This document contains solutions to multiple problems involving determining member forces in truss structures using the method of joints. The problems involve setting up force equilibrium equations at each joint to calculate the forces in each member. For every problem, the document shows the free body diagram of the entire truss and/or individual joints, writes the corresponding equilibrium equations, and solves for the member forces stating whether each member is in tension or compression.
solution manual Vector Mechanics for Engineers:Statics Beer Johnston Mazurek ...MichaelLeigh25
The document contains 18 practice problems involving determining the magnitude and direction of the resultant force of two or more applied forces using trigonometric methods like the law of sines and cosines. The problems involve forces applied to structures like hooks, brackets, stakes pulled from the ground, and tugboats pulling barges. The student must use trigonometry to calculate values like the unknown force magnitude, angle between forces, or magnitude of the resultant force.
This document provides information about trigonometric functions including:
- The objectives are to convert between degrees and radians, recognize trigonometric identities, and solve trigonometric equations.
- Trigonometry has a long history dating back to ancient civilizations for measuring distances and heights. It is now widely used in fields like astronomy.
- It discusses angles, the unit circle, trigonometric ratios, special angle values, identities, conversions between degrees and radians, and solving trigonometric equations.
This document contains 21 problems solving for the moment of inertia of various shapes. The shapes include rectangles, triangles, semicircles, and composite shapes. For each problem, the relevant dimensions are given, a calculation is shown, and the numerical value of the moment of inertia about the specified axis is provided. Formulas for the moment of inertia of common shapes like rectangles and triangles are used.
This document discusses key concepts about circles, including:
- The standard equation of a circle is (x - h)2 + (y - k)2 = r2, where (h, k) are the coordinates of the center and r is the radius.
- Given the equation or properties of a circle, one can determine its center and radius or write the equation in standard form.
- Points can lie inside, outside, or on a circle, which can be determined by comparing distances or substituting into the equation.
- A circle and line can intersect in 0, 1, or 2 points, which can be found using algebraic techniques.
- The equation of a circle can be found given 3
Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.
This document contains 20 multi-part engineering problems involving the calculation of shear and moment diagrams for beams and shafts. The problems include beams under various loading conditions such as point loads, distributed loads, overhanging sections, and compound sections. Shear and moment diagrams are drawn and the shear and moment values are calculated as functions of position along the members.
This document contains solutions to multiple problems involving determining member forces in truss structures using the method of joints. The problems involve setting up force equilibrium equations at each joint to calculate the forces in each member. For every problem, the document shows the free body diagram of the entire truss and/or individual joints, writes the corresponding equilibrium equations, and solves for the member forces stating whether each member is in tension or compression.
solution manual Vector Mechanics for Engineers:Statics Beer Johnston Mazurek ...MichaelLeigh25
The document contains 18 practice problems involving determining the magnitude and direction of the resultant force of two or more applied forces using trigonometric methods like the law of sines and cosines. The problems involve forces applied to structures like hooks, brackets, stakes pulled from the ground, and tugboats pulling barges. The student must use trigonometry to calculate values like the unknown force magnitude, angle between forces, or magnitude of the resultant force.
This document provides information about trigonometric functions including:
- The objectives are to convert between degrees and radians, recognize trigonometric identities, and solve trigonometric equations.
- Trigonometry has a long history dating back to ancient civilizations for measuring distances and heights. It is now widely used in fields like astronomy.
- It discusses angles, the unit circle, trigonometric ratios, special angle values, identities, conversions between degrees and radians, and solving trigonometric equations.
This document contains 21 problems solving for the moment of inertia of various shapes. The shapes include rectangles, triangles, semicircles, and composite shapes. For each problem, the relevant dimensions are given, a calculation is shown, and the numerical value of the moment of inertia about the specified axis is provided. Formulas for the moment of inertia of common shapes like rectangles and triangles are used.
