WBJEE · Maths · Properties of Triangles
If angles \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are in A.P., then \(\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}\) is equal to
- A \(2 \sin \frac{A-C}{2}\)
- B \(2 \cos \frac{A-C}{2}\)
- C \(\cos \frac{\mathrm{A}-\mathrm{C}}{2}\)
- D \(\sin \frac{\mathrm{A}-\mathrm{C}}{2}\)
Answer & Solution
Correct Answer
(B) \(2 \cos \frac{A-C}{2}\)
Step-by-step Solution
Detailed explanation
Hints: \(2 \mathrm{~B}=\mathrm{A}+\mathrm{C}\) \[ =\frac{\sin A+\sin C}{\sin B}=\frac{2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)}{\sin B}=\frac{2 \sin B}{\sin B} \cos \left(\frac{A-C}{2}\right)=2 \cos \left(\frac{A-C}{2}\right) \]
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