AP EAMCET · Maths · Properties of Triangles
In \(\triangle \mathrm{ABC}, \mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are in arithmetic progression and \(a: c=1: 2\). If \(b=4 \sqrt{3} \mathrm{~cm}\), then the area of \(\triangle A B C\) (in sq. \(\mathrm{cm})\) is
- A \(16 \sqrt{3}\)
- B \(12 \sqrt{3}\)
- C \(8 \sqrt{3}\)
- D \(6 \sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(8 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(\because A, B, C\) are in A.P. \(\Rightarrow 2 B=A+C\) Now, \(A+B+C=\pi \Rightarrow 3 \mathrm{~B}=\pi \Rightarrow B=\frac{\pi}{3}\) Now, \(\tan \frac{(C-A)}{2}=\frac{c-a}{c+a} \cot \left(\frac{B}{2}\right)=\frac{\frac{c}{a}-1}{\frac{c}{a}+1} \cot \left(\frac{\pi}{6}\right)\)…
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