Cosine - HiSET

Card 0 of 8

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

Evaluate $\cos N$ in terms of .

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

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

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

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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 .

Tap to see back →

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

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

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

Tap to see back →

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 .

Tap to see back →

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

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

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

Tap to see back →

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 .

Tap to see back →

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

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

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

Tap to see back →

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.