AP EAMCET · Maths · Properties of Triangles
In \(\triangle A B C\), if \((a+c)^2=b^2+3 c a\), then \(\frac{a+c}{2 R}=\)
- A \(\frac{\sqrt{3}}{2}\)
- B \(\sqrt{3} \cdot \cos \left(\frac{A-C}{2}\right)\)
- C \(\cos \left(\frac{A-C}{2}\right)\)
- D \(\sin \left(\frac{A-C}{2}\right)\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3} \cdot \cos \left(\frac{A-C}{2}\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & (a+c)^2=b^2+3 c a \Rightarrow a^2+c^2-b^2=c a \\ & \cos B=\frac{a^2+c^2-b^2}{2 a c}=\frac{1}{2}=\cos 60^{\circ} \Rightarrow \frac{B}{2}=30^{\circ} \\ & \frac{a+c}{2 R}=\sin A+\sin C=2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right) \\ & =2 \sin…
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