Always convert sine A.P. into side relation using sine rule, then apply cosine rule.

Answer 1

The Correct Option is D

Concept: \[ \sin A, \sin B, \sin C \text{ in A.P.} \Rightarrow 2\sin B = \sin A + \sin C \] Using sine rule: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \] \[ \sin A = \frac{a}{k},\quad \sin B = \frac{b}{k},\quad \sin C = \frac{c}{k} \]

Step 1:Apply A.P. condition \[ 2\frac{b}{k} = \frac{a}{k} + \frac{c}{k} \Rightarrow 2b = a + c \]

Step 2:Use cosine rule \[ a^2 = b^2 + c^2 - 2bc \cos A \] Substitute \(a = 2b - c\): \[ (2b - c)^2 = b^2 + c^2 - 2bc \cos A \]

Step 3:Expand \[ 4b^2 - 4bc + c^2 = b^2 + c^2 - 2bc \cos A \] \[ 3b^2 - 4bc = -2bc \cos A \]

Step 4:Solve for \(\cos A\) \[ \cos A = \frac{4bc - 3b^2}{2bc} = \frac{b(4c - 3b)}{2bc} \] \[ \Rightarrow \cos A = \frac{4c - 3b}{2c} \] Conclusion :{4c - 3b}/{2c}