Mathematics · 2022
JEE Main · 26 June 2022, Shift 2 · Q18
16 sin (20^°) sin (40^°) sin (80^°) is equal to:
$\displaystyle 16 \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(80^{\circ}\right)$ is equal to :
Official answer
(2)
$\displaystyle 2 \sqrt{3}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.