Let ABC Be A Triangle Such That AB=6; AC = 5 And BC=7. 1- Show That: Cos BAC = 1/5. 2- A) Show That: AB · AC = 6. B) Deduce That: BA · BC =

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Triangle geometry is a fundamental branch of mathematics, with real-world applications in fields like engineering, architecture, and navigation. This article explores a detailed solution to a triangle problem, highlighting key concepts and techniques in geometric calculations. Let's delve into the intricacies of triangle ABC, where we are given the lengths of its sides: AB = 6, AC = 5, and BC = 7. Our mission is to unravel the geometric properties of this triangle, starting with determining the cosine of angle BAC and then exploring vector relationships within the triangle. This journey will not only reinforce your understanding of geometric principles but also equip you with the problem-solving skills necessary for tackling complex geometric challenges. We'll break down each step, providing clear explanations and insightful observations, ensuring a comprehensive learning experience. So, let's embark on this exploration and uncover the hidden mathematical beauty within this seemingly simple triangle.

1- Proving cos BAC = 1/5

Our first task is to demonstrate that the cosine of angle BAC is equal to 1/5. To achieve this, we'll employ the Law of Cosines, a cornerstone theorem in trigonometry that relates the sides and angles of a triangle. The Law of Cosines states that for any triangle with sides a, b, and c, and angles A, B, and C opposite to those sides respectively, the following equation holds: a² = b² + c² - 2bc * cos(A). In our case, we have the lengths of all three sides of triangle ABC, and we want to find cos(BAC), which corresponds to cos(A) in the Law of Cosines formula. By strategically applying this law, we can establish the relationship between the sides and the angle BAC, ultimately leading to the desired result. This initial step is crucial as it lays the foundation for further analysis of the triangle's properties. The Law of Cosines not only allows us to calculate angles when side lengths are known but also serves as a bridge connecting the geometric and trigonometric aspects of triangles. Let's proceed with the application of this powerful tool and unveil the value of cos(BAC).

To prove that cos(BAC) = 1/5, we'll utilize the Law of Cosines. In triangle ABC, let's denote BC as 'a', AC as 'b', and AB as 'c'. Thus, a = 7, b = 5, and c = 6. The Law of Cosines, when applied to angle BAC (let's call it angle A), gives us:

  • a² = b² + c² - 2bc * cos(A)

Substituting the given values:

  • 7² = 5² + 6² - 2 * 5 * 6 * cos(A)
  • 49 = 25 + 36 - 60 * cos(A)
  • 49 = 61 - 60 * cos(A)

Now, let's isolate the term with cos(A):

  • 60 * cos(A) = 61 - 49
  • 60 * cos(A) = 12

Finally, we solve for cos(A):

  • cos(A) = 12 / 60
  • cos(A) = 1/5

Therefore, we have successfully shown that cos(BAC) = 1/5. This result confirms the relationship between the sides and the angle BAC in triangle ABC, providing a crucial piece of information for further geometric analysis.

2- Exploring Vector Relationships

Now that we have established the value of cos(BAC), we can delve deeper into the vector relationships within triangle ABC. Vectors provide a powerful tool for representing geometric quantities and relationships, allowing us to analyze the triangle from a different perspective. In this section, we will explore the dot product of vectors AB and AC, and subsequently, we will deduce the dot product of vectors BA and BC. Understanding vector relationships not only enhances our comprehension of triangle geometry but also provides a foundation for more advanced concepts in linear algebra and physics. By examining the interactions between vectors within the triangle, we gain insights into its geometric structure and properties. This exploration will further solidify our understanding of how vectors can be used to describe and analyze geometric figures.

2- a) Proving AB · AC = 6

To demonstrate that the dot product of vectors AB and AC is equal to 6, we will utilize the geometric definition of the dot product. The dot product of two vectors, denoted as A · B, is defined as the product of their magnitudes and the cosine of the angle between them. Mathematically, this is expressed as: A · B = |A| |B| cos(θ), where |A| and |B| represent the magnitudes of vectors A and B, respectively, and θ is the angle between them. In our case, we want to find the dot product of vectors AB and AC. We know the magnitudes of AB and AC (6 and 5, respectively), and we have already determined the cosine of the angle between them (cos(BAC) = 1/5). By applying the geometric definition of the dot product, we can directly calculate AB · AC. This calculation will not only provide us with the numerical value of the dot product but also reinforce our understanding of how the geometric properties of the triangle are reflected in its vector representation.

