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Trigonometry : Trigonometry

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6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Law Of Cosines And Law Of Sines

Given the triangle ABC , where a=8 , c=7 , and $\measuredangle B=135^{\circ}$ , calculate the side length to the thousandth decimal point.

Triangle abc

Possible Answers:

b=13.862

b=14.261

b=13.864

b=14.923

b=13.863

Correct answer:

b=13.863

Explanation:

Recall the law of cosines to determine the length of one side of a triangle ABC given the lengths of the other sides and and their included angle $\Theta$ :

$c^2=a^2+b^2-2ab(cos\Theta)$

Here, the unknown side length is denoted , and the other sides and the included angle is given. Substitute these values into the law of cosines and estimate square roots to the nearest thousandth decimal place to determine the side length .

$b^2=8^2+7^2-2(8)(7)(cos135^{\circ})$

$=113-112(-\frac{\sqrt(2)}{2})$

$\approx113+79.184 = 192.184$

$\Rightarrow b \approx 13.863$

Hence, the length of the remaining side of triangle ABC is approximately 13.863 units.

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Example Question #41 : Triangles

Construction

Two sides of a triangle have side lengths 6 and 7, and the angle between them measures 120 degrees. What is the length of the third side, rounded to the nearest hundredth?

Possible Answers:

9.22

11.27

127

Correct answer:

11.27

Explanation:

We will label the side with length 6 as , the side with length 7 as , the unknown side as , and the angle between them as $\theta$ .

The law of cosines relates the three sides and the angle as follows:

$c^2=a^2+b^2-2ab\cos(\theta)$ .

Plugging in 6 for , 7 for , and 120 for $\theta$ yields:

$\begin{align*} c^2&=6^2+7^2-2(6)(7)\cos(120)\\ &=36+49-2(6)(7)\frac{-1}{2}\\ &=36+49-(-42)\\ &=127 \end{align*}$

We then take the square root of both sides:

$c=\sqrt{127}\approx11.27$

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Example Question #42 : Triangles

Find the length of side X using the law of cosines.

Tri2

Possible Answers:

35.68

39.4

30.23

25.5

Correct answer:

35.68

Explanation:

The law of cosines is a generalization of the Pythagorean Theorem:

$a^{2}+b^2-2ab\cos C=c^2$

where a, b, and c are lengths of sides and C is the angle opposite to c.

This problem gives us values for a, b, and C, so we can use the law of cosines.

$25^2 +17^2-2(25)(17)\cos 115^{\circ}=X^2$

After simplifying:

625+289-(-359)=1273=X^2

Taking the square root of both sides:

X=35.68

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Example Question #43 : Triangles

In degrees, find the value of angle B.

Possible Answers:

38.56

36.18

45.29

32.49

Correct answer:

36.18

Explanation:

In order to find the value of the angle, you will need to use the law of cosines. Recall that for any triangle, like the one shown below,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Now, since we want to find the value of B, we will need to use $\cos B=\frac{a^2+c^2-b^2}{2ac}$ .

Plug in the given values of the triangle.

$\cos B=\frac{20^2+14^2-12^2}{2(20)(14)}=0.807$

B=36.18 degrees

Make sure to round to places after the decimal.

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Example Question #11 : Law Of Cosines And Law Of Sines

In degrees, find the value of angle B.

Possible Answers:

17.04

18.94

15.44

16.21

Correct answer:

16.21

Explanation:

In order to find the value of the angle, you will need to use the law of cosines. Recall that for any triangle, like the one shown below,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Now, since we want to find the value of B, we will need to use $\cos B=\frac{a^2+c^2-b^2}{2ac}$ .

Plug in the given values of the triangle.

$\cos B=\frac{11^2+8^2-4^2}{2(11)(8)}=0.960$

B=16.21 degrees

Make sure to round to places after the decimal.

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Example Question #12 : Law Of Cosines And Law Of Sines

In degrees, find the value of angle B.

Possible Answers:

32.48

28.96

30.40

29.51

Correct answer:

28.96

Explanation:

In order to find the value of the angle, you will need to use the law of cosines. Recall that for any triangle, like the one shown below,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Now, since we want to find the value of B, we will need to use $\cos B=\frac{a^2+c^2-b^2}{2ac}$ .

Plug in the given values of the triangle.

$\cos B=\frac{12^2+9^2-6^2}{2(12)(9)}=0.875$

B=28.96 degrees

Make sure to round to places after the decimal.

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Example Question #13 : Law Of Cosines And Law Of Sines

In degrees, find the value of angle C.

Possible Answers:

52.01

41.22

49.46

44.78

Correct answer:

49.46

Explanation:

In order to find the value of the angle, you will need to use the law of cosines. Recall that for any triangle, like the one shown below,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Now, since we want to find the value of C, we will need to use $\cos C=\frac{a^2+b^2-c^2}{2ab}$ .

Plug in the given values of the triangle.

$\cos C=\frac{15^2+6^2-12^2}{2(15)(6)}=0.65$

C=49.46 degrees

Make sure to round to places after the decimal.

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Example Question #301 : Trigonometry

In degrees, find the value of angle C.

Possible Answers:

42.59

40.08

39.51

41.80

Correct answer:

41.80

Explanation:

In order to find the value of the angle, you will need to use the law of cosines. Recall that for any triangle, like the one shown below,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Now, since we want to find the value of C, we will need to use $\cos C=\frac{a^2+b^2-c^2}{2ab}$ .

Plug in the given values of the triangle.

$\cos C=\frac{5^2+11^2-8^2}{2(5)(11)}=0.74545$

C=41.80 degrees

Make sure to round to places after the decimal.

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Example Question #15 : Law Of Cosines And Law Of Sines

Find the length of side .

Possible Answers:

20.92

17.85

19.01

18.44

Correct answer:

17.85

Explanation:

In order to find the value of the angle, you will need to use the law of cosines. Recall that for any triangle, like the one shown below,

$a^2=b^2+c^2-2bc\cos A$

Plug in the given values of the triangle.

$a^2=12^2+13^2-2(12)(13)\cos 91$

a=17.85

Make sure to round to places after the decimal.

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Example Question #52 : Triangles

In degrees, find the value of angle A.

Possible Answers:

66.24

59.63

71.05

67.23

Correct answer:

67.23

Explanation:

In order to find the value of the angle, you will need to use the law of cosines. Recall that for any triangle, like the one shown below,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Now, since we want to find the value of A, we will need to use $\cos A=\frac{b^2+c^2-a^2}{2bc}$ .

Plug in the given values of the triangle.

$\cos A=\frac{17^2+19^2-20^2}{2(17)(19)}=0.387$

A=67.23 degrees

Make sure to round to places after the decimal.

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #2 : Law Of Cosines And Law Of Sines

Example Question #41 : Triangles

Example Question #42 : Triangles

Example Question #43 : Triangles

Example Question #11 : Law Of Cosines And Law Of Sines

Example Question #12 : Law Of Cosines And Law Of Sines

Example Question #13 : Law Of Cosines And Law Of Sines

Example Question #301 : Trigonometry

Example Question #15 : Law Of Cosines And Law Of Sines

Example Question #52 : Triangles

All Trigonometry Resources

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