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    Current Subject
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    Math Deficiency - I
    MD-001
    Progress0 / 38 topics
    Topics
    1. Sets: Definition, Representation, and Operations2. Relation and Function: Graphical Transformation of Functions3. Properties of Functions4. Composition and Inverses of Functions5. Domain and Range of Functions6. Maximum and Minimum Values of Functions7. Increasing and Decreasing Functions8. Zeros and Intercepts of Functions9. Piecewise Functions10. Continuity and Discontinuity of Functions11. Polynomials and Rational Functions12. Polynomial Long Division and Synthetic Division13. Solution of Rational Functions14. Absolute Valued Functions and Their Properties15. Asymptotes: Horizontal, Vertical, and Oblique16. Exponential Functions and Their Properties17. Logarithmic Functions and Their Properties18. Systems of Equations: Two Equations and Two Unknowns19. Systems of Equations: Three Equations and Three Unknowns20. Matrix Algebra: Addition, Subtraction, and Multiplication21. Row Operations and Row Echelon Forms22. Augmented Matrices23. Determinant of Matrices: 2x2 and Higher Order24. Cramer's Rule25. Inverse Matrices26. Series and Sequences27. Trigonometry: Angles in Radians and Degrees28. Right Triangle Trigonometry29. Law of Cosines and Sines30. Area of a Triangle31. Graphs of Trigonometric Functions32. Graphs of Inverse Trigonometric Functions33. Basic Trigonometric Identities34. Trigonometric Equations35. General Form of a Conic: Parabolas, Circles, Ellipses, and Hyperbolas36. Degenerate Conics37. Polar and Parametric Equations38. Polar and Rectangular Coordinates
    MD-001›Law of Cosines and Sines
    Math Deficiency - ITopic 29 of 38

    Law of Cosines and Sines

    10 minread
    1,658words
    Intermediatelevel

    Law of Cosines and Law of Sines

    The Law of Cosines and Law of Sines are crucial tools in trigonometry, used to solve triangles that are not necessarily right-angled. They relate the angles and sides of a triangle, and they are especially useful for solving oblique triangles (triangles that are not right-angled).

    1. Law of Cosines

    The Law of Cosines relates the sides and angles of a triangle by expressing the length of a side in terms of the other two sides and the cosine of the included angle.

    For any triangle with sides aaa, bbb, and ccc, and corresponding angles AAA, BBB, and CCC, the Law of Cosines states:

    c2=a2+b2−2ab⋅cos⁡(C)c^2 = a^2 + b^2 - 2ab \cdot \cos(C)c2=a2+b2−2ab⋅cos(C) b2=a2+c2−2ac⋅cos⁡(B)b^2 = a^2 + c^2 - 2ac \cdot \cos(B)b2=a2+c2−2ac⋅cos(B) a2=b2+c2−2bc⋅cos⁡(A)a^2 = b^2 + c^2 - 2bc \cdot \cos(A)a2=b2+c2−2bc⋅cos(A)

    Where:

    • aaa, bbb, and ccc are the lengths of the sides of the triangle.
    • AAA, BBB, and CCC are the angles opposite to the sides aaa, bbb, and ccc, respectively.

    Uses of the Law of Cosines:

    • Finding a side when two sides and the included angle are known.
    • Finding an angle when all three sides are known (using the inverse cosine function).

    Example 1: Finding the third side of a triangle.

    Given a triangle with sides a=5a = 5a=5, b=7b = 7b=7, and the included angle C=60∘C = 60^\circC=60∘, find side ccc.

    Using the Law of Cosines formula:

    c2=a2+b2−2ab⋅cos⁡(C)c^2 = a^2 + b^2 - 2ab \cdot \cos(C)c2=a2+b2−2ab⋅cos(C)

    Substitute the known values:

    c2=52+72−2(5)(7)⋅cos⁡(60∘)c^2 = 5^2 + 7^2 - 2(5)(7) \cdot \cos(60^\circ)c2=52+72−2(5)(7)⋅cos(60∘)

    Since cos⁡(60∘)=0.5\cos(60^\circ) = 0.5cos(60∘)=0.5, we have:

    c2=25+49−70×0.5c^2 = 25 + 49 - 70 \times 0.5c2=25+49−70×0.5 c2=25+49−35c^2 = 25 + 49 - 35c2=25+49−35 c2=39c^2 = 39c2=39 c=39≈6.24c = \sqrt{39} \approx 6.24c=39​≈6.24

    So, side c≈6.24c \approx 6.24c≈6.24.


    2. Law of Sines

    The Law of Sines relates the sides of a triangle to the sines of its angles. It is used to solve triangles when two angles and one side are known, or two sides and a non-included angle are known.

