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Trigonometry : Triangles

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Example Questions

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:

$-29.5^{\circ}}$

$60^{\circ}$

$5^{\circ}$

$29.5^{\circ}$

Undefined

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:

$\sin B = 0.5\sin100$

Finally taking the inverse sine to obtain the 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

1.83

16.07

5.88

5.21

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 #2 : Law Of Sines

Triangle

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

Possible Answers:

38.7

62.1

28.1

67.3

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 #71 : Triangles

Screen_shot_2015-03-07_at_5.09.32_pm

By what factor is larger than in the triangle pictured above.

Possible Answers:

1.51

0.66

It isn't

1.8

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 #1 : Law Of Sines

Solve for $\theta$ :
Sines 1

Possible Answers:

0.15^o

45.49^o

68.18 ^o

7.24 ^ o

47.5 ^ 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 #1 : Law Of Sines

Evaluate using law of sines:

Sines 2

Possible Answers:

2.96

7.72

6.34

16.56

3.26

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.

In 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 #1 : 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:

$18.2^\circ$

$47.8^\circ$

$20.0^\circ$

$16.6^\circ$

$0.3^\circ$

Correct answer:

$16.6^\circ$

Explanation:

The law of sines tells us that $\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$ , where a, b, and c are the sides opposite of angles A, B, and C. In $\triangle DEF$ , 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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Example Question #1 : Law Of Sines

Find the length of the line segment $\overline{AB}$ in the triangle below.

Round to the nearest hundredth of a centimeter.

Triangle

Possible Answers:

$10.83\ cm$

$14.66\ cm$

$15.60\ cm$

$12.21\ cm$

$13.84\ cm$

Correct answer:

$13.84\ cm$

Explanation:

The law of sines states that

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$ .

In this triangle, we are looking for the side length c, and we are given angle A, angle B, and side b. The sum of the interior angles of a triangle is $180^{\circ}$ ; using subtraction we find that angle C = $57^{\circ}$ .

We can now form a proportion that includes only one unknown, c:

$\frac{sin52^{\circ}}{13}=\frac{sin57^{\circ}}{c}$

Solving for c, we find that

$c=\frac{13sin57^{\circ}}{sin52^{\circ}}\approx13.84$ .

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Example Question #1 : Law Of Sines

In the triangle below, $m\angle A=78^\circ$ , $m\angle B=40^\circ$ , and c=14 . What is the length of side to the nearest tenth?

Triangle abc

Possible Answers:

10.2

12.6

15.5

9.2

19.2

Correct answer:

10.2

Explanation:

First, find $m\angle C$ . The sum of the interior angles of a triangle is $180^\circ$ , so $m\angle C = 180^\circ - 78^\circ - 40^\circ$ , or $62^\circ$ .

Using this information, you can set up a proportion to find side b:

$\frac{sinC}{c}=\frac{sinB}{b}$

$\frac{sin62^\circ}{14}=\frac{sin40^\circ}{b}$

$b=\frac{14sin40^\circ}{sin62^\circ}$

$b\approx10.2$

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Example Question #1 : Law Of Sines

In the triangle below, $m\angle A=75^\circ$ , $m\angle B=21^\circ$ , and c=19.3 .

Triangle abc

What is the length of side a to the nearest tenth?

Possible Answers:

7.0

7.2

19.9

52.0

18.7

Correct answer:

18.7

Explanation:

To use the law of sines, first you must find the measure of $\angle C$ . Since the sum of the interior angles of a triangle is $180^\circ$ , $m\angle C=84^\circ$ .

Law of sines:

$\frac{\sin A}{a}=\frac{\sin C}{c}$

$\frac{\sin 75^\circ}{a}=\frac{\sin 84^\circ}{19.3}$

$a=\frac{19.3 \sin 75^\circ}{\sin 84^\circ}$

$a\approx 18.7$

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Trigonometry : Triangles

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Law Of Sines

Example Question #2 : Law Of Sines

Example Question #2 : Law Of Sines

Example Question #71 : Triangles

Example Question #1 : Law Of Sines

Example Question #1 : Law Of Sines

Example Question #1 : Law Of Sines

Example Question #1 : Law Of Sines

Example Question #1 : Law Of Sines

Example Question #1 : Law Of Sines

All Trigonometry Resources

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