29.1 Law of Sines and Cosines Problem Set copy
LAW OF SINES AND COSINES | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: Given a general π΄π΅πΆ triangle, side βπβ and angle βπ΅β are opposite from one another. If side βπβ has a length of 14, angle βπ΅β is 40Β°, and side βπβ has a length of 12, then the length of side βπβ is most close to A. 18.4 B. 19. 2 C. 20.8 D. 22.4
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PROBLEM 2: Given the triangle below, where π = 55, π = 20, and π΄ = 110Β°, the measure of angle βπΆβ is most close to:
A. 20Β° B. 40Β° C. 60Β° D. 80Β°
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PROBLEM 3: !
Given a standard PQR Triangle, the ratio of side βπβ to side βπβ, if sin π = ! , sin π = ! !
, and π = 110Β°, is most close to:
A. 1: 3 B. 2: 3 C. 3: 4 D. 5: 3
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PROBLEM 4: The interior angles of a particular triangle measure at 62Β°, 70Β° and 48Β°. If the length of the the side opposite to the 70Β° angle is 52 cm, then the length of the side opposite the 62Β° angle is most close to: A. 39.2 cm B. 48.9 cm C. 54.6 cm D. 57.3 cm
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PROBLEM 5: Given a general ABC triangle, if Angle A is between sides b and c, and angle π΄ = 50Β°, side b is 4 ft, and side c is 7 ft, the length of side a is most close to: A. 29 ft B. 7 ft C. 5 ft D. 15 ft
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PROBLEM 6: The interior angles of a particular triangle measure at 62Β°, 70Β° and 48Β°. If the length of the the side opposite to the 70Β° angle is 52 cm, then the length of the side opposite the 48Β° angle is most close to: A. 48.9 cm B. 41.1 cm C. 54.6 cm D. 45.7 cm
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LAW OF SINES AND COSINES | SOLUTIONS SOLUTION 1: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Recall that the LAW OF SINES states that for any triangle, the ratio of a side length to the sine of its opposite angle is the same for all three sides and corresponding angles opposite, such that: π π π = = sin π΄ sin π΅ sin πΆ We know both side βπβ and angle βπ΅β, which therefore allows us to define the ratio: 14 = 21.8 sin 40 We are asked to dertermine the length of Side A, but we donβt have any specifically geometrical data defined for it or itβs angle opposite. However, we do know the length of side C, which allows us to continue to make our towards knowing some actionable information.
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Knowing that the length of side C is 12 units, we can write the following relationship: 12 = 21.8 sin πΆ Remember, that the LAW OF SINES relates the sides of triangles to their opposite interior angles, providing for us a ratio that is consistent as we move our way around itβs geometrical layout. Therefore, rearranging this relationship to solve for the interior angle C, we find that:
πΆ = sin!!
12 = 33.3Β° 21.8
We now know the following ANGLES: ANGLE B: 40Β° ANGLE C: 33.3Β° We also know that there are three interior angles within a triangle, and that all of them added up equal 180Β°, which means: ANGLE A = 180Β° β 33.3Β° β 40Β° = 106.7Β° With ANGLE A now defined, we have enough information to deploy the LAW OF SINES once more to determine the length of side βπ":
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π = 21.8 sin 106.7 Rearranging to isolate and solve for the length of side βπ": π = 21.8 sin 106.7 = 20.8 The correct answer choice is C. ππ. π
SOLUTION 2: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given that: π = 55 π΄ = 110Β° π = 20 Since we have a defined side with an associated opposite interior angle, we are able to develop the LAW OF SINES relationship to quickly and seamlessly solve for Angle βπΆβ
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Recall that the LAW OF SINES states: π π π = = sin π΄ sin π΅ sin πΆ We will use this portion of the relationship: π π = sin π΄ sin πΆ Plugging in the given values provided in our problem statement, we have: 54 20 = sin 110 sin πΆ Rearranging we get:
sin πΆ =
20 sin 110 55
And solving for the ANGLE βπΆβ:
πΆ = sin!!
20 sin 110 = 20Β° 55
The correct answer choice is A. ππΒ°
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SOLUTION 3: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. !
To determine the ratio of side βπβ to side βπβ, we only need to know that sin π = ! and !
sin π = . !
With this information we can deploy our knowledge of the LAW OF SINES and determine the set side to opposite interior angle relationship that holds true around the geometry of this particular triangle. The LAW OF SINES states that: π π π = = sin π΄ sin π΅ sin πΆ Our general triangle doesnβt use the variables ABC but PQR, which does nothing to change anything, other than how we display our general formula. We can rewrite the LAW OF SINES to read: π π π = = sin π sin π sin π
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We are only concerned with the ratio involving side βπβ to side βπβ, therefore we will focus in on: π π = sin π sin π Plugging in the information we are given in the problem statement, we find: π π = 1 1 3 4 Which tells us that the ratio of side βπβ to side βπβ is 3:4. The correct answer choice is C. 3:4
SOLUTION 4: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. To determine the unknown side, we will apply the LAW OF SINES by matching each side with the sine of the angle opposite to that side.
