Problem 31 - Calculus

Question

Draw the base of a triangle from the origin \(O=(0,0)\) to \(P=(a, 0) .\) The third corner is at \(Q=(b \cos \theta, b \sin \theta)\). What are the side lengths \(O P\) and \(O Q\) ? From the distance formula (1) show that the side \(P Q\) has length \(d^{2}=a^{2}+b^{2}-2 a b \cos \theta\) (law of cosines).

Step-by-Step Solution

Answer

The side lengths are \(OP = a\) and \(OQ = b\). Using the distance formula, the side \(PQ\) has length \(\sqrt{a^2 + b^2 - 2ab \cos \theta}\).

Step 1: Understanding the Problem

We have a triangle with vertices at points \(O = (0, 0)\), \(P = (a, 0)\), and \(Q = (b \cos \theta, b \sin \theta)\). We need to determine the side lengths \(OP\) and \(OQ\), and show that the side \(PQ\) has the length given by the law of cosines.

Step 2: Calculate Side Length OP

Since \(O\) and \(P\) lie on the x-axis, the distance \(OP\) is simply the distance along the x-axis from \((0,0)\) to \((a,0)\). Thus, the length \(OP\) is \(a\).

Step 3: Calculate Side Length OQ

Using the distance formula, the distance \(OQ\) is calculated as follows:\[OQ = \sqrt{(b \cos \theta - 0)^2 + (b \sin \theta - 0)^2} = \sqrt{b^2 (\cos^2 \theta + \sin^2 \theta)} = \sqrt{b^2} = b.\]So, the length \(OQ\) is \(b\).

Step 4: Setting up the Distance Formula for PQ

Using the distance formula, the length \(PQ\) is determined by computing the distance between points \(P(a, 0)\) and \(Q(b \cos \theta, b \sin \theta)\). The formula is:\[PQ = \sqrt{(b \cos \theta - a)^2 + (b \sin \theta - 0)^2}.\]

Step 5: Simplifying the Expression for PQ Squared

Substitute the expressions inside the square root from the previous step and simplify:\[PQ^2 = (b \cos \theta - a)^2 + (b \sin \theta)^2.\]Expanding the terms, we get:\[(b \cos \theta - a)^2 + (b \sin \theta)^2 = b^2 \cos^2 \theta - 2ab \cos \theta + a^2 + b^2 \sin^2 \theta.\]Combining \(b^2 \cos^2 \theta\) and \(b^2 \sin^2 \theta\) gives \(b^2 (\cos^2 \theta + \sin^2 \theta) = b^2\). Thus, \[PQ^2 = b^2 - 2ab \cos \theta + a^2.\]

Step 6: Conclusion with Law of Cosines

Thus, we have shown that the squared length \(PQ^2\) satisfies:\[PQ^2 = a^2 + b^2 - 2ab \cos \theta,\]which is the statement of the law of cosines for this specific triangle configuration.

Key Concepts

Understanding the Distance Formula

The distance formula is a staple in geometry and is invaluable when dealing with triangle problems. It allows us to calculate the distance between two points in a coordinate plane. For two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the distance \( d \) between them is given by:

• \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

This formula is derived from the Pythagorean theorem by considering the horizontal and vertical differences between the two points. In the context of our triangle, point \( O \) is the origin \( (0, 0) \), point \( P \) at \( (a, 0) \), and point \( Q \) at \( (b \cos \theta, b \sin \theta) \). The formula helps calculate the exact lengths of sides \( OQ, PQ, \) and \( OP \). In this case, \( OP \) is simply \( a \), and for \( OQ \), it simplifies neatly to \( b \), utilizing trigonometric identities to further simplify calculations.

Exploring Triangle Geometry

Triangles are fundamental shapes in geometry and consist of three sides and three angles. The position and relation of these sides and angles are central to many theorems and formulas, like the Pythagorean theorem and the law of cosines. In our problem, we are working with a triangle defined by its vertices \( O \), \( P \), and \( Q \).

• The special property here is that one side of the triangle is along the x-axis, making calculations simpler.
• This setup makes it easy to apply trigonometric principles as points \( O \) and \( P \) align horizontally.
• The third corner \( Q \) introduces an angle \( \theta \) which affects side lengths and the calculation of angles within the triangle.

Understanding this setup helps students grasp the spatial arrangement and the application of formulas like the distance formula more effectively.

Introducing Trigonometry into Triangle Calculations

Trigonometry deals with the relationships between the sides and angles of triangles. It is especially powerful in solving problems involving right triangles, but also extends to other triangle types through laws such as the sine and cosine rules. In the triangle in question, trigonometry is used to express coordinates of point \( Q \) using \( b \cos \theta \) and \( b \sin \theta \), representing the horizontal and vertical distances respectively.

• Using \( \cos \theta \) and \( \sin \theta \), we can break down complex geometrical problems into more manageable computations.
• This approach allows the step from side calculations to angle relations.

In fact, trigonometric identities like \( \cos^2 \theta + \sin^2 \theta = 1 \) simplify expressions, as seen in simplifying \( OQ \) to \( b \). Overall, trigonometry gives us tools to go beyond simple distance calculations, enabling deeper analysis of triangles.

Demonstrating a Mathematical Proof

In mathematics, a proof serves as a logical argument demonstrating the truth of a statement. In the context of triangle geometry, proofs often validate the relationships between side lengths and angles—such as the law of cosines. For the triangle in the exercise, we start with the distance formula and aim to express \( PQ^2 \). Through a series of algebraic steps, you expand and rearrange terms. These use trigonometric identities like \( \cos^2 \theta + \sin^2 \theta = 1 \), eventually simplifying to reveal:

• \( PQ^2 = a^2 + b^2 - 2ab \cos \theta \)

This is a concrete derivation of the law of cosines for the specific triangle configuration. The effectiveness of mathematical proof lies in its ability to show, without doubt, how and why a particular formula or theorem is true, forming the basis of much of mathematical reasoning. Understanding these proofs helps in applying concepts like the law of cosines correctly in varied contexts.