Problems involving right triangle trigonometry - HiSET

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$\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.

$\cos N = $\frac{x}{3}$$

Evaluate $\sin N$ 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 cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = $\frac{A}{H}$ = $\frac{x}{3}$$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = $$\sqrt{H^{2}$$- $A^{2}$} = $$\sqrt{x^{2}$$$-3^{2}$} = $$\sqrt{x^{2}$$-9 }$

The sine of the angle is equal to the ratio of the length of the opposite leg to that of the hypotenuse, so

$\sin N = $\frac{O}{H}$ = $$\frac{\sqrt{x^{2}$$-9 }}{3}$ .

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

Evaluate $\sin N$ 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 sine of the angle is equal to the ratio of the length opposite leg to that of the hypotenuse, so

$\sin N = $\frac{O}{H}$ = $\frac{\sqrt{y }$}{ $\sqrt{y+25}$} =$\sqrt{ \frac{y}{y+25}$}$ .

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

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The area of a right triangle is half the product of the lengths of its legs, which here are $\overline{AB }$ and $\overline{AC}$ - that is,

$a= $\frac{1}{2}$(AB)(AC)$

We are given that . $\overline{AB }$ is the leg opposite the angle $\angle C$ and $\overline{AC}$ is its adjacent leg, we can find using the tangent ratio:

$\tan (m \angle C) = $\frac{AB}{AC}$$

Setting and $m \angle C = 35 ^{\circ }$ , we get

$\tan $35^{\circ }$ = $\frac{AB}{12}$$

Solve for by multiplying both sides by 12:

$12 \cdot \tan $35^{\circ }$ =12 \cdot $\frac{AB}{12}$$

$12 \tan $35^{\circ }$ =AB$

Now, set $AB = 12 \tan $35^{\circ }$$ and in the area formula:

$a= $\frac{1}{2}$(AB)(AC)$

$= $\frac{1}{2}$(12 \tan $35^{\circ }$)\left (12\right )$

$=72 \tan $35^{\circ }$$ ,

the correct choice.

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

Tap to see back →

The area of a right triangle is half the product of the lengths of its legs, which here are $\overline{AB }$ and $\overline{AC}$ - that is,

$a= $\frac{1}{2}$(AB)(AC)$

We are given that . $\overline{AB }$ is the leg opposite the angle $\angle C$ and $\overline{AC}$ is its adjacent leg, we can find using the tangent ratio:

$\tan (m \angle C) = $\frac{AB}{AC}$$

Setting and $m \angle C = 35 ^{\circ }$ , we get

$\tan $35^{\circ }$ = $\frac{8}{AC}$$

Solve for by first, finding the reciprocal of both sides:

$\frac{1}{\tan $35^{\circ }$$} = $\frac{AC}{8}$

Now, multiply both sides by 8:

$$\frac{1}{\tan $35^{\circ }$$} \cdot 8 = $\frac{AC}{8}$ \cdot 8$

$$\frac{8}{\tan $35^{\circ }$$} = AC$

Now, set and $AC = $\frac{8}{\tan $35^{\circ }$$}$ in the area formula:

$a= $\frac{1}{2}$(AB)(AC)$

$= $\frac{1}{2}$(8)\left ( $\frac{8}{\tan $35^{\circ}$$} \right)$

$= $\frac{1}{2}$ \left($\frac{8}{1}$ \right)\left( $\frac{8}{\tan $35^{\circ}$$} \right )$

$= $\frac{32}{\tan $35^{\circ}$$}$ ,

the correct choice.

$\cos N = $\frac{x}{3}$$

Evaluate $\tan N$ 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 cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = $\frac{A}{H}$ = $\frac{x}{3}$$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = $$\sqrt{H^{2}$$- $A^{2}$} = $$\sqrt{x^{2}$$$-3^{2}$} = $$\sqrt{x^{2}$$-9 }$

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

$\tan N = $\frac{O}{A}$ = $$\frac{\sqrt{x^{2}$$-9}}{x}$ .

$\sin $a^{\circ }$ = $\frac{4}{7}$$

$\tan $a^{\circ }$ = ?$

Tap to see back →

The sine of an angle is defined to be the ratio of the length of the opposite leg of a right triangle to the length of its hypotenuse. Therefore, we can set . By the Pythagorean Theorem:
the adjacent leg of the triangle has measure

$A = $$\sqrt{H^{2}$$$-O^{2}$} = $$\sqrt{7^{2}$$$-4^{2}$} = $\sqrt{49-16}$ = $\sqrt{33}$$

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

$\tan a ^{\circ }= $\frac{O}{A}$ = $\frac{4}{\sqrt{33}$}$ ,

the correct response.

$\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.

$\cos N = $\frac{x}{3}$$

Evaluate $\sin N$ 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 cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = $\frac{A}{H}$ = $\frac{x}{3}$$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = $$\sqrt{H^{2}$$- $A^{2}$} = $$\sqrt{x^{2}$$$-3^{2}$} = $$\sqrt{x^{2}$$-9 }$

The sine of the angle is equal to the ratio of the length of the opposite leg to that of the hypotenuse, so

$\sin N = $\frac{O}{H}$ = $$\frac{\sqrt{x^{2}$$-9 }}{3}$ .

