\(sin30^o+sin60^o>sin90^o\) । Madhyamik 2019


বিবৃতিটি মিথ্যা

\(sin30^o+sin60^o\)
\(=\cfrac{1}{2}+\cfrac{\sqrt3}{2}\)
\(\gt \cfrac{1}{2}+\cfrac{1}{2} [\because \cfrac{\sqrt3}{2}\gt \cfrac{1}{2}]\)
\(\gt 1\)
\(\gt sin 90^o\)

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