In the study of trigonometry within the Cambridge IGCSE syllabus, a critical area of focus is the calculation of the area of triangles, not limited to right-angled triangles. This section explores the formula $\frac{1}{2} ab \text{ sin } C$, which is universally applicable to any triangle, providing a powerful tool for understanding and solving problems related to the areas of these geometric shapes.

Introduction to the Area Formula

The formula for calculating the area of a triangle, $\frac{1}{2} ab \sin C$, is not just limited to right-angled triangles but applies to all types of triangles. This formula uses the lengths of two sides of the triangle (a and b) and the sine of the included angle $C$ to find the area.

• a and b are the lengths of any two sides of the triangle.
• C is the angle between sides a and b.
• Area is calculated as $\frac{1}{2} \times a \times b \times \sin C$.

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Importance of the Formula

• Versatility: Works for all types of triangles, whether they are right-angled, obtuse-angled, or acute-angled.
• Application: Essential for solving real-world problems involving land area, architectural design, and in various fields of engineering.

Worked Examples

Example 1: Calculating the Area of an Acute-Angled Triangle

Question: Determine the area of a triangle with side lengths 8 cm and 15 cm, and an included angle of 30 degrees.

Solution:

• Given: $a = 8$ cm, $b = 15$ cm, and $C = 30^\circ$.
• Formula: $\text{Area} = ( \frac{1}{2} \times a \times b \times \sin C )$.

Calculation:

$\text{Area} = \frac{1}{2} \times 8 \times 15 \times \sin(30^\circ)$$\text {Area} = 30 \, \text{cm}^2$

Example 2: Area of an Obtuse Triangle

Question: Find the area of a triangle with sides 7 m and 10 m, with the included angle of 120 degrees.

Solution:

• Given: $a = 7$ m, $b = 10$ m, and $C = 120^\circ$.
• Formula: $\text{Area} = \frac{1}{2} \times a \times b \times \sin C$.

Calculation:

$\text{Area} = \frac{1}{2} \times 7 \times 10 \times \sin(120^\circ) $<p></p>$\text{Area} = 17.5 \sqrt{3} \, \text{m}^2 $<p></p>$\text{Area} \approx 30.31 \, \text{m}^2$<p></p><h2 id="applying-the-formula"><strong>Applying the Formula</strong></h2><h3><strong>Tips for Effective Application</strong></h3><ul><li><strong>Measurements:</strong> Ensure all measurements are in the same unit before using them in the formula.</li><li><strong>Angle Precision:</strong> Use a scientific calculator to find accurate sine values for the angles.</li><li><strong>Practice:</strong> Solve various problems to gain confidence in using the formula across different scenarios.</li></ul><p></p><h2 id="practice-problem">Practice Problem</h2><p><strong>Problem:</strong> Determine the area of a triangle with side lengths of 9 cm and 13 cm, and an included angle of 45 degrees.</p><img src="https://tutorchase-production.s3.eu-west-2.amazonaws.com/e4b3b265-b92b-48e6-857b-c2da17d5c6b8-file.png" alt="Area of a Triangle" style="width: 500px; height: 336px" width="500" height="336"><p></p><h4><strong>Solution:</strong></h4><ul><li>Start by identifying the known values:$a = 9$cm,$b = 13$cm, and$C = 45$degrees.</li></ul><p></p><ul><li>Apply the area formula:</li></ul><p></p>$\text{Area} = \frac{1}{2} \times a \times b \times \sin C$<p></p><ul><li>Solve the area of the triangle:</li></ul><p></p>$\text{Area} = \frac{1}{2} \times 9 \times 13 \times \sin 45^\circ$<p></p>$\text {Area} = \frac{117 \sqrt 2}{4}$<p></p>$\text {Area} \approx 41.37 \text { cm}$