TS EAMCET · Maths · Properties of Triangles
In a triangle ABC, if \(c^2-a^2=b(\sqrt{3} c-b)\) and \(b^2-a^2=c(c-a)\), then \(\lfloor\mathrm{ACB}=\)
- A \(30^{\circ}\)
- B \(60^{\circ}\)
- C \(45^{\circ}\)
- D \(90^{\circ}\)
Answer & Solution
Correct Answer
(D) \(90^{\circ}\)
Step-by-step Solution
Detailed explanation
\(c^2-a^2=b(\sqrt{3} c-b)\) \(c^2+b^2-a^2=\sqrt{3}bc\) \(2bc \cos A = \sqrt{3}bc\) \(\cos A = \frac{\sqrt{3}}{2} \implies A = 30^{\circ}\) \(b^2-a^2=c(c-a)\) \(b^2=a^2+c^2-ac\) \(2ac \cos B = ac\) \(\cos B = \frac{1}{2} \implies B = 60^{\circ}\) \(A+B+C = 180^{\circ}\)…
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