Problem 79 - Precalculus

Question

Show that the area for an SAA triangle is given by $$ A=\frac{a^{2} \sin \beta \sin \gamma}{2 \sin \alpha} $$ Assume that \(\alpha, \beta,\) and \(a\) are given.

Step-by-Step Solution

Answer

We've shown the formula is correct by simplifying the expression using the sine rule.

Step 1: Understand the Given Formula

We need to prove that the area \(A\) of an SAA (Side-Angle-Angle) triangle is \(A=\frac{a^{2} \sin \beta \sin \gamma}{2 \sin \alpha}\), where \(a\) is the given side and \(\alpha, \beta, \gamma\) are the angles. In a triangle, the sum of the angles is always \(180^\circ\). Thus, \(\gamma = 180^\circ - \alpha - \beta\).

Step 2: Use the Sine Rule

The sine rule states that \(\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}\). From this, we can express side \(b\) as \(b = \frac{a \sin \beta}{\sin \alpha}\), and side \(c\) as \(c = \frac{a \sin \gamma}{\sin \alpha}\).

Step 3: Derive the Area Using Known Formulas

The area of a triangle can also be calculated using the formula \(A = \frac{1}{2}bc\sin\alpha\). Substitute \(b\) and \(c\) from Step 2 into this formula: \[A = \frac{1}{2} \cdot \frac{a \sin \beta}{\sin \alpha} \cdot \frac{a \sin \gamma}{\sin \alpha} \cdot \sin \alpha\]

Step 4: Simplify the Expression

Simplify the expression from Step 3:\[A = \frac{a^2 \sin \beta \sin \gamma \sin \alpha}{2 \sin^2 \alpha}\]Canceling \(\sin \alpha\) from the numerator and denominator gives:\[A = \frac{a^2 \sin \beta \sin \gamma}{2 \sin \alpha}\]This proves the given formula.

Key Concepts

Sine Rule

The Sine Rule is a very useful tool in trigonometry, especially when dealing with triangles where two angles and a side are known, such as an SAA triangle. The rule states that for any given triangle, the ratio of each side length to the sine of its opposite angle is constant. This can be written as:

• \( \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} \)

Here, \(a, b, \) and \(c\) are the sides of the triangle, and \(\alpha, \beta,\) and \(\gamma\) are the opposite angles. The Sine Rule helps us find unknown sides or angles. For example, given side \(a\) and angle \(\alpha\), along with angle \(\beta\), we can find side \(b\) as:

• \(b = \frac{a \sin \beta}{\sin \alpha} \)

This makes the Sine Rule extremely practical for calculating the sides of triangles when angles are known.

Angle Sum in a Triangle

In any triangle, the sum of the angles is always \(180^\circ\). This is a fundamental property of triangles and it helps us find a missing angle when the other two are known. For an SAA triangle, knowing two angles allows us to calculate the third angle easily:

• \( \gamma = 180^\circ - \alpha - \beta \)

This foundational principle is essential in triangle problems, as it provides the third angle necessary to apply the Sine Rule or other trigonometric laws. Understanding this can simplify complicated trigonometric calculations, converting them into much simpler arithmetic.

Triangle Area Formula Derivation

The area of a triangle can be calculated through various means, and for an SAA triangle, one effective method involves using trigonometry. Knowing side \(a\) and angles \(\alpha, \beta\) allows us to derive an expression for the area \(A\) using known sides and angles. The general formula for the area using two sides and the sine of the included angle is:

• \( A = \frac{1}{2}bc\sin\alpha \)

Given the expressions for sides \(b\) and \(c\) from the Sine Rule:

• \( b = \frac{a \sin \beta}{\sin \alpha} \)
• \( c = \frac{a \sin \gamma}{\sin \alpha} \)

We substitute these into the area formula:

• \( A = \frac{1}{2} \cdot \frac{a \sin \beta}{\sin \alpha} \cdot \frac{a \sin \gamma}{\sin \alpha} \cdot \sin \alpha \)
• \( A = \frac{a^2 \sin \beta \sin \gamma \sin \alpha}{2 \sin^2 \alpha} \)

Canceling out \(\sin \alpha\) gives:

• \( A = \frac{a^2 \sin \beta \sin \gamma}{2 \sin \alpha} \)

This establishes the specific area formula for an SAA triangle, using the intuitive approach of leveraging known sides and trigonometric relationships.