Cosine

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$\sin a^{\circ } = \frac{2}{5}$

$\cos (90-a)^{\circ } = ?$

None of the other choices gives the correct response.

$\frac{5}{\sqrt{21}}$

$\frac{2}{\sqrt{29}}$

$\frac{5}{\sqrt{29}}$

$\frac{2}{\sqrt{21}}$

Explanation

An identity of trigonometry is

$\sin a^{\circ} = \cos (90-a)^{\circ}$

for any value of .

Since $\sin a^{\circ } = \frac{2}{5}$ , it immediately follows that $\cos (90-a)^{\circ} =\frac{2}{5}$ .

This response is not among the given choices.

$\tan N = \frac{\sqrt{y}}{5}$

Evaluate $\cos N$ in terms of .

$\frac{5}{ \sqrt{y+25}}$

$\sqrt{ \frac{y}{y+25}}$

$\sqrt{ \frac{y+25}{y}}$

$\frac{ \sqrt{y+25}}{5}$

$\frac{5} {\sqrt{y}}$

Explanation

Suppose we allow be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring .

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, so

$\tan N = \frac{O}{A} =\frac{\sqrt{y}}{5}$

We can set the lengths of the opposite and adjacent legs to $\sqrt{y}$ and 5, respectively. The length of the hypotenuse can be determined using the Pythagorean Theorem:

$H = \sqrt{O^{2}+A^{2}} = \sqrt{( \sqrt{y})^{2}+5^{2}} = \sqrt{y+25}$

The cosine of the angle is equal to the ratio of the length of the adjacent leg to that of the hypotenuse, so

$\cos N = \frac{A}{H} = \frac{5}{ \sqrt{y+25}}$ .

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