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Precalculus : Use the laws of cosines and sines

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Example Questions

Example Question #11 : Law Of Cosines And Sines

Use the Law of Cosines to solve for the specified variable.

Pcq1

A=36.870

b=4

c=5

Solve for . Round to the nearest tenth.

Possible Answers:

None of these answers are correct.

$\frac{-32}{-40}$

Correct answer:

Explanation:

Law of Cosines

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

Therefore...

$a^2=16+25-2(4)(5)(\cos 36.9)$

$a^2=41-40(\cos 36.9)$

a^2=41-31.987

a^2=9.01

a=3.00

After rounding...

a=3

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Example Question #11 : Use The Laws Of Cosines And Sines

Use the Law of Sines to find in the following triangle.

(Not drawn to scale.)

Possible Answers:

c=13.29

c=14.4

c=15.95

c=9.03

c=7.52

Correct answer:

c=9.03

Explanation:

We use the Law of Sines to solve this problem

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

we plug in $b=12, B=70^{\circ}, C=45^{\circ}$

$\frac{12}{\sin 70^{\circ}}=\frac{c}{\sin 45^{\circ}}$

solving for we get:

$c=\frac{12\times \sin 45^{\circ}}{\sin 70^{\circ}}=9.03$

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Example Question #12 : Use The Laws Of Cosines And Sines

Use the Law of Sines to find .

(Not drawn to scale.)

Possible Answers:

b=7.17

b=23.56

b=6.52

b=14.29

b=2.9

Correct answer:

b=7.17

Explanation:

We use the Law of Sines to solve this problem:

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

where $A=65^{\circ}, B=30^{\circ}, C=85^{\circ}, a=13$

We plug in the values that we will need:

$\frac{13}{\sin 65^{\circ}}=\frac{b}{\sin 30^{\circ}}$

Notice that we did not use .

Solve for we get:

$b=\frac{13\times \sin 30^{\circ}}{\sin 65^{\circ}}=7.17$

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Example Question #13 : Use The Laws Of Cosines And Sines

Which of the following are the missing sides of the triangle?

$\text{I. } 13.95$

$\text{II. }6.55$

$\text{III. }4.59$

$\text{IV. }9.77$

Possible Answers:

$\text{II and IV}$

$\text{I and III}$

None of the above

$\text{II and III}$

Correct answer:

$\text{II and III}$

Explanation:

In order to solve this problem, we need to find . We do so by remembering the sum of the angles in a triangle is $180^{\circ}$ :

$C=180^{\circ}-55^{\circ}-35^{\circ}=90^{\circ}$

We can now use the Law of Sines to find the missing sides.

$\frac{a}{\sin 55}=\frac{8}{\sin 90}$

$a=\frac{8\times \sin 55}{\sin 90}=6.55$ which is II.

$\frac{b}{\sin 35}=\frac{8}{\sin 90}$

$b=\frac{8\times \sin 35}{\sin 90}=4.59$ which is III.

Our answers are then II and III

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Example Question #1444 : Pre Calculus

Which of the following are the missing sides of the triangle?

$\text{I. }4.55$

$\text{II. }5.49$

$\text{III. }1.72$

$\text{IV. }14.56$

Possible Answers:

$\text{I and III}$

$\text{II and IV}$

None of the above

$\text{I and II}$

Correct answer:

$\text{I and III}$

Explanation:

In order to solve this problem, we need to find

Since all the angles of a triangle add to $180^{\circ}$ , we can easily find it:

$180^{\circ}-65^{\circ}-20^{\circ}=95^{\circ}$

We can now use the Law of Sines to find the missing sides:

$\frac{a}{\sin 65}=\frac{5}{\sin 95}$

$a=\frac{5\times \sin 65}{\sin 95}=4.55$ which is I.

$\frac{b}{\sin 20}=\frac{5}{\sin 95}$

$b=\frac{5\times \sin 20}{\sin 95}=1.72$ which is III.

Our answers are then I and III.

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Example Question #14 : Use The Laws Of Cosines And Sines

Solve the triangle using the Law of Sines:

Law_of_sines__aas_

Possible Answers:

$a=6.01, \measuredangle A=31^\circ, b=12$

$a = 4.37, \measuredangle A = 20^{\circ}, b = 9.78$

$a = 5.81, \measuredangle A = 20^{\circ}, b = 10.34$

$a = 5.64, \measuredangle A = 30^{\circ}, b = 8.55$

None of the other answers

Correct answer:

$a = 4.37, \measuredangle A = 20^{\circ}, b = 9.78$

Explanation:

First we need to know what the Law of Sines is:

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

Looking at the triangle, we know c, C, and B. We can either solve for side b, using the law, or angle A using our knowledge that the interior angles of a triangle must add up to be 180.