This document discusses key concepts about circles, including:
- The standard equation of a circle is (x - h)2 + (y - k)2 = r2, where (h, k) are the coordinates of the center and r is the radius.
- Given the equation or properties of a circle, one can determine its center and radius or write the equation in standard form.
- Points can lie inside, outside, or on a circle, which can be determined by comparing distances or substituting into the equation.
- A circle and line can intersect in 0, 1, or 2 points, which can be found using algebraic techniques.
- The equation of a circle can be found given 3
Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.
This document contains 20 multi-part engineering problems involving the calculation of shear and moment diagrams for beams and shafts. The problems include beams under various loading conditions such as point loads, distributed loads, overhanging sections, and compound sections. Shear and moment diagrams are drawn and the shear and moment values are calculated as functions of position along the members.
(1) The document presents several bending problems involving the determination of stress at various points on beams subjected to bending couples.
(2) Solutions are provided that calculate the stress based on the couple magnitude, beam geometry, and bending axis.
(3) Stresses are determined at points A, B, C, D, and E on different beams and range from -136 MPa to 91.7 MPa depending on the couple magnitude and distance from the beam's neutral axis.
Vector mechanics for engineers statics 7th chapter 5 Nahla Hazem
This problem involves locating the centroid of a plane area shown in multiple problems. The solution provides the area (A) of each section, the x and y coordinates, the moment of area about the x-axis (xA), and the moment of area about the y-axis (yA). It then calculates the x and y coordinates of the centroid by taking the sum of the xA and yA values, respectively, and dividing each by the total area.
The document describes the process of analyzing a truss structure using the method of joints. It provides two examples of solving for the forces in each member of a truss given applied loads. In both examples, the document first calculates the support reactions, then analyzes the force in each member by examining the equilibrium of forces at each joint. It is able to determine the force magnitude and whether each member is in tension or compression.
The document provides information about finding the coordinates of points, plotting pairs of points, calculating the slope of lines between points, and obtaining the equation of a line using the two-point form. It gives examples of finding the slope and equation of lines passing through various pairs of points. It also includes examples for students to practice finding the equations of lines from given points.
The document contains solutions to example problems from a mechanics of materials textbook. Solution 1 resolves a 90 N force into components and calculates the moment MB as -13.02 N⋅m. Solution 2 similarly resolves a force and calculates the moment MB. Solution 3 calculates the moment MA for a 3 lb force as 16.03 lb⋅in.
Aula 30 relacoes mericas no triangulo retangulojatobaesem
Este documento apresenta as relações métricas no triângulo retângulo e como determinar medidas desconhecidas. Ele define os elementos do triângulo retângulo e apresenta quatro relações métricas principais: (1) o quadrado da hipotenusa é igual à soma dos quadrados das catetas, (2) o quadrado da cateta é igual ao produto da outra cateta pela altura, (3) o quadrado da cateta é igual ao produto da hipotenusa pela projeção da cateta, e (4) o quadrado da altura é igual ao
The document contains solutions to problems involving the determination of reaction forces in beams, trusses, and other structures using equations of equilibrium. The solutions involve writing force and moment equilibrium equations and solving the resulting simultaneous equations to determine the unknown reaction forces at supports. Components of the solutions are summarized as follows:
1) Free body diagrams are drawn showing the external forces acting on the structure.
2) Equilibrium equations are written for the forces and moments acting on the structure.
3) The equations are solved to determine values for the unknown reaction forces.
The document discusses the Law of Sines, which can be used to solve oblique triangles (non-right triangles) when given two angles and the side between them (AAS), two sides and the angle opposite one of the sides (SSA), or one angle, one adjacent side, and one opposite side. It provides examples of using the Law of Sines to solve triangles in these cases. It also discusses a special case where the opposite side could form two possible triangles, requiring consideration of both solutions.
This document contains 15 solutions to problems from Chapter 3 of the textbook "Vector Mechanics for Engineers: Statics and Dynamics" by Ferdinand P. Beer et al. Each solution involves resolving forces into components, calculating moments, or determining the minimum force required. Key steps and results are shown for each multi-part problem, with final answers provided in the format of "12.54 N mD = ⋅M".
