AP EAMCET · Maths · Properties of Triangles
In \(\triangle \mathrm{ABC}\), if the sides \(a, b, c\) are in geometric progression and the largest angle exceeds the smallest angle by \(60^{\circ}\), then \(\cos B\) is equal to
- A \(\frac{\sqrt{13}+1}{4}\)
- B \(\frac{1-\sqrt{13}}{4}\)
- C \(1\)
- D \(\frac{\sqrt{13}-1}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{13}-1}{4}\)
Step-by-step Solution
Detailed explanation
In \(\triangle \mathrm{ABC}\), sides \(a, b\) and \(c\) are in GP. \(\therefore \quad b^2=a c \quad \ldots .(\mathrm{i})\) Given, largest angle exceeds the smallest angle by \(60^{\circ}\).…
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