Problem 32 - Calculus for Biology and Medicine

Question

A triangle has vertices at coordinates \(P=(0,0), Q=(0,3)\), and \(R=(2,0)\). (a) Use basic trigonometry to compute the lengths of all three sides and the measures of all three angles. (b) Use the results of this section to repeat (a).

Step-by-Step Solution

Answer

Side lengths are 3, \(\sqrt{13}\), and 2. Angles are approximately \(90^\circ\), \(\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)\), and \(180^\circ - 90^\circ - \cos^{-1}\left(\frac{3}{\sqrt{13}}\right)\)."}

Step 1: Calculate Side Lengths

For a triangle with vertices at coordinates, we calculate the side lengths using the distance formula, which is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).\- Length of \(PQ\): The coordinates for \(P\) and \(Q\) are \((0,0)\) and \((0,3)\) respectively. This gives: \\[ PQ = \sqrt{(0-0)^2 + (3-0)^2} = \sqrt{9} = 3 \]- Length of \(QR\): The coordinates for \(Q\) and \(R\) are \((0,3)\) and \((2,0)\) respectively. This gives: \\[ QR = \sqrt{(2-0)^2 + (0-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]- Length of \(PR\): The coordinates for \(P\) and \(R\) are \((0,0)\) and \((2,0)\) respectively. This gives: \\[ PR = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{4} = 2 \]

Step 2: Calculate Angles Using Trigonometry

We'll use the Law of Cosines to find the angles:- For angle \(\angle PQR\): - Use the formula: \( \cos(\angle PQR) = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR} \) - Substituting: \( \cos(\angle PQR) = \frac{9 + 13 - 4}{2 \cdot 3 \cdot \sqrt{13}} = \frac{18}{6\sqrt{13}} = \frac{3}{\sqrt{13}} \) - \( \angle PQR \approx \cos^{-1}\left(\frac{3}{\sqrt{13}}\right) \)- The right angle at \(\angle QPR\) with coordinates being vertical and horizontal gives: \(\angle QPR = 90^\circ\).- Use the angle sum property to find \(\angle PRQ\): - \( \angle PRQ = 180^\circ - 90^\circ - \cos^{-1}\left(\frac{3}{\sqrt{13}}\right) \)

Key Concepts

Distance Formula

Understanding the distance formula is essential when working with the distance between two points on a coordinate plane. The formula is represented as \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). It helps us find the length of any side of a triangle given its endpoints' coordinates.

- To find the distance between points \(P(0,0)\) and \(Q(0,3)\), plug the values into the formula:
\[ PQ = \sqrt{(0-0)^2 + (3-0)^2} = \sqrt{9} = 3 \]
- Similarly, for points \(Q(0,3)\) and \(R(2,0)\):
\[ QR = \sqrt{(2-0)^2 + (0-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]
- Lastly, between \(P(0,0)\) and \(R(2,0)\):
\[ PR = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{4} = 2 \]

This simple formula is powerful in calculating distances, which forms the basis for many other calculations in trigonometry and geometry.

Law of Cosines

The Law of Cosines is a crucial tool when dealing with triangles, especially when you need to find an unknown angle or side. It is particularly useful for calculating angles when you have all three sides of a triangle.

The Law of Cosines states:
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
where \(c\) is the side opposite angle \(C\).

For our triangle with sides \(PQ = 3\), \(QR = \sqrt{13}\), and \(PR = 2\), we can find angle \(\angle PQR\) using:

• \(\cos(\angle PQR) = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR} = \frac{9 + 13 - 4}{2 \cdot 3 \cdot \sqrt{13}} = \frac{3}{\sqrt{13}}\)
• To find the angle itself, solve \(\angle PQR \approx \cos^{-1}\left(\frac{3}{\sqrt{13}}\right)\)

The Law of Cosines effectively allows for calculating any unknown angle in a non-right triangle based on given side lengths.

Triangle Angles

Knowing the angles in a triangle is essential because the sum of all angles in any triangle is always \(180^\circ\). This basic property is useful in finding missing angles.

For a right triangle, one angle is \(90^\circ\). In our case, angle \(\angle QPR\) is a right angle because segment \(PQ\) is vertical, and segment \(PR\) is horizontal, creating a direct intersection.

Once we know \(\angle QPR\) is \(90^\circ\) and we have found \(\angle PQR\) using the Law of Cosines, we use the angle sum property for the third angle:

• \(\angle PRQ = 180^\circ - 90^\circ - \cos^{-1}\left(\frac{3}{\sqrt{13}}\right)\)

This step ensures all angles sum up to \(180^\circ\), verifying our calculations.

Coordination Geometry

Coordination geometry involves expressing geometric relationships using a coordinate system, providing a powerful way to solve geometry problems using algebra and trigonometry.

In triangular problems like this, the coordinates are used to determine distances and angles within the triangle, leveraging:

• The Distance Formula for side lengths
• The Law of Cosines for angles

With vertices \(P(0,0)\), \(Q(0,3)\), and \(R(2,0)\), you see that coordination geometry simplifies complex calculations into manageable formulas.

This technique extends beyond simple triangles, offering a framework to approach various geometric situations with precision and clarity, and is foundational for more advanced studies.