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Trigonometry - Sine, Cosine, and Tangent ratios

Grade 11IGCSE

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

🔑Concepts

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Definition of Sine, Cosine, and Tangent in right-angled triangles using SOH CAH TOA.

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Applying trigonometric ratios to find unknown side lengths and unknown angles.

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The relationship between the sides and angles in non-right-angled triangles (Sine Rule and Cosine Rule).

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Calculating the area of a triangle using the sine ratio.

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Trigonometric ratios for obtuse angles: sin(180-x) = sin x and cos(180-x) = -cos x.

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Solving problems in 3D contexts by identifying right-angled triangles within 3D shapes.

📐Formulae

sin⁡(θ)=OppositeHypotenuse\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}

cos⁡(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}

tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}

asin⁡A=bsin⁡B=csin⁡C (Sine Rule)\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \text{ (Sine Rule)}

a2=b2+c2−2bccos⁡A (Cosine Rule)a^2 = b^2 + c^2 - 2bc \cos A \text{ (Cosine Rule)}

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

💡Examples

Problem 1:

In a right-angled triangle ABC, the hypotenuse AC is 12 cm and angle BAC is 35°. Calculate the length of the side BC.

Solution:

BC = 12×sin⁡(35∘)≈6.8812 \times \sin(35^\circ) \approx 6.88 cm

Explanation:

Identify that BC is the opposite side to the given angle and AC is the hypotenuse. Use the sine ratio: sin⁡(35∘)=BC12\sin(35^\circ) = \frac{BC}{12}. Multiply both sides by 12 to solve for BC.

Problem 2:

In triangle PQR, PQ = 7 cm, QR = 10 cm, and PR = 8 cm. Find the size of angle QPR.

Solution:

cos⁡(P)=72+82−1022×7×8=49+64−100112=13112\cos(P) = \frac{7^2 + 8^2 - 10^2}{2 \times 7 \times 8} = \frac{49 + 64 - 100}{112} = \frac{13}{112}; P=cos⁡−1(13112)≈83.3∘P = \cos^{-1}(\frac{13}{112}) \approx 83.3^\circ

Explanation:

Since all three sides of a non-right triangle are known, use the Cosine Rule rearranged for the angle: cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}.

Problem 3:

Calculate the area of a triangle where two sides are 5 cm and 9 cm, and the included angle is 42°.

Solution:

Area = 12×5×9×sin⁡(42∘)≈15.06\frac{1}{2} \times 5 \times 9 \times \sin(42^\circ) \approx 15.06 cm²

Explanation:

Use the formula for the area of a triangle when two sides and the included angle (SAS) are known: 12absin⁡C\frac{1}{2}ab \sin C.