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Trigonometry - Trigonometric Ratios

Grade 9ICSE

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

🔑Concepts

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Definition of Trigonometric Ratios: Trigonometry is the study of the relationship between the sides and angles of a right-angled triangle. In a right-angled triangle ABCABC with ∠B=90∘\angle B = 90^\circ, if we consider ∠C=θ\angle C = \theta, the side ABAB (opposite to θ\theta) is the Perpendicular (pp), BCBC (adjacent to θ\theta) is the Base (bb), and ACAC (opposite to the right angle) is the Hypotenuse (hh).

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Primary Ratios: There are three primary trigonometric ratios: Sine (sin⁡\sin), Cosine (cos⁡\cos), and Tangent (tan⁡\tan). Imagine a vertical line rising from a horizontal base to meet a diagonal slope; sin⁡θ\sin \theta measures the height relative to the slope length, cos⁡θ\cos \theta measures the base relative to the slope length, and tan⁡θ\tan \theta measures the height relative to the base.

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Reciprocal Ratios: Every primary ratio has a reciprocal. Cosecant (csc⁡\csc) is the reciprocal of sin⁡\sin, Secant (sec⁡\sec) is the reciprocal of cos⁡\cos, and Cotangent (cot⁡\cot) is the reciprocal of tan⁡\tan. Visually, if sin⁡θ\sin \theta is ph\frac{p}{h}, then csc⁡θ\csc \theta is the ratio of the hypotenuse to the vertical side, hp\frac{h}{p}.

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Pythagorean Theorem Relation: Since these ratios are based on right-angled triangles, they follow the Pythagorean property p2+b2=h2p^2 + b^2 = h^2. This relationship allows us to derive the fundamental identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, which represents a circle of radius 11 on a coordinate plane.

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Quotient Property: The tangent of an angle can be expressed as the ratio of sine to cosine. Geometrically, this means the slope of the hypotenuse (tangent) is equal to the vertical displacement (sine) divided by the horizontal displacement (cosine).

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Values of Specific Angles: Trigonometric ratios for standard angles like 0∘0^\circ, 30∘30^\circ, 45∘45^\circ, 60∘60^\circ, and 90∘90^\circ are constant. For example, in a 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangle, the perpendicular and base are equal, making tan⁡45∘=1\tan 45^\circ = 1 and sin⁡45∘=12\sin 45^\circ = \frac{1}{\sqrt{2}}.

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Effect of Angle Variation: As the angle θ\theta increases from 0∘0^\circ to 90∘90^\circ, the perpendicular side grows while the base shrinks relative to a fixed hypotenuse. Consequently, sin⁡θ\sin \theta increases from 00 to 11, while cos⁡θ\cos \theta decreases from 11 to 00.

📐Formulae

sin⁡θ=PerpendicularHypotenuse=ph\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{p}{h}

cos⁡θ=BaseHypotenuse=bh\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{b}{h}

tan⁡θ=PerpendicularBase=pb\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{p}{b}

csc⁡θ=1sin⁡θ=hp\csc \theta = \frac{1}{\sin \theta} = \frac{h}{p}

sec⁡θ=1cos⁡θ=hb\sec \theta = \frac{1}{\cos \theta} = \frac{h}{b}

cot⁡θ=1tan⁡θ=bp\cot \theta = \frac{1}{\tan \theta} = \frac{b}{p}

tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}

cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta

1+cot⁡2θ=csc⁡2θ1 + \cot^2 \theta = \csc^2 \theta

💡Examples

Problem 1:

In a right-angled triangle ABCABC, right-angled at BB, if AB=5 cmAB = 5\text{ cm} and BC=12 cmBC = 12\text{ cm}, find the values of sin⁡A\sin A and cos⁡A\cos A.

Solution:

  1. Identify sides relative to ∠A\angle A: Perpendicular (pp) is the side opposite to AA, which is BC=12BC = 12. Base (bb) is the side adjacent to AA, which is AB=5AB = 5.
  2. Find the Hypotenuse (hh) using Pythagoras Theorem: h2=p2+b2h^2 = p^2 + b^2 h2=122+52=144+25=169h^2 = 12^2 + 5^2 = 144 + 25 = 169 h=169=13 cmh = \sqrt{169} = 13\text{ cm}
  3. Calculate sin⁡A\sin A: sin⁡A=ph=1213\sin A = \frac{p}{h} = \frac{12}{13}
  4. Calculate cos⁡A\cos A: cos⁡A=bh=513\cos A = \frac{b}{h} = \frac{5}{13}

Explanation:

To find trigonometric ratios, we first ensure all three sides of the right-angled triangle are known. Using the side lengths relative to the specific angle AA, we apply the standard ratio definitions.

Problem 2:

If sin⁡θ=35\sin \theta = \frac{3}{5}, evaluate the expression 4tan⁡θ−5cos⁡θsec⁡θ+cot⁡θ\frac{4 \tan \theta - 5 \cos \theta}{\sec \theta + \cot \theta}.

Solution:

  1. Given sin⁡θ=35\sin \theta = \frac{3}{5}, let p=3kp = 3k and h=5kh = 5k.
  2. Find bb using b=h2−p2b = \sqrt{h^2 - p^2}: b=(5k)2−(3k)2=25k2−9k2=16k2=4kb = \sqrt{(5k)^2 - (3k)^2} = \sqrt{25k^2 - 9k^2} = \sqrt{16k^2} = 4k
  3. Find required ratios: cos⁡θ=bh=45,tan⁡θ=pb=34,sec⁡θ=hb=54,cot⁡θ=bp=43\cos \theta = \frac{b}{h} = \frac{4}{5}, \quad \tan \theta = \frac{p}{b} = \frac{3}{4}, \quad \sec \theta = \frac{h}{b} = \frac{5}{4}, \quad \cot \theta = \frac{b}{p} = \frac{4}{3}
  4. Substitute values into the expression: Numerator=4(34)−5(45)=3−4=−1\text{Numerator} = 4\left(\frac{3}{4}\right) - 5\left(\frac{4}{5}\right) = 3 - 4 = -1 Denominator=54+43=15+1612=3112\text{Denominator} = \frac{5}{4} + \frac{4}{3} = \frac{15 + 16}{12} = \frac{31}{12}
  5. Result: Value=−13112=−1231\text{Value} = \frac{-1}{\frac{31}{12}} = -\frac{12}{31}

Explanation:

When one ratio is given, we use the Pythagorean theorem to find the missing side (Base in this case). Once all sides are known in terms of a ratio, we calculate the remaining trigonometric functions and substitute them into the given algebraic expression.