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3 years ago

How do you know that 21 over 30 is greater than 2 over 3

1 answer:

7 0

The problem deals with fractions comparison, lets do it:
21/30 > 2/3
we begin solving:
21 > (2/3)*30
21 > 2*10
21 > 20
therefore the proposed inequality is true, 21/30 > 2/3

You can solve as well getting same denominator for both fractions and comparing directly, in this case we need to get 2/3 to be divided by 30:
2/3 = (10/10)(2/3) = 20/30
So we have:
21/30 > 2/3
which is equal to:
21/30 > 20/30
and we compare directly because both fractions are divided by the same number, and we can see that the inequality is true.

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$\displaystyle\frac{\sin A}{BC} \ = \ \displaystyle\frac{\sin B}{AC} \\ \\ \\ \sin A \ = \ \displaystyle\frac{\sin B \ \times \ BC}{AC} \\ \\ \\\sin A \ = \ \displaystyle\frac{\sin \left(90^{\circ}\right) \ \times \ \sqrt{\left(10\right)^{2} \ - \ \left(\sqrt{78}\right)^{2}}}{10} \\ \\ \\ \sin A \ = \ \dsiplaystyle\frac{\sqrt{22}}{10} \\ \\ \\ \sin A \ = \ 0.47 \ \ \ \ (\text{nearest hundredth})$

Alternatively, you can solve this question using the definition of the trigonometric function sine. For the angle $\theta$ in the figure below,

$\sin \theta \ = \ \displaystyle\frac{\text{opposite}}{\text{hypothenuse}}$ .

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