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Circular Measurement of Angle

The approxima...

Question

The approximate value of $sin 31^{o}$ is

> 0.5

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> 0.6

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< 0.5

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< 0.4

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Solution

The correct option is C $> 0.5$
As we know that sine function is increasing in the interval $(0, 45^{o})$
$∴ sin 30^{o} < s i n 31^{o} < s i n 45^{o}$
$0.5 < s i n 31^{o} < 0.7$
it will be just greater than $0.5$ as $31^{o}$ is just greater than $30^{o}$
Therefore Answer is $A$

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\frac{0.6 \times 0.6 \times 0.6 + 0.5 \times 0.5 \times 0.5 + 0.1 \times 0.1 - 0.09}{0.6 \times 0.6 + 0.5 \times 0.5 + 0.1 \times 0.1 - 0.41}

sin x = 0.5

tan x

P (A)

P (B)

P (A) = 0.5, P (B) = 0.7

P (A \cap B) = 0.6

P (A) = 0.5, P (B) = 0.4

P (A \cup B) = 0.8

A \to B + C,

0.001 M/s

1

M

A

B

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