This document contains solutions to physics problems involving statics. Problem 4.1 calculates the reactions at the rear and front wheels of a tractor lifting a load. Problem 4.2 determines the force needed on each handle of a wheelbarrow transporting a bag of fertilizer. Problem 4.3 finds the maximum distance from the wheelbarrow axle that a second bag can be placed. The remaining problems involve calculating reactions, tensions, and distances for various beams and loads. Free body diagrams and static equilibrium equations are used to solve for the unknown forces and distances in each problem.
This document discusses loci and how to write equations representing loci given certain conditions. It begins by defining a locus as a set of points traced by a moving point based on a specific rule or condition. Examples are then given of writing equations for loci of points equidistant from two given points, finding the midpoint, or a given distance from one point. The key is setting up an equation relating the distances defined by the condition and solving for the relationship between x and y.
Here are the steps to find the centroid of each given plane region:
1. Region bounded by y = 10x - x^2, x-axis, x = 2, x = 5:
- Set up the integral to find the area A: ∫2^5 (10x - x^2) dx
- Evaluate the integral: A = 96
- Set up the integrals to find the x- and y-moments: ∫2^5 x(10x - x^2) dx and ∫2^5 (10x - x^2)x dx
- Evaluate the integrals: Mx = 192, My = 960
- Use the formulas for centroid: C
The document is a chapter from an engineering mechanics textbook covering statics. It provides 11 example problems involving drawing free body diagrams to represent physical systems. The problems include spheres, beams, cranes, rods, and other objects, and require identifying the relevant forces and calculating reactions. Solutions are provided for each problem, with diagrams and step-by-step working. The chapter demonstrates how to set up and solve static equilibrium problems using free body diagrams.
easy step on how to solve slope deflectionAlmasdan Alih
The document describes the displacement method of analysis using slope-deflection equations. It discusses general cases, derivation of stiffness coefficients, fixed-end moments, and analysis of beams and frames. Key steps include:
1) Developing slope-deflection equations relating displacements (slopes and translations) to applied loads and support reactions using beam stiffness properties.
2) Assembling equations into a matrix formulation relating displacement vectors to applied load vectors and fixed-end moment vectors using a stiffness matrix.
3) Solving the matrix equation to determine member displacements and internal forces for given loads.
This document contrasts vector mechanics and variational formulations for structural analysis. It uses the example of a simply supported beam under a uniformly distributed load to illustrate the approaches.
[1] Using vector mechanics, the beam is modeled as discrete elements and equilibrium equations are written for each element and solved. [2] Alternatively, the variational approach introduces a potential energy function for the system and finds the shape that minimizes this function. [3] For the beam example, an approximate solution is first found using a simple shape function, then a more accurate solution is obtained using a shape function with more degrees of freedom that matches the exact solution.
1) Special right triangles have specific angle measurements (30-60-90 or 45-45-90) that result in consistent side length ratios.
2) The Pythagorean theorem, a2 + b2 = c2, always applies to right triangles and relates the sides.
3) Key properties of 30-60-90 and 45-45-90 triangles include specific ratios between short, medium, and long sides that remain consistent regardless of the triangle's size.
The document defines conic sections and describes parabolas. It provides specific objectives related to defining conic sections, identifying different types, describing parabolas, and converting between general and standard forms of parabola equations. It then gives details on the focus, directrix, vertex, latus rectum, and eccentricity of parabolas. Examples of problems involving finding parabola equations and properties from conditions are also provided.
This document defines angles and angle measure in geometry and trigonometry. It explains that an angle is formed by two rays with a common endpoint, and can be measured in degrees from 0 to 360 degrees. The document discusses angle terminology like initial side, terminal side, standard position, coterminal angles, quadrantal angles, and locating angles by quadrant. It provides examples of finding coterminal angles and sketching angles in standard position. Exercises at the end have the reader practice finding coterminal angles, sketching angles, and determining angle locations.
