Problem 12

Question

If \(B=40^{\circ}, C=70^{\circ}\), and \(c=82 \mathrm{~km}\), find \(A\) and then find \(a\).

Step-by-Step Solution

Verified
Answer
Angle \(A\) is \(70^{\circ}\), and side \(a\) is \(82 \mathrm{~km}\).
1Step 1: Find the Sum of Angles in the Triangle
The sum of the internal angles in any triangle is always 180 degrees. Given that angles \(B\) and \(C\) are \(40^{\circ}\) and \(70^{\circ}\) respectively, we can find angle \(A\) using the equation \(A + B + C = 180^{\circ}\).
2Step 2: Calculate Angle A
Substitute the known angles into the equation:\[A + 40^{\circ} + 70^{\circ} = 180^{\circ}\]Solve for \(A\):\[A = 180^{\circ} - 40^{\circ} - 70^{\circ} = 70^{\circ}\]Therefore, angle \(A\) is \(70^{\circ}\).
3Step 3: Use the Law of Sines to Find Side a
The Law of Sines states:\[\frac{a}{\sin(A)} = \frac{c}{\sin(C)}\]We will use this formula to find \(a\).
4Step 4: Set Up the Equation Using Known Values
Substitute the known values \(A=70^{\circ}\), \(C=70^{\circ}\), and \(c=82 \mathrm{~km}\) into the equation:\[\frac{a}{\sin(70^{\circ})} = \frac{82}{\sin(70^{\circ})}\]Since \(\sin(70^{\circ})\) is the same on both sides, \(a = 82\).
5Step 5: Calculate a
Given that both angles \(A\) and \(C\) are equal and their corresponding sides \(a\) and \(c\) have to be equal:\[a = 82 \mathrm{~km}\]Thus, side \(a\) is \(82 \mathrm{~km}\).

Key Concepts

Triangle Angle SumSolving TrianglesTrigonometric Ratios
Triangle Angle Sum
A triangle is a three-sided polygon that has certain distinct properties. One of the fundamental principles governing triangles is that the sum of their internal angles is always 180 degrees. This is known as the "triangle angle sum". This property is crucial when working with any type of triangle as it allows us to determine an unknown angle given the other two.
For example, if you know two angles, you can easily find the third by subtracting the sum of the known angles from 180.

In the given exercise, you were provided with two angles: \( B = 40^{\circ} \) and \( C = 70^{\circ} \). By applying the triangle angle sum, you can determine the third angle \( A \):
  • Add the given angles: \( 40^{\circ} + 70^{\circ} = 110^{\circ} \)
  • Subtract from 180 to find \( A \): \( A = 180^{\circ} - 110^{\circ} = 70^{\circ} \)
This application not only helps to solve for angle \( A \) in this context but is a valuable tool in a wide variety of geometry problems.
Solving Triangles
Solving a triangle means finding all the unknown angles and side lengths within the triangle. There are several methods to solve triangles, depending on the information given.
Generally, you need to know at least one side length with either two angles (ASA or AAS) or another side length (SSA, SAS, or SSS) to solve it.

In the context of the exercise provided, once you know all three angles, you can move to solving for the unknown side lengths. The known side, \( c = 82 \text{ km} \), along with the known angles \( A \), \( B \), and \( C \), enable the application of a trigonometric approach to find side \( a \). This process not only makes use of angle measurements but efficiently utilizes trigonometric relationships.
Trigonometric Ratios
Trigonometric ratios are critical in connecting angles and side lengths within a triangle. The Law of Sines is an essential trigonometric identity often used when solving triangles. It states that the ratio of the length of a side to the sine of its opposite angle is constant for all three sides of a triangle.
The formula for the Law of Sines is:
  • \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \)
In the provided solution, the exercise utilizes this law to find side \( a \). Given \( \sin(A) = \sin(70^{\circ}) \) and \( \sin(C) = \sin(70^{\circ}) \), and the corresponding side \( c = 82 \text{ km} \), you set up the equation:
  • \( \frac{a}{\sin(70^{\circ})} = \frac{82}{\sin(70^{\circ})} \)
Since \( \sin(70^{\circ}) \) is the same on both sides, it simplifies the calculation, showing that \( a = 82 \text{ km} \). This application highlights the symmetry and usefulness of trigonometric ratios in solving triangles.