AP EAMCET · Maths · Properties of Triangles
In a \(\triangle A B C, A-B=120^{\circ}, R=8 r\), then \(\frac{1+\cos C}{1-\cos C}=\)
- A 16
- B 14
- C 15
- D 10
Answer & Solution
Correct Answer
(C) 15
Step-by-step Solution
Detailed explanation
\(\frac{1+\cos C}{1-\cos C} = \cot^2(C/2)\) \(R=8r \implies 1 = 32\sin(A/2)\sin(B/2)\sin(C/2)\) \(1 = 32 \cdot \frac{1}{2}[\cos((A-B)/2) - \cos((A+B)/2)]\sin(C/2)\) \(1 = 16[\cos(60^{\circ}) - \cos(90^{\circ}-C/2)]\sin(C/2)\) \(1 = 16[1/2 - \sin(C/2)]\sin(C/2)\)…
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