1. The document describes solving physics problems using kinematics, impulse and momentum principles.
2. It involves calculating the speed, time, force or impulse in problems where objects move under the influence of forces or collisions.
3. The solutions show applying equations of motion, drawing free body diagrams, and setting up and solving impulse-momentum equations to determine the requested variables.
(1) The document presents several bending problems involving the determination of stress at various points on beams subjected to bending couples.
(2) Solutions are provided that calculate the stress based on the couple magnitude, beam geometry, and bending axis.
(3) Stresses are determined at points A, B, C, D, and E on different beams and range from -136 MPa to 91.7 MPa depending on the couple magnitude and distance from the beam's neutral axis.
Vector mechanics for engineers statics 7th chapter 5 Nahla Hazem
This problem involves locating the centroid of a plane area shown in multiple problems. The solution provides the area (A) of each section, the x and y coordinates, the moment of area about the x-axis (xA), and the moment of area about the y-axis (yA). It then calculates the x and y coordinates of the centroid by taking the sum of the xA and yA values, respectively, and dividing each by the total area.
The document describes the process of analyzing a truss structure using the method of joints. It provides two examples of solving for the forces in each member of a truss given applied loads. In both examples, the document first calculates the support reactions, then analyzes the force in each member by examining the equilibrium of forces at each joint. It is able to determine the force magnitude and whether each member is in tension or compression.
The document provides information about finding the coordinates of points, plotting pairs of points, calculating the slope of lines between points, and obtaining the equation of a line using the two-point form. It gives examples of finding the slope and equation of lines passing through various pairs of points. It also includes examples for students to practice finding the equations of lines from given points.
The document contains solutions to example problems from a mechanics of materials textbook. Solution 1 resolves a 90 N force into components and calculates the moment MB as -13.02 N⋅m. Solution 2 similarly resolves a force and calculates the moment MB. Solution 3 calculates the moment MA for a 3 lb force as 16.03 lb⋅in.
Aula 30 relacoes mericas no triangulo retangulojatobaesem
Este documento apresenta as relações métricas no triângulo retângulo e como determinar medidas desconhecidas. Ele define os elementos do triângulo retângulo e apresenta quatro relações métricas principais: (1) o quadrado da hipotenusa é igual à soma dos quadrados das catetas, (2) o quadrado da cateta é igual ao produto da outra cateta pela altura, (3) o quadrado da cateta é igual ao produto da hipotenusa pela projeção da cateta, e (4) o quadrado da altura é igual ao
The document contains solutions to problems involving the determination of reaction forces in beams, trusses, and other structures using equations of equilibrium. The solutions involve writing force and moment equilibrium equations and solving the resulting simultaneous equations to determine the unknown reaction forces at supports. Components of the solutions are summarized as follows:
1) Free body diagrams are drawn showing the external forces acting on the structure.
2) Equilibrium equations are written for the forces and moments acting on the structure.
3) The equations are solved to determine values for the unknown reaction forces.
The document discusses the Law of Sines, which can be used to solve oblique triangles (non-right triangles) when given two angles and the side between them (AAS), two sides and the angle opposite one of the sides (SSA), or one angle, one adjacent side, and one opposite side. It provides examples of using the Law of Sines to solve triangles in these cases. It also discusses a special case where the opposite side could form two possible triangles, requiring consideration of both solutions.
This document contains 15 solutions to problems from Chapter 3 of the textbook "Vector Mechanics for Engineers: Statics and Dynamics" by Ferdinand P. Beer et al. Each solution involves resolving forces into components, calculating moments, or determining the minimum force required. Key steps and results are shown for each multi-part problem, with final answers provided in the format of "12.54 N mD = ⋅M".
