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HiSET: Math : Problems involving right triangle trigonometry

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Example Questions

Example Question #151 : Hi Set: High School Equivalency Test: Math

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

Evaluate $\sin N$ in terms of .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Suppose we allow O, A, H 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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Example Question #2 : Problems Involving Right Triangle Trigonometry

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

Evaluate $\sin N$ in terms of .

Possible Answers:

$\frac{x}{\sqrt{x^{2}-9}}$

$\frac{3}{x}$

$\frac{\sqrt{x^{2}-9}}{3}$

$\frac{3}{\sqrt{x^{2}-9}}$

$\frac{\sqrt{x^{2}-9}}{x}$

Correct answer:

$\frac{\sqrt{x^{2}-9}}{3}$

Explanation:

Suppose we allow O, A, H 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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Example Question #1 : Problems Involving Right Triangle Trigonometry

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

Evaluate $\cos N$ in terms of .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Suppose we allow O, A, H 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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Example Question #1 : Problems Involving Right Triangle Trigonometry

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

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

Possible Answers:

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

None of the other choices gives the correct response.

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

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

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

Correct answer:

None of the other choices gives the correct response.

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.

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Example Question #3 : Problems Involving Right Triangle Trigonometry

Right

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

Possible Answers:

$\frac{144}{ \tan 35^{\circ }}$

$144 \tan 35^{\circ }$

$\frac{72}{ \tan 35^{\circ }}$

$72 \tan 35^{\circ }$

None of the other choices gives the correct response.

Correct answer:

$72 \tan 35^{\circ }$

Explanation:

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 AC=12 . $\overline{AB }$ 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 AC=12 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 AC=12 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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Example Question #1 : Tangent

Right

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

Possible Answers:

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

None of the other choices gives the correct response.

$32 \tan 35^{\circ}$

$64 \tan 35^{\circ}$

$\frac{64}{\tan 35^{\circ}}$

Correct answer:

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

Explanation:

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 AB = 8 . $\overline{AB }$ 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 AB = 8 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 AB = 8 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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Example Question #1 : Problems Involving Right Triangle Trigonometry

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

Evaluate $\tan N$ in terms of .

Possible Answers:

$\frac{\sqrt{x^{2}-9}}{3}$

$\frac{3}{x}$

$\frac{3}{\sqrt{x^{2}-9}}$

$\frac{\sqrt{x^{2}-9}}{x}$

$\frac{x}{\sqrt{x^{2}-9}}$

Correct answer:

$\frac{\sqrt{x^{2}-9}}{x}$

Explanation:

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

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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Example Question #101 : Measurement And Geometry

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

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

Possible Answers:

$\frac{\sqrt{33}}{7}$

$\frac{4}{\sqrt{33}}$

$\frac{7}{4}$

$\frac{\sqrt{33}}{4}$

$\frac{7}{\sqrt{33}}$

Correct answer:

$\frac{4}{\sqrt{33}}$

Explanation:

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 O = 4, H = 7 . 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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HiSET: Math : Problems involving right triangle trigonometry

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #151 : Hi Set: High School Equivalency Test: Math

Example Question #2 : Problems Involving Right Triangle Trigonometry

Example Question #1 : Problems Involving Right Triangle Trigonometry

Example Question #1 : Problems Involving Right Triangle Trigonometry

Example Question #3 : Problems Involving Right Triangle Trigonometry

Example Question #1 : Tangent

Example Question #1 : Problems Involving Right Triangle Trigonometry

Example Question #101 : Measurement And Geometry

All HiSET: Math Resources

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