Problem 47 - Trigonometry

Question

Bike Frame Geometry Given \(B C=51 \mathrm{~cm}, B D=61 \mathrm{~cm}, C D=78 \mathrm{~cm}, \angle A B C=52^{\circ}\), and \(\angle A C B=65^{\circ}\), find the following. a. The length of the chainstay, \(A C\) b. \(\angle B C D\)

Step-by-Step Solution

Answer

a. AC is approximately 52.24 cm. b. \( \angle BCD \) is approximately 66.77°.

Step 1: Understand the Problem

We are given a triangle with points B, C, and D where the sides and some angles are known. We need to solve for another triangle involving point A and line segments BC, AC, and BD. Our task is to find lengths and angles related to the bike frame geometry problem.

Step 2: Apply the Law of Sines

For triangle ABC, use the Law of Sines: \( \frac{AC}{\sin \angle ACB} = \frac{BC}{\sin \angle BAC} \). First, find \( \angle BAC \) since the angles in a triangle add up to 180 degrees: \( \angle BAC = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 52^\circ - 65^\circ = 63^\circ \). Then solve for AC.

Step 3: Calculate Length of AC

Substitute the known values into the Law of Sines equation: \( \frac{AC}{\sin 65^\circ} = \frac{51}{\sin 63^\circ} \). Solving for AC gives: \( AC = \frac{51 \times \sin 65^\circ}{\sin 63^\circ} \), which approximately equals 52.24 cm.

Step 4: Find Angle BCD with Cosine Rule

For triangle BCD, use the Law of Cosines to find angle \( \angle BCD \): \[ CD^2 = BC^2 + BD^2 - 2 \times BC \times BD \times \cos \angle BCD \]. Plug in the known values: \( 78^2 = 51^2 + 61^2 - 2 \times 51 \times 61 \times \cos \angle BCD \). Solve for \( \cos \angle BCD \) and then find \( \angle BCD \).

Step 5: Solve for Cosine and Angle

Calculate: \( 6084 = 2601 + 3721 - 2 \times 51 \times 61 \times \cos \angle BCD \). Simplifying, find \( \cos \angle BCD = \frac{2601 + 3721 - 6084}{6222} \approx 0.3939 \). Thus, \( \angle BCD = \cos^{-1}(0.3939) \), which is approximately 66.77 degrees.

Key Concepts

Law of Sines

The Law of Sines is a fundamental principle in trigonometry that relates the sides of a triangle to its angles. It is very useful for solving problems in triangle geometry, especially when two angles and a side are known, or two sides and a non-enclosed angle are given.
Here's the key relationship:

• In any triangle, the ratio of the length of a side to the sine of its opposite angle is the same for all three sides.

For triangle ABC: \[ \frac{AC}{\sin \angle ACB} = \frac{BC}{\sin \angle BAC} = \frac{AB}{\sin \angle ABC} \]
In the given problem, to find the length of the chainstay, AC, we use the relationship: \[ \frac{AC}{\sin 65^\circ} = \frac{51}{\sin 63^\circ} \] This step is crucial, especially because it shows how we can directly solve for a missing side when the opposite angle is given. By substituting known values and solving, you determine that the length of AC is approximately 52.24 cm. This reveals the interconnected nature of side lengths and angles in triangles.

Law of Cosines

The Law of Cosines is another essential tool in trigonometry, often used when dealing with non-right triangles. It expresses the relationship between the lengths of a triangle's sides and the cosine of one of its angles. This is particularly helpful when you know two sides and the angle between them or all three sides.For any triangle, the formula is:

• \[ c^2 = a^2 + b^2 - 2ab \times \cos(C) \]

In the problem under discussion, we're tasked with finding the angle BCD in triangle BCD, which is accessed through the Law of Cosines:\[ CD^2 = BC^2 + BD^2 - 2 \times BC \times BD \times \cos \angle BCD \]
Given side lengths, 51 cm, 61 cm, and 78 cm, we plug these into the equation to find \( \angle BCD \). By rearranging and simplifying, \[ 78^2 = 51^2 + 61^2 - 2 \times 51 \times 61 \times \cos \angle BCD \] we isolate \( \cos \angle BCD \) to eventually solve for \( \angle BCD \). Using the cosine inverse, \( \angle BCD \) is approximately 66.77 degrees. This calculation highlights how the Law of Cosines is vital for determining angles when side lengths are mostly known.

Triangle Geometry

Understanding the basics of triangle geometry is a cornerstone for applying trigonometric laws like the Law of Sines and the Law of Cosines effectively. Triangles are two-dimensional shapes with three sides, three angles, and play a crucial role in geometry and trigonometry. Here's a quick overview of fundamental concepts:

• Triangles can be classified by their sides or angles: Equilateral, Isosceles, or Scalene (by sides) and Acute, Right, or Obtuse (by angles).
• The angle sum property states that the sum of the interior angles in a triangle is always 180 degrees, which helps compute unknown angles when two angles are known.
• Understanding side-side-angle (SSA) or side-angle-side (SAS) helps identify which trigonometric rule or approach best suits solving the triangle.

In the context of the exercise, knowing two angles (\( \angle ABC \) and \( \angle ACB \)) allowed us to find the third angle (\( \angle BAC \)). This understanding of triangle geometry was combined with the Law of Sines to solve the unknown side length efficiently. For triangle BCD, the knowledge of side lengths and the application of the Law of Cosines allowed us to find a specific angle, demonstrating the power and necessity of geometric understanding in solving complex problems.