This document contains solutions to physics problems involving statics. Problem 4.1 calculates the reactions at the rear and front wheels of a tractor lifting a load. Problem 4.2 determines the force needed on each handle of a wheelbarrow transporting a bag of fertilizer. Problem 4.3 finds the maximum distance from the wheelbarrow axle that a second bag can be placed. The remaining problems involve calculating reactions, tensions, and distances for various beams and loads. Free body diagrams and static equilibrium equations are used to solve for the unknown forces and distances in each problem.
This document discusses loci and how to write equations representing loci given certain conditions. It begins by defining a locus as a set of points traced by a moving point based on a specific rule or condition. Examples are then given of writing equations for loci of points equidistant from two given points, finding the midpoint, or a given distance from one point. The key is setting up an equation relating the distances defined by the condition and solving for the relationship between x and y.
Here are the steps to find the centroid of each given plane region:
1. Region bounded by y = 10x - x^2, x-axis, x = 2, x = 5:
- Set up the integral to find the area A: ∫2^5 (10x - x^2) dx
- Evaluate the integral: A = 96
- Set up the integrals to find the x- and y-moments: ∫2^5 x(10x - x^2) dx and ∫2^5 (10x - x^2)x dx
- Evaluate the integrals: Mx = 192, My = 960
- Use the formulas for centroid: C
The document is a chapter from an engineering mechanics textbook covering statics. It provides 11 example problems involving drawing free body diagrams to represent physical systems. The problems include spheres, beams, cranes, rods, and other objects, and require identifying the relevant forces and calculating reactions. Solutions are provided for each problem, with diagrams and step-by-step working. The chapter demonstrates how to set up and solve static equilibrium problems using free body diagrams.
easy step on how to solve slope deflectionAlmasdan Alih
The document describes the displacement method of analysis using slope-deflection equations. It discusses general cases, derivation of stiffness coefficients, fixed-end moments, and analysis of beams and frames. Key steps include:
1) Developing slope-deflection equations relating displacements (slopes and translations) to applied loads and support reactions using beam stiffness properties.
2) Assembling equations into a matrix formulation relating displacement vectors to applied load vectors and fixed-end moment vectors using a stiffness matrix.
3) Solving the matrix equation to determine member displacements and internal forces for given loads.
This document contrasts vector mechanics and variational formulations for structural analysis. It uses the example of a simply supported beam under a uniformly distributed load to illustrate the approaches.
[1] Using vector mechanics, the beam is modeled as discrete elements and equilibrium equations are written for each element and solved. [2] Alternatively, the variational approach introduces a potential energy function for the system and finds the shape that minimizes this function. [3] For the beam example, an approximate solution is first found using a simple shape function, then a more accurate solution is obtained using a shape function with more degrees of freedom that matches the exact solution.
1) Special right triangles have specific angle measurements (30-60-90 or 45-45-90) that result in consistent side length ratios.
2) The Pythagorean theorem, a2 + b2 = c2, always applies to right triangles and relates the sides.
3) Key properties of 30-60-90 and 45-45-90 triangles include specific ratios between short, medium, and long sides that remain consistent regardless of the triangle's size.
The document defines conic sections and describes parabolas. It provides specific objectives related to defining conic sections, identifying different types, describing parabolas, and converting between general and standard forms of parabola equations. It then gives details on the focus, directrix, vertex, latus rectum, and eccentricity of parabolas. Examples of problems involving finding parabola equations and properties from conditions are also provided.
This document defines angles and angle measure in geometry and trigonometry. It explains that an angle is formed by two rays with a common endpoint, and can be measured in degrees from 0 to 360 degrees. The document discusses angle terminology like initial side, terminal side, standard position, coterminal angles, quadrantal angles, and locating angles by quadrant. It provides examples of finding coterminal angles and sketching angles in standard position. Exercises at the end have the reader practice finding coterminal angles, sketching angles, and determining angle locations.
1. The document describes solving physics problems using kinematics, impulse and momentum principles.
2. It involves calculating the speed, time, force or impulse in problems where objects move under the influence of forces or collisions.
