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Trigonometry : Law of Cosines

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Example Question #11 : Law Of Cosines

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 #11 : Law Of Cosines And Law Of Sines

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

Tri2

Possible Answers:

25.5

39.4

35.68

30.23

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 #41 : Triangles

In degrees, find the value of angle B.

Possible Answers:

45.29

38.56

36.18

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:

18.94

17.04

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 #11 : Law Of Cosines And Law Of Sines

In degrees, find the value of angle B.

Possible Answers:

30.40

28.96

29.51

32.48

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 #11 : Law Of Cosines And Law Of Sines

In degrees, find the value of angle C.

Possible Answers:

44.78

41.22

52.01

49.46

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 #11 : Law Of Cosines And Law Of Sines

In degrees, find the value of angle C.

Possible Answers:

41.80

40.08

42.59

39.51

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 #51 : Triangles

Find the length of side .

Possible Answers:

18.44

17.85

20.92

19.01

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 #11 : Law Of Cosines

In degrees, find the value of angle A.

Possible Answers:

59.63

67.23

71.05

66.24

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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Example Question #311 : Trigonometry

In degrees, find the value of angle A.

Possible Answers:

Correct answer:

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.

$\cos A=\frac{15^2+20^2-25^2}{2(15)(20)}=0$

A=90 degrees

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Trigonometry : Law of Cosines

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #11 : Law Of Cosines

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

Example Question #41 : Triangles

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

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

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

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

Example Question #51 : Triangles

Example Question #11 : Law Of Cosines

Example Question #311 : Trigonometry

All Trigonometry Resources

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