The dot product of two vectors is a scalar quantity that captures the extent to which the two vectors point in the same direction. A positive dot product indicates that the vectors have a component in the same direction, while a negative dot product suggests that they point in generally opposite directions. A dot product of zero implies that the vectors are orthogonal (perpendicular) to each other. By calculating AB · AC, we are essentially quantifying the alignment between these two vectors within triangle ABC.

To prove that AB · AC = 6, we will use the geometric definition of the dot product:

  • AB · AC = |AB| |AC| cos(BAC)

We are given that |AB| = 6 and |AC| = 5. From the previous part, we know that cos(BAC) = 1/5. Substituting these values into the formula:

  • AB · AC = 6 * 5 * (1/5)
  • AB · AC = 30 * (1/5)
  • AB · AC = 6

Thus, we have successfully shown that the dot product of vectors AB and AC is equal to 6. This result provides valuable information about the relationship between these two vectors within triangle ABC. The positive value of the dot product indicates that AB and AC have a component in the same direction.

2- b) Deducing BA · BC

Having calculated AB · AC, we now turn our attention to deducing the dot product of vectors BA and BC. This step requires us to leverage our understanding of vector properties and how they relate to geometric configurations. Recall that the dot product is commutative, meaning that A · B = B · A. Also, note that the vector BA is simply the negative of vector AB, i.e., BA = -AB. By utilizing these properties and the information we have already gathered about triangle ABC, we can strategically manipulate the dot product expression to arrive at the desired result. This process highlights the interconnectedness of vector operations and their ability to reveal hidden relationships within geometric figures.

The challenge in deducing BA · BC lies in expressing it in terms of quantities we already know. We have calculated AB · AC, and we know the side lengths of the triangle. By carefully applying vector identities and geometric principles, we can bridge the gap between the known and the unknown, ultimately unveiling the value of BA · BC. This exercise not only reinforces our understanding of vector algebra but also demonstrates the power of deductive reasoning in solving geometric problems.

To deduce the value of BA · BC, we can use the following vector identity:

  • BA · BC = (AC - BC) · BC

This identity expresses BA as the difference between AC and BC, allowing us to relate BA · BC to vectors we already have information about. Now, we can expand the dot product:

  • BA · BC = AC · BC - BC · BC

We know that BC · BC is simply the square of the magnitude of BC, which is 7² = 49. So, we have:

  • BA · BC = AC · BC - 49

To find AC · BC, we can use the Law of Cosines again, but this time focusing on angle C. The Law of Cosines gives us:

  • AB² = AC² + BC² - 2(AC)(BC)cos(C)
  • 6² = 5² + 7² - 2(5)(7)cos(C)
  • 36 = 25 + 49 - 70cos(C)
  • 70cos(C) = 38
  • cos(C) = 19/35

Now we can calculate AC · BC:

  • AC · BC = |AC| |BC| cos(C)
  • AC · BC = 5 * 7 * (19/35)
  • AC · BC = 19

Finally, we substitute this value back into our equation for BA · BC:

  • BA · BC = 19 - 49
  • BA · BC = -30

Therefore, we have deduced that BA · BC = -30. This result provides further insights into the vector relationships within triangle ABC. The negative value indicates that vectors BA and BC have a significant component pointing in opposite directions.

In conclusion, this article has provided a comprehensive solution to a triangle geometry problem, demonstrating the application of key concepts such as the Law of Cosines and vector dot products. We successfully proved that cos(BAC) = 1/5, calculated AB · AC = 6, and deduced that BA · BC = -30. These results highlight the interconnectedness of geometric and algebraic principles in solving triangle problems. By understanding and applying these concepts, one can effectively analyze and solve a wide range of geometric challenges. The journey through this problem not only reinforces mathematical skills but also fosters a deeper appreciation for the elegance and power of geometric reasoning.