3. The solutions show applying equations of motion, drawing free body diagrams, and setting up and solving impulse-momentum equations to determine the requested variables.
4. 單元3 三角函數的應用
3 三角函數的應用
47
2 2 2 2
sin cosa b a b a bθ θ− + ≤ ± ≤ +
極值
正弦定理
2
sin sin sin
a b c
R
A B C
= = =
: : sin : sin : sina b c A B C=
ABC△ 面積 1 1 1
sin sin sin
2 2 2
ab C ac B bc A= = =
和差角公式
正弦 sin( ) sin cos cos sinα β α β α β± = ±
餘弦 cos( ) cos cos sin sinα β α β α β± = ∓
正切 tan tan
tan( )
1 tan tan
α βα β
α β
±
± =
∓
P.48
P.51
P.52
P.53
P.53
餘弦定理
2 2 2
2 cosa b c bc A= + − ×
2 2 2
2 cosb a c ac B= + − ×
2 2 2
2 cosc a b ab C= + − ×
海龍公式 ABC△ 面積 ( )( )( )s s a s b s c= − − − , 2
a b c
s
+ +
= P.54
二倍角公式
正弦 sin 2 2sin cosθ θ θ=
餘弦 2 2
cos2 cos sinθ θ θ= −
正切 2
2tan
tan2
1 tan
θ
θ
θ
=
−
2
cos2 2cos 1θ θ= −
2
cos 2 1 2sinθ θ= −
P.50
9. 單元3 三角函數的應用52
3-2 正弦與餘弦定理
焦點主題1
正弦定理:
在 ABC△ 中,設a 、b 、c 分別表示 A∠ 、 B∠ 、 C∠ 的對邊長,
Δ表示 ABC△ 的面積,且R 表示 ABC△ 的外接圓半徑。
2
sin sin sin
a b c
R
A B C
= = = ⇒ sin sin sina b c A B C=: : : : 。
已 知 ABC△ 中 , 12BC = , 75B∠ = ° ,
60C∠ = °,試求(1) AB (2) ABC△ 外接圓的
半徑。
【答:(1) 6 6 (2) 6 2 】
∵ 180 45A B C∠ = °−∠ −∠ = °
由正弦定理知
2
sin sin
a c
R
A C
= =
⇒ 12
2
sin45 sin60
c
R= =
° °
⇒ 12
2
2 3
2 2
c
R= =
∴ (1) 6 6c AB= =
(2) 6 2R =
已知 ABC△ 中, 2 3BC = , 2AC = ,且
120A∠ = °,試求(1) B∠ (2) AB (3)外接圓
半徑。
【答:(1) 30° (2) 2 (3) 2 】
(1) 由正弦定理知
2
sin sin sin
a b c
R
A B C
= = =
⇒ 2 3 2
2
sin120 sin sin
c
R
B C
= = =
°
⇒ 2 3 2
sin3
2
B
=
∴ 1
sin
2
B = ⇒ 30B∠ = ° 或150°(不合)
(2) 180 30C A B B∠ = °−∠ −∠ = °=∠
得知 ABC△ 為等腰三角形,即 2AB AC= =
(3)
2 3
2
3
2
R= ⇒ 2R =
2
sin sin sin
a b c
R
A B C
= = = 。
1
10. 3
單元3 三角函數的應用 53
已知 ABC△ 中, : : 1 : 2 : 3A B C∠ ∠ ∠ = ,
求 : :a b c 。
【答:1 : 3 : 2 】
∵ 1
180 30
1 2 3
A∠ = °× = °
+ +
2
180 60
1 2 3
B∠ = °× = °
+ +
3
180 90
1 2 3
C∠ = °× = °
+ +
∴ : : sin : sin : sina b c A B C=
1 3
: : 1
2 2
=
1 : 3 : 2=
ABC△ 中, : : 1 : 2 : 1A B C∠ ∠ ∠ = ,
求 : :a b c 。
【答:1 : 2 : 1】
∵ 1
180 45
1 2 1
A C∠ = °× = °=∠
+ +
且 2
180 90
1 2 1
B∠ = °× = °
+ +
∴ : : sin : sin : sina b c A B C=
2 2
: 1 :
2 2
=
2 : 2 : 2=
1 : 2 : 1=
焦點主題2
三角形面積公式:
1 1 1
sin sin sin
2 2 2
a b C a c B b c AΔ = × × × = × × × = × × × 。
ABC△ 中,若 4AB = , 10AC = , 60A∠ = °,
則 ABC△ 面積為何?
