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In $\triangle ABC$, $\angle B$ = $90^{0}$ and AB : BC = 2 : 1. The value of sin A + cot C is1). $3+\sqrt{5}$2). $\frac{2+\sqrt{5}}{2\sqrt{5}}$3). $2+\sqrt{5}$4). $3\sqrt{5}$ |
Answer» OPTION 2 : - $\FRAC{2+\sqrt{5}}{2\sqrt{5}}$ | |