    The Law of Sines states that:

    asin⁡(A)=bsin⁡(B)=csin⁡(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}sin(A)a​=sin(B)b​=sin(C)c​

    Where:

    • aaa, bbb, and ccc are the sides of the triangle.
    • AAA, BBB, and CCC are the angles opposite those sides.

    Uses of the Law of Sines:

    • Finding an angle when two sides and an angle (non-included) are known (called the SSA case).
    • Finding a side when two angles and one side are known.

    Example 2: Finding an angle when two sides and an opposite angle are known.

    Given a triangle with sides a=8a = 8a=8, b=10b = 10b=10, and angle A=30∘A = 30^\circA=30∘, find angle BBB.

    Using the Law of Sines:

    asin⁡(A)=bsin⁡(B)\frac{a}{\sin(A)} = \frac{b}{\sin(B)}sin(A)a​=sin(B)b​

    Substitute the known values:

    8sin⁡(30∘)=10sin⁡(B)\frac{8}{\sin(30^\circ)} = \frac{10}{\sin(B)}sin(30∘)8​=sin(B)10​

    Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5sin(30∘)=0.5, we have:

    80.5=10sin⁡(B)\frac{8}{0.5} = \frac{10}{\sin(B)}0.58​=sin(B)10​ 16=10sin⁡(B)16 = \frac{10}{\sin(B)}16=sin(B)10​

    Solve for sin⁡(B)\sin(B)sin(B):

    sin⁡(B)=1016=0.625\sin(B) = \frac{10}{16} = 0.625sin(B)=1610​=0.625

    Now, take the inverse sine:

    B=sin⁡−1(0.625)≈38.68∘B = \sin^{-1}(0.625) \approx 38.68^\circB=sin−1(0.625)≈38.68∘

    So, angle B≈38.68∘B \approx 38.68^\circB≈38.68∘.


    3. Ambiguous Case (SSA Case)

    The Ambiguous Case occurs when you are given two sides and a non-included angle (SSA case). In this case, there may be:

    1. One solution (a unique triangle).
    2. Two solutions (two possible triangles).
    3. No solution (no possible triangle).

    For example, given side aaa, side bbb, and angle AAA, the Law of Sines might result in two possible values for angle BBB. Depending on the situation, this could lead to two possible triangles.

    Example of the Ambiguous Case: Given a=8a = 8a=8, b=10b = 10b=10, and A=30∘A = 30^\circA=30∘, let's find angle BBB. Using the Law of Sines:

    8sin⁡(30∘)=10sin⁡(B)\frac{8}{\sin(30^\circ)} = \frac{10}{\sin(B)}sin(30∘)8​=sin(B)10​ 80.5=10sin⁡(B)\frac{8}{0.5} = \frac{10}{\sin(B)}0.58​=sin(B)10​ 16=10sin⁡(B)16 = \frac{10}{\sin(B)}16=sin(B)10​ sin⁡(B)=1016=0.625\sin(B) = \frac{10}{16} = 0.625sin(B)=1610​=0.625

    So, sin⁡(B)=0.625\sin(B) = 0.625sin(B)=0.625, and taking the inverse sine:

    B=sin⁡−1(0.625)≈38.68∘B = \sin^{-1}(0.625) \approx 38.68^\circB=sin−1(0.625)≈38.68∘

    Now, since B=38.68∘B = 38.68^\circB=38.68∘ is possible, we can also find a second solution by noting that B′=180∘−BB' = 180^\circ - BB′=180∘−B, which gives:

    B′=180∘−38.68∘≈141.32∘B' = 180^\circ - 38.68^\circ \approx 141.32^\circB′=180∘−38.68∘≈141.32∘

    This means that there are two possible values for angle BBB, leading to two possible triangles.


    4. Summary

    • Law of Cosines: Useful when you know two sides and the included angle (SAS) or all three sides (SSS).

      • Formula: c2=a2+b2−2ab⋅cos⁡(C)c^2 = a^2 + b^2 - 2ab \cdot \cos(C)c2=a2+b2−2ab⋅cos(C)
    • Law of Sines: Useful when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA).

      • Formula: asin⁡(A)=bsin⁡(B)=csin⁡(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}sin(A)a​=sin(B)b​=sin(C)c​
    • Ambiguous Case (SSA): When given two sides and a non-included angle, there may be one, two, or no solutions.

    Both laws are powerful tools for solving non-right triangles, and they allow us to find unknown sides or angles in a triangle using only the information about the other sides and angles.

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    Right Triangle Trigonometry
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    Area of a Triangle

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