【答:10 3 】
1
sin
2
ABC b c AΔ = × × ×
1
10 4 sin60
2
= × × × °
10 3=
ABC△ 中,若 6AB = , 8BC = , 30B∠ = °,
則 ABC△ 面積為何?
【答:12 】
1
sin
2
ABC a c BΔ = × × × 1
8 6 sin30
2
= × × × ° 12=
焦點主題3
餘弦定理:
在 ABC△ 中,若a 、b 、c 分別表示 A∠ 、 B∠ 、 C∠ 的對邊長,則
2 2 2
2 2 2
2 2 2
2 cos
2 cos
2 cos
a b c bc A
b a c ac B
c a b ab C
⎧ = + − ×
⎪
= + − ×⎨
⎪ = + − ×⎩
。
1
sin
2
ABC ac BΔ = (兩邊及其夾角)。
3
2
11. 單元3 三角函數的應用54
ABC△ 中,已知 60A∠ = °, 6AC = , 8AB = ,
試求BC 的長度。
【答:2 13 】
由餘弦定理知
∵ 2 2 2
2 cosa b c bc A= + − ×
⇒ 2 2 2
6 8 2 6 8 cos60a = + − × × × °
36 64 48 52= + − =
∴ 52 2 13BC a= = =
ABC△ 中,已知 120C∠ = ° , 4AC = ,
3BC = ,試求AB 的長度。
【答: 37 】
由餘弦定理知
∵ 2 2 2
2 cosc a b ab C= + − ×
⇒ 2 2 2
3 4 2 3 4 cos120c = + − × × × °
1
9 16 2 3 4 ( )
2
= + − × × × − 37=
∴ 37AB c= =
設 ABC△ 中, 8AB = , 5BC = , 7CA = ,求 B∠
之值。
【答:60° 】
由餘弦定理知
∵ 2 2 2
2 cosb a c ac B= + − ×
⇒ 2 2 2
7 5 8 2 5 8 cosB= + − × × ×
⇒ 49 25 64 80 cosB= + − ×
∴ 25 64 49 1
cos
80 2
B
+ −
= =
⇒ 60B∠ = °
設 ABC△ 的三邊長之比為3 5 7: : ,求最大內
角之值。
【答:120° 】
設三邊長為3k ,5k ,7k ( 0k )
∵ 大邊對大角
∴ 2 2 2
(7 ) (3 ) (5 ) 2 3 5 cosk k k k k θ= + − × × ×
⇒
2 2 2
9 25 49 1
cos
2 3 5 2
k k k
k k
θ
+ −
= = −
× ×
,故最大內角為120°
焦點主題4
海龍公式:
在 ABC△ 中,設a、b 、c 分別表示 A∠ 、 B∠ 、 C∠ 的對邊長,s 表示 ABC△ 周長的一半,
且r 表示 ABC△ 的內切圓半徑。
公式 已知 2
a b c
s
+ +
= ,則 ABC△ 面積 ( )( )( )s s a s b s c−= − − 。
內切圓半徑
已知 2
a b c
s
+ +
= , ABC△ 面積為Δ,
則內切圓半徑r
s
Δ
= 。
2 2 2
2 cosc a b ab C= + − 。
大邊對大角,小邊對小角。
5
4