TS EAMCET · Maths · Properties of Triangles
The angle of \(\triangle A B C\) are in an arithmetic progression. If the larger sides \(a, b\) satisfy the relation \(\frac{\sqrt{3}}{2} < \frac{b}{a} < 1\), then the possible values of the smallest side are
- A \(\frac{a \pm \sqrt{4 b^2-3 a^2}}{2 a}\)
- B \(\frac{a \pm \sqrt{4 b^2-3 a^2}}{2 b}\)
- C \(\frac{a \pm \sqrt{4 b^2-3 a^2}}{2 c}\)
- D \(\frac{a \pm \sqrt{4 b^2-3 a^2}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{a \pm \sqrt{4 b^2-3 a^2}}{2}\)
Step-by-step Solution
Detailed explanation
Since, \(A, B, C\) are in an AP, so \(A=90^{\circ}, B=60^{\circ}\) and \(C=30^{\circ}\). Now, \(\quad \cos B=\frac{a^2+c^2-b^2}{2 a c}\)…
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