Problem 63 - Thomas Calculus

Question

A triangle has side \(c=2\) and angles \(A=\pi / 4\) and \(B=\pi / 3\) . Find the length \(a\) of the side opposite \(A .\)

Step-by-Step Solution

Answer

The length of side \(a\) is \(2\sqrt{3} - 2\).

Step 1: Identify the Relationship

We are dealing with a triangle where two angles are given. Use the fact that the sum of angles in a triangle is \(\pi\). Therefore, the remaining angle \(C\) can be found by subtracting \(A\) and \(B\) from \(\pi\).

Step 2: Calculate Angle C

Calculate \(C\) by subtracting \(A\) and \(B\) from \(\pi\) as follows: \[ C = \pi - A - B = \pi - \frac{\pi}{4} - \frac{\pi}{3} \]. Convert fractions over a common denominator to simplify: \[ C = \pi - \frac{3\pi}{12} - \frac{4\pi}{12} = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} \].

Step 3: Apply the Law of Sines

The Law of Sines relates the lengths of the sides of a triangle to the sines of its angles: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]. We need to find \(a\), so we use: \[ \frac{a}{\sin A} = \frac{c}{\sin C} \]. Plug in the values: \[ \frac{a}{\sin \frac{\pi}{4}} = \frac{2}{\sin \frac{5\pi}{12}} \].

Step 4: Solve for Side a

Solve the equation \( \frac{a}{\frac{\sqrt{2}}{2}} = \frac{2}{\sin \frac{5\pi}{12}} \) for \(a\): \[ a = \frac{2 \cdot \frac{\sqrt{2}}{2}}{\sin \frac{5\pi}{12}} \]. Find \( \sin \frac{5\pi}{12} \) using angle subtraction formulas: \[ \sin \frac{5\pi}{12} = \sin (\frac{\pi}{4} + \frac{\pi}{6}) = \sin \frac{\pi}{4} \cdot \cos \frac{\pi}{6} + \cos \frac{\pi}{4} \cdot \sin \frac{\pi}{6} \]. Calculate: \[ \sin \frac{5\pi}{12} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \]. Now substitute back: \[ a = \frac{\sqrt{2} \cdot 2}{\frac{\sqrt{6} + \sqrt{2}}{4}} \]. Simplify this equation: \[ a = \frac{4\sqrt{2}}{\sqrt{6} + \sqrt{2}} \]. To simplify, multiply by the conjugate to rationalize the denominator: \[ a = \frac{4\sqrt{2}(\sqrt{6} - \sqrt{2})}{6 - 2} = \frac{4\sqrt{2}(\sqrt{6} - \sqrt{2})}{4} \]. Solve: \[ a = \sqrt{2}(\sqrt{6} - \sqrt{2}) = \sqrt{12} - \sqrt{4} = 2\sqrt{3} - 2 \].

Step 5: Check Consistency

Verify calculations by revisiting angle relationships and the Law of Sines, ensuring rationale and calculations align. Verify each inverse trigonometric relation used as well.

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics dealing with the relationships between the angles and sides of triangles, particularly right-angled triangles. It introduces functions such as sine, cosine, and tangent, which are crucial in calculating unknown parts of a triangle when certain measurements are known. These trigonometric functions enable us to explore the properties and measurements of non-right triangles as well, like in the example problem.In the given problem, the Law of Sines, a significant rule in trigonometry, is employed. This law states that in any triangle, the ratio of the length of a side to the sine of the angle opposite it is constant.

• Equation: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)
• It applies to any type of triangle—right, acute, or obtuse.
• Allows solving for unknown side lengths or angles when two angles and a side (AAS or ASA) are known.

Understanding trigonometry and these fundamental identities allows for solving complex triangle-based problems, as seen in the solution where side \(a\) was found using known angles and one side.

Triangle Angles

A very important aspect of studying triangles is understanding their angles. Every triangle has three angles, and a fundamental property is that their sum always equals \( \pi \) radians (or \(180^\circ\)). This is known as the Angle Sum Property of triangles.In our problem, two angles, \(A\) and \(B\), are given, and we must find the third angle, \(C\). To find \(C\):

• Sum of angles formula: \( A + B + C = \pi \)
• Subtract \(A\) and \(B\) from \(\pi\) to find \(C\).
• The calculation involves basic fraction operations to achieve a common denominator.

For our example, organizing the calculations clarifies how \(C\) was reduced to \(\frac{5\pi}{12}\). A solid grasp of angle relationships is essential for correctly solving for unknown elements in triangle problems.

Angle Subtraction Formula

The angle subtraction formula is a powerful tool in trigonometry, used to find the sine, cosine, or tangent of a given angle obtained by subtracting two known angles. It is particularly helpful when the angle is not a special angle, such as \(30^\circ\), \(45^\circ\), or \(60^\circ\).In the problem at hand, we need to find \( \sin \frac{5\pi}{12} \), a non-special angle. We can express this angle as the sum of two other angles whose sine values are known, like \( \sin \left(\frac{\pi}{4} + \frac{\pi}{6}\right)\). Using the sine angle addition formula:

• \( \sin(a + b) = \sin a \cdot \cos b + \cos a \cdot \sin b \)
• Substituting \(a = \frac{\pi}{4}\) and \(b = \frac{\pi}{6}\), resolve the equation step-by-step.
• Calculate each sine and cosine value based on known angles and substitute into the formula.

The breakdown calculated \( \sin \frac{5\pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4} \), thereby assisting in solving for side \(a\). Thorough knowledge of such trigonometric identities enhances problem-solving efficiency and accuracy.