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The general formula for the LAW OF SINES reads: π π π = = sin π΄ sin π΅ sin πΆ Letβs say that the information we have defined is for side a and the interior angle opposite of that, ANGLE A, and that we are looking to determine the length of side b. With that, we are given: Side a = 52 ANGLE A = 70Β° ANGLE B = 62Β° Plugging in these values we have: 52 π = sin 70Β° sin 62Β° Rearranging and solving for βπβ we get:
π=
52 sin 62Β° = 48.9 cm sin 70Β°
The correct answer choice is B. ππ. π ππ¦
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SOLUTION 5: The LAW OF COSINES develops a unique relationship given specific geometrical information of a triangle. It states that the square of the length of any one side of a triangle is the sum of the squares of the other two sides minus twice the product of those two sides and the cosine of the angle between them. Come exam day, one common layout we may be presented is an oblique triangle with two sides and the angle between those two sides is defined. This is the exact case for this particular problem, therefore, the LAW OF COSINES will be our go-to tool in determining the other remaining unknown data. In FORMULAIC TERMS, the LAW OF COSINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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It states that given a triangle any triangle:
The LAW OF COSINES can be written as: π! = π ! + π ! β 2ππ COS π΄ π ! = π! + π ! β 2ππ COS π΅ π ! = π! + π ! β 2ππ COS πΆ In the problem statement, we are given: A = 50Β° b=4 c=7 With this information known, we can use the following form of the LAW OF COSINES: π! = π ! + π ! β 2ππ cos (π΄)
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Plugging in the appropriate values, we find: π! = 4! + 7! β 2 4 7 cos 50 = 29 And further solving for the length of side a, we get: π = 5.39 NOTE: Be sure to have your calculator in DEGREE MODE. The correct answer choice is C. π
SOLUTION 6: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. To determine the unknown side, we will apply the LAW OF SINES by matching each side with the sine of the angle opposite to that side. The general formula for the LAW OF SINES reads: π π π = = sin π΄ sin π΅ sin πΆ
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Letβs say that the information we have defined is for side a and the interior angle opposite of that, ANGLE A, and that we are looking to determine the length of side c. With that, we are given: Side a = 52 ANGLE A = 70Β° ANGLE C = 48Β° Plugging in these values we have: 52 π = sin 70Β° sin 48Β° Rearranging and solving for βπβ we get:
π=
52 sin 48Β° = 41.1 ππ sin 70Β°
The correct answer choice is B. ππ. π ππ¦
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PROBLEM 1: Given a general π΄π΅πΆ triangle, side βπβ and angle βπ΅β are opposite from one another. If side βπβ has a length of 14, angle βπ΅β is 40Β°, and side βπβ has a length of 12, then the length of side βπβ is most close to A. 18.4 B. 19. 2 C. 20.8 D. 22.4
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PROBLEM 2: Given the triangle below, where π = 55, π = 20, and π΄ = 110Β°, the measure of angle βπΆβ is most close to:
A. 20Β° B. 40Β° C. 60Β° D. 80Β°
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PROBLEM 3: !
Given a standard PQR Triangle, the ratio of side βπβ to side βπβ, if sin π = ! , sin π = ! !
, and π = 110Β°, is most close to:
A. 1: 3 B. 2: 3 C. 3: 4 D. 5: 3
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PROBLEM 4: The interior angles of a particular triangle measure at 62Β°, 70Β° and 48Β°. If the length of the the side opposite to the 70Β° angle is 52 cm, then the length of the side opposite the 62Β° angle is most close to: A. 39.2 cm B. 48.9 cm C. 54.6 cm D. 57.3 cm
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PROBLEM 5: Given a general ABC triangle, if Angle A is between sides b and c, and angle π΄ = 50Β°, side b is 4 ft, and side c is 7 ft, the length of side a is most close to: A. 29 ft B. 7 ft C. 5 ft D. 15 ft
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PROBLEM 6: The interior angles of a particular triangle measure at 62Β°, 70Β° and 48Β°. If the length of the the side opposite to the 70Β° angle is 52 cm, then the length of the side opposite the 48Β° angle is most close to: A. 48.9 cm B. 41.1 cm C. 54.6 cm D. 45.7 cm
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LAW OF SINES AND COSINES | SOLUTIONS SOLUTION 1: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Recall that the LAW OF SINES states that for any triangle, the ratio of a side length to the sine of its opposite angle is the same for all three sides and corresponding angles opposite, such that: π π π = = sin π΄ sin π΅ sin πΆ We know both side βπβ and angle βπ΅β, which therefore allows us to define the ratio: 14 = 21.8 sin 40 We are asked to dertermine the length of Side A, but we donβt have any specifically geometrical data defined for it or itβs angle opposite. However, we do know the length of side C, which allows us to continue to make our towards knowing some actionable information.
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Knowing that the length of side C is 12 units, we can write the following relationship: 12 = 21.8 sin πΆ Remember, that the LAW OF SINES relates the sides of triangles to their opposite interior angles, providing for us a ratio that is consistent as we move our way around itβs geometrical layout. Therefore, rearranging this relationship to solve for the interior angle C, we find that:
πΆ = sin!!