$\frac{b}{sin50^{\circ}} = \frac{12}{sin110^{\circ}}$

$b = \frac{12\cdot sin{50^{\circ}}}{sin110^{\circ}} = 9.78$

$A = 180^{\circ} - 110^{\circ} - 50^{\circ} = 20^{\circ}$

Now all that's left is to find side a:

$\frac{a}{sin20^{\circ}} = \frac{12}{sin110^{\circ}}$

$a = \frac{12\cdot sin{20^{\circ}}}{sin110^{\circ}} = 4.37$

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Example Question #17 : Law Of Cosines And Sines

Use the Law of Sines to solve for the specified variable.

Pcq2

A=60

B=70

a=6

Solve for . Round to the nearest tenth.

Possible Answers:

4.9

5.6

0.1

None of these answers are correct.

6.5

Correct answer:

6.5

Explanation:

Law of Sines

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

Therefore...

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

$\frac{\sin 60}{6}= \frac{\sin 70}{b}$

$\sin 60 = \frac{6 \cdot \sin 70}{b}$

$b=\frac{6 \cdot \sin 70}{\sin 60}$

b=6.526

After rounding...

b=6.5

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Example Question #18 : Use The Laws Of Cosines And Sines

Pcq2

Solve for c using Law of Sines, given:

B=70

C=50

b=6.5

Round to the nearest tenth.

Possible Answers:

None of these answers are correct.

5.3

0.7

5.0

1.1

Correct answer:

5.3

Explanation:

Law of Sines

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

Therefore...

$\frac{\sin 70}{6.5}=\frac{\sin 50}{c}$

$\sin 70 = \frac{6.5 \cdot \sin 50}{c}$

$c= \frac{6.5 \cdot \sin 50}{\sin 70}$

c=5.320

After rounding...

c=5.3

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Example Question #15 : Use The Laws Of Cosines And Sines

Triangle_700

Given b=4ft, c=6ft, and $B=36^\circ$ , what is the measurement of to the nearest degree?

Possible Answers:

$82^\circ$

$82.15^\circ$

$83^\circ$

$82.2^\circ$

Correct answer:

$82^\circ$

Explanation:

Using the information we have, we can solve for $\angle C$ :

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

Plugging in what we know, we have:

$\frac{sin(36)}{4}=\frac{sinC}{6}$ .

Then, solve for sinC :

$6\cdot\frac{sin(36)}{4}=sinC$ .

Simplify, then solve for : 0.8816778784=sinC which means $sin^{-1}(.8816778784)=C$ .

Therefore, after rounding to the nearest degree, $C=62^\circ$ .

To solve for $\angle A$ , subtract $\angle B$ and $\angle C$ from $180^\circ$ : $180^\circ-36^\circ-62^\circ=82^\circ$ .

Therefore, $\angle A = 82^\circ$ .

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Example Question #15 : Law Of Cosines And Sines

Find the length of the missing side, .

Possible Answers:

26.77

24.51

22.36

29.01

Correct answer:

22.36

Explanation:

First, use the Law of Sines to find the measurement of angle

$\frac{15}{\sin 41}=\frac{20}{\sin B}$

$15\sin B = 13.12$

$\sin B=0.87$

B= 61.01

Recall that all the angles in a triangle need to add up to 180 degrees.

41+61.01+A=180

A=77.99

Now, use the Law of Sines again to find the length of .

$\frac{d}{\sin 77.99}=\frac{15}{\sin 41}$

$d \sin 41= 14.67$

d=22.36

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Precalculus : Use the laws of cosines and sines

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #11 : Law Of Cosines And Sines

Example Question #11 : Use The Laws Of Cosines And Sines

Example Question #12 : Use The Laws Of Cosines And Sines

Example Question #13 : Use The Laws Of Cosines And Sines

Example Question #1444 : Pre Calculus

Example Question #14 : Use The Laws Of Cosines And Sines

Example Question #17 : Law Of Cosines And Sines

Example Question #18 : Use The Laws Of Cosines And Sines

Example Question #15 : Use The Laws Of Cosines And Sines

Example Question #15 : Law Of Cosines And Sines

All Precalculus Resources

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