Problem 32 - Precalculus Mathematics for Calculus

Question

Show that, given the three angles \(A, B, C\) of a triangle and one side, say \(a\), the area of the triangle is $$\text { area }=\frac{a^{2} \sin B \sin C}{2 \sin A}$$

Step-by-Step Solution

Answer

The triangle's area is \(\frac{a^2 \sin B \sin C}{2 \sin A}\).

Step 1: Understanding the Problem

We are given a triangle with angles \(A, B, C\) and a side \(a\). We are to prove the area of the triangle is \(\text{area} = \frac{a^2 \sin B \sin C}{2 \sin A}\).

Step 2: Area of Triangle Formula Using Two Sides and an Angle

The area of any triangle can be calculated using the formula: \( \text{Area} = \frac{1}{2} ab \sin C \), where \(a\) and \(b\) are two sides, and \(C\) is the included angle.

Step 3: Utilizing the Law of Sines

Apply the Law of Sines: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \). This allows us to express sides \(b\) and \(c\) in terms of side \(a\) and the corresponding angles.

Step 4: Express Side b Using Sin B

From the Law of Sines, \( b = a \cdot \frac{\sin B}{\sin A} \). This expression relates side \(b\) to side \(a\).

Step 5: Express Side c Using Sin C

Similarly, \( c = a \cdot \frac{\sin C}{\sin A} \), giving us the relationship between side \(c\) and side \(a\).

Step 6: Substitute b and c in Area Formula

Substitute \(b\) and \(c\) into the formula \( \text{Area} = \frac{1}{2} bc \sin A \):\[ \text{Area} = \frac{1}{2} \left(a \cdot \frac{\sin B}{\sin A}\right) \left(a \cdot \frac{\sin C}{\sin A}\right) \sin A \]

Step 7: Simplification

Simplify the expression:\[ \text{Area} = \frac{1}{2} \cdot \frac{a^2 \sin B \sin C \sin A}{\sin^2 A} \] Thus, \( \text{Area} = \frac{a^2 \sin B \sin C}{2 \sin A} \). This confirms the original equation is correct.

Key Concepts

Law of Sines

The Law of Sines is a fundamental concept in triangle geometry. It relates the lengths of the sides of a triangle to the sines of its angles.
This law is expressed as follows:

• \( \frac{a}{\sin A} \)
• \( \frac{b}{\sin B} \)
• \( \frac{c}{\sin C} \)

All these ratios are equal for any given triangle. This means, no matter the shape of the triangle, these ratios help us find missing sides or angles when we know one complete pair of angle and side.
To apply the Law of Sines, you need at least one side length paired with its opposite angle. This allows us to compute other unknown sides or angles of the triangle, providing an essential tool for triangle geometry.

Triangle Geometry

Triangle geometry involves understanding the properties and relationships within a triangle, a three-sided polygon.
Triangles are defined by three angles and three sides, and there are several types, including:

• Equilateral: all sides and angles are equal
• Isosceles: two sides and the opposite angles are equal
• Scalene: all sides and angles are different

To compute the area of any triangle, you can use the formula: \( \text{Area} = \frac{1}{2}ab\sin C \), where \(a\) and \(b\) are two sides, and \(C\) is the included angle.
Understanding how angles and sides interact is crucial for calculating measures such as area, perimeter, and even more complex properties like the triangle’s center of mass.

Trigonometric Identities

Trigonometric identities are mathematical equations that relate the trigonometric functions \( \sin, \cos, \tan, \) etc., to one another.
These identities help simplify complex trigonometric expressions and solve equations. Some of the fundamental trigonometric identities include:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)
• \( 1 + \tan^2 \theta = \sec^2 \theta \)
• \( \sin (A + B) = \sin A \cos B + \cos A \sin B \)

These identities are particularly useful in transforming and simplifying trigonometric functions, which is an integral part of solving triangle problems.
By utilizing these identities, you can tackle various aspects of triangle geometry, such as confirming properties or deriving equations that relate to triangle dimensions and angles.