Problems involving right triangle trigonometry - HiSET

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Question

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

Evaluate $\sin N$ in terms of .

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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 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}$ .

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Question

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

Evaluate $\sin N$ in terms of .

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

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

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Question

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

Evaluate $\cos N$ in terms of .

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

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Question

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

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

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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.

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Question

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

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Answer

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.

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Question

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

Tap to reveal answer

Answer

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.

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Question

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

Evaluate $\tan N$ 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 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}$ .

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Question

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

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

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Answer

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.

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

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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.

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Question

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

Tap to reveal answer

Answer

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.

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Question

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

Tap to reveal answer

Answer

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.

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Question

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

Evaluate $\tan N$ 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 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}$ .

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Question

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

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

Tap to reveal answer

Answer

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.

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Question

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

Evaluate $\sin N$ 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 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}$ .

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Question

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

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

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

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Question

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

Tap to reveal answer

Answer

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.

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Question

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

Tap to reveal answer

Answer

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.

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Question

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

Evaluate $\tan N$ 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 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}$ .

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Question

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

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

Tap to reveal answer

Answer

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.

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