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Trigonometry : Law of Cosines and Law of Sines

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Example Question #31 :Law Of Cosines

If $\small a=8.4$ , $\small b=11.2$ , $\small \angle C$ = $\small 22^o$ , find the length of side $\small c$ to the nearest tenth of a degree.

Possible Answers:

$\small 9.8$

$\small 7.2$

$\small 4.6$

$\small 5.3$

Correct answer:

$\small 4.6$

Explanation:

Since we are give the length of two sides of a triangle and the corresponding angle of the third side, we can use the Law of Cosines to find the length of the third side. Because we are looking for, we use the equation

$\small c^2 = a^2 + b^2 - 2ab\cos\angle C$

Taking the square root of both sides we isolate

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

Substituting in the values from the problem gives

$\small c = \sqrt{8.4^2 + 11.2^2 - 2(8.4)(11.2)\cos\angle 22^o}$

$\small \small c = 4.6$

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Example Question #31 :Law Of Cosines And Law Of Sines

If $\small \angle A$ = $\small 20^o$ , $\small b$ = $\small 9$ , $\small c$ = $\small 20$ , find the length of side $\small a$ to the nearest whole number.

Possible Answers:

$\small 12$

$\small 10$

$\small 5$

$\small 7$

Correct answer:

$\small 12$

Explanation:

Because the problem gives the length of two of the three sides of a triangle and the corresponding angle of the third side, we can use the Law of Cosines to find the length of the third side. This gives use the equation

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

We can take the square root of both sides in order to give the equation

$\small a= \sqrt{b^2+c^2-2bc\cos\angle A}$

Substituting in the values given from the problem gives

$\small a = \sqrt{9^2+20^2-2(9)(20)\cos20^o}$

$\small a = 12$

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Example Question #33 :Law Of Cosines And Law Of Sines

Suppose there was a triangle with side lengths 8,7, 14. What is the measure of the largest angle, rounded to the nearest degree?

Possible Answers:

158^o

123^o

98^o

138^o

52^o

Correct answer:

138^o

Explanation:

The largest angle will be opposite the largest of the three side lengths, 14. We can find its measure using the law of cosines:

$14^2 = 7^2 + 8^2 - 2(7)(8)(cos\ C)$

$196 = 49 + 64 - 112cos \ C = 113 - 112 cos \ C$

$83 = -112 cos \ C$

$-0.74107 = cos \ C$

From here take the inverse cosine of each side to find the degress value that is angle C.

C = 138^o

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Example Question #1 :Law Of Sines

Given sides a = 10 , b = 5 and angle $\angle A = 100^{\circ}$ determine the corresponding value for $\angle B$

Possible Answers:

Undefined

$60^{\circ}$

$5^{\circ}$

$29.5^{\circ}$

$-29.5^{\circ}}$

Correct answer:

$29.5^{\circ}$

Explanation:

The Law of Sines is used here since we have Side - Angle - Side. We setup our equation as follows:

$\frac{a}{\sin A} = \frac{b}{\sin B}$

Next, we substitute the known values:

$\frac{10}{\sin 100} = \frac{5}{\sin B}$

Now we cross multiply:

$10\sin B = 5\sin 100$

Divide by 10 on both sides:

$\sin B = 0.5\sin100$

最后反正弦获取desired angle:

$B = sin^{-1}\left(0.5\sin100\right ) = 29.5^{\circ}$

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Example Question #2 :Law Of Sines

Let $\angle A = 17^{\circ}$ , $\angle C = 110^{\circ}$ and a = 5 , determine the length of side。

Possible Answers:

13.66

5.21

16.07

1.83

5.88

Correct answer:

13.66

Explanation:

We have two angles and one side, however we do not have $\angle B$ 。We can determine the angle using the property of angles in a triangle summing to $180^{\circ}$ :

$\angle B = 180^{\circ} - 17^{\circ} - 110^{\circ} = 53^{\circ}$

Now we can simply utilize the Law of Sines:

$\frac{5}{\sin 17} = \frac{b}{\sin 53}$

Cross multiply and divide:

$\frac{5\sin 53}{\sin 17} = b$

Reducing to obtain the final solution:

b = 13.66

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Example Question #1 :Law Of Sines

Triangle

在the above triangle, $a = 35^{\circ}$ and $c = 93^{\circ}$ 。If X = 49 , what isto the nearest tenth? (note: triangle not to scale)

Possible Answers:

28.1

67.3

38.7

62.1

35.7

Correct answer:

38.7

Explanation:

