O Level MathematicsE6.3 Area of a triangle (using $\frac{1}{2}ab\sin C$).

🔺 Half-Product & a Sine! Conquering Triangle Areas

Edudent Academy
19 Jan 26

Struggling to find the area of a triangle when the height isn’t obvious? The **sine area formula** comes to the rescue! By using two sides and the sine of their included angle, we can tackle non-right-angled triangles quickly—a vital skill for O Level exams and real-life applications like surveying and design.

Main Concept: The Sine Area Formula

For any triangle with sides aa and bb and included angle CC, the area is
Area=12absin⁡C.\text{Area}=\tfrac12 ab\sin C.
**Why does it work?** Drop a perpendicular from one vertex to create a right-angled triangle; the height becomes bsin⁡Cb\sin C, turning the familiar 12×base×height\tfrac12 \times \text{base} \times \text{height} into this elegant expression.

Worked Example: Surveyor’s Triangle

  • Problem: Two sides of a plot measure 7 cm7\text{ cm} and 10 cm10\text{ cm} with an included angle of 35∘35^\circ. Find the area of the plot.
  • Step 1: Identify variables — here a=7a=7, b=10b=10, C=35∘C=35^\circ.
  • Step 2: Substitute into the formula:
    Area=12×7×10×sin⁡35∘.\text{Area}=\tfrac12 \times 7 \times 10 \times \sin 35^\circ.
  • Step 3: Calculate — sin⁡35∘≈0.574\sin 35^\circ \approx 0.574. Therefore
    Area≈35×0.574=20.1 cm2.\text{Area}\approx 35 \times 0.574 = 20.1\text{ cm}^2.
  • Exam Tip: Always keep your calculator in degree mode and round answers sensibly (3 s.f. unless stated). Practice with different side-angle combinations to build speed!