12 = 33.3Β° 21.8
We now know the following ANGLES: ANGLE B: 40Β° ANGLE C: 33.3Β° We also know that there are three interior angles within a triangle, and that all of them added up equal 180Β°, which means: ANGLE A = 180Β° β 33.3Β° β 40Β° = 106.7Β° With ANGLE A now defined, we have enough information to deploy the LAW OF SINES once more to determine the length of side βπ":
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π = 21.8 sin 106.7 Rearranging to isolate and solve for the length of side βπ": π = 21.8 sin 106.7 = 20.8 The correct answer choice is C. ππ. π
SOLUTION 2: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given that: π = 55 π΄ = 110Β° π = 20 Since we have a defined side with an associated opposite interior angle, we are able to develop the LAW OF SINES relationship to quickly and seamlessly solve for Angle βπΆβ
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Recall that the LAW OF SINES states: π π π = = sin π΄ sin π΅ sin πΆ We will use this portion of the relationship: π π = sin π΄ sin πΆ Plugging in the given values provided in our problem statement, we have: 54 20 = sin 110 sin πΆ Rearranging we get:
sin πΆ =
20 sin 110 55
And solving for the ANGLE βπΆβ:
πΆ = sin!!
20 sin 110 = 20Β° 55
The correct answer choice is A. ππΒ°
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SOLUTION 3: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. !
To determine the ratio of side βπβ to side βπβ, we only need to know that sin π = ! and !
sin π = . !
With this information we can deploy our knowledge of the LAW OF SINES and determine the set side to opposite interior angle relationship that holds true around the geometry of this particular triangle. The LAW OF SINES states that: π π π = = sin π΄ sin π΅ sin πΆ Our general triangle doesnβt use the variables ABC but PQR, which does nothing to change anything, other than how we display our general formula. We can rewrite the LAW OF SINES to read: π π π = = sin π sin π sin π
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We are only concerned with the ratio involving side βπβ to side βπβ, therefore we will focus in on: π π = sin π sin π Plugging in the information we are given in the problem statement, we find: π π = 1 1 3 4 Which tells us that the ratio of side βπβ to side βπβ is 3:4. The correct answer choice is C. 3:4
SOLUTION 4: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. To determine the unknown side, we will apply the LAW OF SINES by matching each side with the sine of the angle opposite to that side.
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The general formula for the LAW OF SINES reads: π π π = = sin π΄ sin π΅ sin πΆ Letβs say that the information we have defined is for side a and the interior angle opposite of that, ANGLE A, and that we are looking to determine the length of side b. With that, we are given: Side a = 52 ANGLE A = 70Β° ANGLE B = 62Β° Plugging in these values we have: 52 π = sin 70Β° sin 62Β° Rearranging and solving for βπβ we get:
π=
52 sin 62Β° = 48.9 cm sin 70Β°
The correct answer choice is B. ππ. π ππ¦
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SOLUTION 5: The LAW OF COSINES develops a unique relationship given specific geometrical information of a triangle. It states that the square of the length of any one side of a triangle is the sum of the squares of the other two sides minus twice the product of those two sides and the cosine of the angle between them. Come exam day, one common layout we may be presented is an oblique triangle with two sides and the angle between those two sides is defined. This is the exact case for this particular problem, therefore, the LAW OF COSINES will be our go-to tool in determining the other remaining unknown data. In FORMULAIC TERMS, the LAW OF COSINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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It states that given a triangle any triangle:
The LAW OF COSINES can be written as: π! = π ! + π ! β 2ππ COS π΄ π ! = π! + π ! β 2ππ COS π΅ π ! = π! + π ! β 2ππ COS πΆ In the problem statement, we are given: A = 50Β° b=4 c=7 With this information known, we can use the following form of the LAW OF COSINES: π! = π ! + π ! β 2ππ cos (π΄)
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Plugging in the appropriate values, we find: π! = 4! + 7! β 2 4 7 cos 50 = 29 And further solving for the length of side a, we get: π = 5.39 NOTE: Be sure to have your calculator in DEGREE MODE. The correct answer choice is C. π
SOLUTION 6: The GENERAL FORMULA for the LAW OF SINES can be referenced under the topic of MATHEMATICS on page 23 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. To determine the unknown side, we will apply the LAW OF SINES by matching each side with the sine of the angle opposite to that side. The general formula for the LAW OF SINES reads: π π π = = sin π΄ sin π΅ sin πΆ
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Letβs say that the information we have defined is for side a and the interior angle opposite of that, ANGLE A, and that we are looking to determine the length of side c. With that, we are given: Side a = 52 ANGLE A = 70Β° ANGLE C = 48Β° Plugging in these values we have: 52 π = sin 70Β° sin 48Β° Rearranging and solving for βπβ we get:
π=
52 sin 48Β° = 41.1 ππ sin 70Β°
The correct answer choice is B. ππ. π ππ¦
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