If we solve for, we can use the Law of Sines to find。

Since the sum of angles in a triangle equals $180^{\circ}$ ,

$a + b + c = 180^{\circ}$

$35^{\circ} + b + 93^{\circ} = 180^{\circ}$

$b = 52^{\circ}$

Now, using the Law of Sines:

$\frac{\textup{side opposite a}}{\sin\left ( a\right )} = \frac{\textup{side opposite b}}{\sin\left ( b\right )} = \frac{\textup{side opposite c}}{\sin\left ( c\right )}$

$\frac{{X}}{\sin\left ( c\right )} = \frac{{Y}}{\sin\left ( b\right )}$

$\frac{{49}}{\sin\left ( 93^{\circ}\right )} = \frac{{Y}}{\sin\left ( 52^{\circ}\right )}$

$\frac{49}{0.999} \approx \frac{Y}{0.788}$

$Y \approx \frac{49 \cdot 0.788 }{0.999}$

$Y \approx 38.7$

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Example Question #1 :Law Of Sines

Screen_shot_2015-03-07_at_5.09.32_pm

By what factor islarger thanin the triangle pictured above.

Possible Answers:

1.8

0.66

It isn't

1.51

Correct answer:

1.51

Explanation:

The Law of Sines states

$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$

so for a and b, that sets up

$\frac{a}{\sin(35)}=\frac{b}{\sin(60)}, b=a\frac{\sin(60)}{\sin(35)}=1.51a$

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Example Question #5 :Law Of Sines

Solve for $\theta$ :
Sines 1

Possible Answers:

7.24 ^ o

47.5 ^ o

45.49^o

0.15^o

68.18 ^o

Correct answer:

68.18 ^o

Explanation:

To solve, use the law of sines, $\frac{ \sin A }{a } = \frac{ \sin B }{b}$ where a is the side across from the angle A, and b is the side across from the angle B.

$\frac{ \sin \theta }{ 19 } = \frac{ \sin ( 20^o )}{ 7 }$ cross-multiply

$7 \sin \theta = 19 \cdot \sin (20^o )$ evaluate the right side

$7 \sin \theta = 6.4984$ divide by 7

$\sin \theta = 0.9283$ take the inverse sine

$\theta = \sin ^ {-1 } (0.9283 ) \approx 68.18 ^o$

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Example Question #6 :Law Of Sines

Evaluate using law of sines:

Sines 2

Possible Answers:

3.26

6.34

7.72

16.56

2.96

Correct answer:

7.72

Explanation:

To solve, use law of sines, $\frac{ \sin A }{a} = \frac{ \sin B }{b}$ where side a is across from angle A, and side b is across from angle B.

在this case, we have a 90-degree angle across from x, but we don't currently know the angle across from the side length 7. We can figure out this angle by subtracting 25^o from 90^o :

90^o - 25 ^o = 65^o

Now we can set up and solve using law of sines:

$\frac{ \sin (65^o )}{ 7 } = \frac{ \sin (90 ^o )} { x }$ cross-multiply

$\sin (90 ^o ) \cdot 7 = \sin (65^o ) \cdot x$ evaluate the sines

$1 \cdot 7 = (0.9063) x$ divide by 0.9063

7.72 = x

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Example Question #7 :Law Of Sines

What is the measure of $\angle F$ in $\triangle DEF$ below? Round to the nearest tenth of a degree.

Triangle def

Possible Answers:

$18.2^\circ$

$20.0^\circ$

$16.6^\circ$

$0.3^\circ$

$47.8^\circ$

Correct answer:

$16.6^\circ$

Explanation:

The law of sines tells us that $\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$ , wherea,b, andcare the sides opposite of anglesA,B, andC。在 $\triangle DEF$ , these ratios can be used to find $\angle F$ :

$\frac{sinD}{d}=\frac{sinF}{f}$

$\frac{sin114^\circ}{16}=\frac{sin F}{5}$

$sinF=\frac{5sin114^\circ}{16}$

$F=sin^{-1}\left ( \frac{5sin114^\circ}{16}\right )$

$F\approx16.6^\circ$

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Trigonometry : Law of Cosines and Law of Sines

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #31 :Law Of Cosines

Example Question #31 :Law Of Cosines And Law Of Sines

Example Question #33 :Law Of Cosines And Law Of Sines

Example Question #1 :Law Of Sines

Example Question #2 :Law Of Sines

Example Question #1 :Law Of Sines

Example Question #1 :Law Of Sines

Example Question #5 :Law Of Sines

Example Question #6 :Law Of Sines

Example Question #7 :Law Of Sines

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