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Trigonometry - Area of a triangle (1/2 ab sin C)

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The formula is used to find the area of any triangle when two sides and the included angle (SAS - Side-Angle-Side) are known.

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The included angle is the angle located between the two known side lengths.

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Labeling convention: Side 'a' is opposite angle 'A', side 'b' is opposite angle 'B', and side 'c' is opposite angle 'C'.

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The formula works for acute, obtuse, and right-angled triangles.

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If the triangle is right-angled, sin⁡(90∘)=1\sin(90^\circ) = 1, which reduces the formula to the standard 12×base×height\frac{1}{2} \times \text{base} \times \text{height}.

📐Formulae

Area=12absin⁡C\text{Area} = \frac{1}{2} ab \sin C

Area=12bcsin⁡A\text{Area} = \frac{1}{2} bc \sin A

Area=12acsin⁡B\text{Area} = \frac{1}{2} ac \sin B

💡Examples

Problem 1:

In triangle ABC, side a=8 cma = 8\text{ cm}, side b=11 cmb = 11\text{ cm}, and the included angle C=35∘C = 35^\circ. Calculate the area of the triangle correct to 3 significant figures.

Solution:

Area=12×8×11×sin⁡(35∘)≈25.237...≈25.2 cm2\text{Area} = \frac{1}{2} \times 8 \times 11 \times \sin(35^\circ) \approx 25.237... \approx 25.2\text{ cm}^2

Explanation:

Substitute the known values a=8a=8, b=11b=11, and C=35C=35 directly into the formula 12absin⁡C\frac{1}{2}ab \sin C and evaluate using a calculator.

Problem 2:

The area of a triangle PQR is 40 cm240\text{ cm}^2. Given that PQ=10 cmPQ = 10\text{ cm} and QR=12 cmQR = 12\text{ cm}, find the size of the acute angle QQ.

Solution:

40=12×10×12×sin⁡Q⇒40=60sin⁡Q⇒sin⁡Q=4060=23⇒Q=arcsin⁡(23)≈41.8∘40 = \frac{1}{2} \times 10 \times 12 \times \sin Q \Rightarrow 40 = 60 \sin Q \Rightarrow \sin Q = \frac{40}{60} = \frac{2}{3} \Rightarrow Q = \arcsin(\frac{2}{3}) \approx 41.8^\circ

Explanation:

Rearrange the area formula to solve for the missing angle: sin⁡Q=2×Areap×r\sin Q = \frac{2 \times \text{Area}}{p \times r}.

Problem 3:

Calculate the area of an equilateral triangle with side length 6 cm6\text{ cm}.

Solution:

Area=12×6×6×sin⁡(60∘)=18×32=93≈15.6 cm2\text{Area} = \frac{1}{2} \times 6 \times 6 \times \sin(60^\circ) = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3} \approx 15.6\text{ cm}^2

Explanation:

In an equilateral triangle, all sides are equal (a=b=6a=b=6) and all angles are 60∘60^\circ.