Cosine - HiSET

Card 1 of 8

Didn't Know

Knew It

1 of 87 left

Question

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

Evaluate $\cos N$ in terms of .

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

Evaluate $\cos N$ in terms of .

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

Evaluate $\cos N$ in terms of .

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

Evaluate $\cos N$ in terms of .

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Knew It

Cosine - HiSET

Evaluate in terms of .

An identity of trigonometry is for any value of . Since , it immediately follows that . This response is not among the given choices.

Evaluate in terms of .

An identity of trigonometry is for any value of . Since , it immediately follows that . This response is not among the given choices.

Evaluate in terms of .

An identity of trigonometry is for any value of . Since , it immediately follows that . This response is not among the given choices.

Evaluate in terms of .

An identity of trigonometry is for any value of . Since , it immediately follows that . This response is not among the given choices.

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

Evaluate $\cos N$ in terms of .

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 .

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 .

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 .

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.