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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 #62 : Triangles

If $\small a=10$ , $\small b=6.7$ , $\small \angle C$ = $\small 45^o$ find $\small c$ to the nearest tenth.

Possible Answers:

$\small 4.0$

$\small 8.7$

$\small 7.1$

$\small 2.3$

Correct answer:

$\small 7.1$

Explanation:

Because we are given 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 side $\small c$ . To find side $\small c$ we use

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

Taking the square root of both sides gives us

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

Substituting in the values from the problem

$\small c = \sqrt{10^2 +6.7^2 - 2(10)(6.7)\cos45^o}$

$\small c = 7.1$

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Example Question #63 : Triangles

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:

$\small 9.8$

$\small 5.3$

$\small 7.2$

$\small 4.6$

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

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:

$\small 12$

$\small 10$

$\small 7$

$\small 5$

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:

123^o

98^o

52^o

158^o

138^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:

$-29.5^{\circ}}$

$60^{\circ}$

$5^{\circ}$

$29.5^{\circ}$

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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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #62 : Triangles

Example Question #63 : Triangles

Example Question #31 : Law Of Cosines

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

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

All Trigonometry Resources

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