If $x = sin 1, y = sin 2; z = sin 3$ then

x < y < z

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x > y > z

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y < z < x

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z < x < y

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Solution

The correct option is A $x < y < z$
We have, $x = sin 1 \Rightarrow x = 0.017$
$x = sin 1 \Rightarrow x = 0.017$
$y = sin 2 \Rightarrow y = 0.035$
$z = sin 3 \Rightarrow z = 0.052$
Comparing the above value, $x < y < z$

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