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

Evaluate $\sin N$ 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 sine of the angle is equal to the ratio of the length opposite leg to that of the hypotenuse, so

$\sin N = $\frac{O}{H}$ = $\frac{\sqrt{y }$}{ $\sqrt{y+25}$} =$\sqrt{ \frac{y}{y+25}$}$ .

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

Tap to see back →

The area of a right triangle is half the product of the lengths of its legs, which here are $\overline{AB }$ and $\overline{AC}$ - that is,

$a= $\frac{1}{2}$(AB)(AC)$

We are given that . $\overline{AB }$ is the leg opposite the angle $\angle C$ and $\overline{AC}$ is its adjacent leg, we can find using the tangent ratio:

$\tan (m \angle C) = $\frac{AB}{AC}$$

Setting and $m \angle C = 35 ^{\circ }$ , we get

$\tan $35^{\circ }$ = $\frac{AB}{12}$$

Solve for by multiplying both sides by 12:

$12 \cdot \tan $35^{\circ }$ =12 \cdot $\frac{AB}{12}$$

$12 \tan $35^{\circ }$ =AB$

Now, set $AB = 12 \tan $35^{\circ }$$ and in the area formula:

$a= $\frac{1}{2}$(AB)(AC)$

$= $\frac{1}{2}$(12 \tan $35^{\circ }$)\left (12\right )$

$=72 \tan $35^{\circ }$$ ,

the correct choice.

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

Tap to see back →

The area of a right triangle is half the product of the lengths of its legs, which here are $\overline{AB }$ and $\overline{AC}$ - that is,

$a= $\frac{1}{2}$(AB)(AC)$

We are given that . $\overline{AB }$ is the leg opposite the angle $\angle C$ and $\overline{AC}$ is its adjacent leg, we can find using the tangent ratio:

$\tan (m \angle C) = $\frac{AB}{AC}$$

Setting and $m \angle C = 35 ^{\circ }$ , we get

$\tan $35^{\circ }$ = $\frac{8}{AC}$$

Solve for by first, finding the reciprocal of both sides:

$\frac{1}{\tan $35^{\circ }$$} = $\frac{AC}{8}$

Now, multiply both sides by 8:

$$\frac{1}{\tan $35^{\circ }$$} \cdot 8 = $\frac{AC}{8}$ \cdot 8$

$$\frac{8}{\tan $35^{\circ }$$} = AC$

Now, set and $AC = $\frac{8}{\tan $35^{\circ }$$}$ in the area formula:

$a= $\frac{1}{2}$(AB)(AC)$

$= $\frac{1}{2}$(8)\left ( $\frac{8}{\tan $35^{\circ}$$} \right)$

$= $\frac{1}{2}$ \left($\frac{8}{1}$ \right)\left( $\frac{8}{\tan $35^{\circ}$$} \right )$

$= $\frac{32}{\tan $35^{\circ}$$}$ ,

the correct choice.

$\cos N = $\frac{x}{3}$$

Evaluate $\tan N$ 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 cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = $\frac{A}{H}$ = $\frac{x}{3}$$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = $$\sqrt{H^{2}$$- $A^{2}$} = $$\sqrt{x^{2}$$$-3^{2}$} = $$\sqrt{x^{2}$$-9 }$

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

$\tan N = $\frac{O}{A}$ = $$\frac{\sqrt{x^{2}$$-9}}{x}$ .

$\sin $a^{\circ }$ = $\frac{4}{7}$$

$\tan $a^{\circ }$ = ?$

Tap to see back →

The sine of an angle is defined to be the ratio of the length of the opposite leg of a right triangle to the length of its hypotenuse. Therefore, we can set . By the Pythagorean Theorem:
the adjacent leg of the triangle has measure

$A = $$\sqrt{H^{2}$$$-O^{2}$} = $$\sqrt{7^{2}$$$-4^{2}$} = $\sqrt{49-16}$ = $\sqrt{33}$$

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

$\tan a ^{\circ }= $\frac{O}{A}$ = $\frac{4}{\sqrt{33}$}$ ,

the correct response.

$\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.

$\cos N = $\frac{x}{3}$$

Evaluate $\sin N$ 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 cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = $\frac{A}{H}$ = $\frac{x}{3}$$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = $$\sqrt{H^{2}$$- $A^{2}$} = $$\sqrt{x^{2}$$$-3^{2}$} = $$\sqrt{x^{2}$$-9 }$

The sine of the angle is equal to the ratio of the length of the opposite leg to that of the hypotenuse, so

$\sin N = $\frac{O}{H}$ = $$\frac{\sqrt{x^{2}$$-9 }}{3}$ .

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

Evaluate $\sin N$ 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 sine of the angle is equal to the ratio of the length opposite leg to that of the hypotenuse, so

$\sin N = $\frac{O}{H}$ = $\frac{\sqrt{y }$}{ $\sqrt{y+25}$} =$\sqrt{ \frac{y}{y+25}$}$ .