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Introduction to Trigonometry - Trigonometric Ratios of Some Specific Angles

Grade 10CBSE

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

🔑Concepts

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The values of trigonometric ratios for specific angles (0∘,30∘,45∘,60∘,0^\circ, 30^\circ, 45^\circ, 60^\circ, and 90∘90^\circ) are derived using geometric properties of equilateral triangles and isosceles right-angled triangles.

Isosceles right triangle showing side ratios for 45 degrees
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For an angle of 45∘45^\circ in a right triangle, the base and perpendicular are equal (aa), and the hypotenuse is 2a\sqrt{2}a. Thus, sin⁡45∘=12\sin 45^\circ = \frac{1}{\sqrt{2}}, cos⁡45∘=12\cos 45^\circ = \frac{1}{\sqrt{2}}, and tan⁡45∘=1\tan 45^\circ = 1.

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In a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle (derived from an equilateral triangle), the sides follow the ratio 1:3:21 : \sqrt{3} : 2 for the side opposite 30∘30^\circ, 60∘60^\circ, and the hypotenuse respectively.

30-60-90 degree triangle showing side ratios
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As the angle θ\theta increases from 0∘0^\circ to 90∘90^\circ, the value of sin⁡θ\sin \theta increases from 00 to 11, while the value of cos⁡θ\cos \theta decreases from 11 to 00.

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The ratio tan⁡θ\tan \theta increases from 00 to infinity (undefined at 90∘90^\circ) as θ\theta increases from 0∘0^\circ to 90∘90^\circ.

📐Formulae

0110\begin{array}{r} 01 \\ \hline 10 \end{array}

csc⁡θ=1sin⁡θ,sec⁡θ=1cos⁡θ,cot⁡θ=1tan⁡θ\csc \theta = \frac{1}{\sin \theta}, \sec \theta = \frac{1}{\cos \theta}, \cot \theta = \frac{1}{\tan \theta}

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

💡Examples

Problem 1:

Evaluate the expression: sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ.

Solution:

Substitute the values from the trigonometric table: sin⁡60∘=32,cos⁡30∘=32,sin⁡30∘=12,cos⁡60∘=12\sin 60^\circ = \frac{\sqrt{3}}{2}, \cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 30^\circ = \frac{1}{2}, \cos 60^\circ = \frac{1}{2} Substituting these into the expression: (32×32)+(12×12)\left( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{2} \times \frac{1}{2} \right) =34+14=44=1= \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1

Explanation:

We replace each trigonometric ratio with its specific numerical value for the given angle and then simplify the resulting arithmetic expression.

Problem 2:

If tan⁡(A+B)=3\tan(A + B) = \sqrt{3} and tan⁡(A−B)=13\tan(A - B) = \frac{1}{\sqrt{3}}, where 0∘<A+B≤90∘0^\circ < A + B \leq 90^\circ and A>BA > B, find AA and BB.

Solution:

We know that tan⁡60∘=3\tan 60^\circ = \sqrt{3} and tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}}. Therefore: A+B=60∘— (1)A + B = 60^\circ \quad \text{--- (1)} A−B=30∘— (2)A - B = 30^\circ \quad \text{--- (2)} Adding equation (1) and (2): A+B=60+A−B=302A=90\begin{array}{r} A + B = 60 \\ + A - B = 30 \\ \hline 2A = 90 \end{array} A=90∘2=45∘A = \frac{90^\circ}{2} = 45^\circ Substituting A=45∘A = 45^\circ in (1): 45∘+B=60∘45^\circ + B = 60^\circ B=60∘−45∘=15∘B = 60^\circ - 45^\circ = 15^\circ So, A=45∘A = 45^\circ and B=15∘B = 15^\circ.

Explanation:

By identifying which specific angles correspond to the given tangent values, we form a system of linear equations to solve for the unknown angles AA and BB.

Problem 3:

In △ABC\triangle ABC, right-angled at BB, AB=5 cmAB = 5\text{ cm} and ∠ACB=30∘\angle ACB = 30^\circ. Determine the lengths of the sides BCBC and ACAC.

Triangle ABC with angle C = 30 degrees and side AB = 5cm

Solution:

We are given AB=5 cmAB = 5\text{ cm} and ∠C=30∘\angle C = 30^\circ. To find BCBC (adjacent side): tan⁡30∘=ABBC\tan 30^\circ = \frac{AB}{BC} 13=5BC\frac{1}{\sqrt{3}} = \frac{5}{BC} BC=53 cmBC = 5\sqrt{3}\text{ cm} To find ACAC (hypotenuse): sin⁡30∘=ABAC\sin 30^\circ = \frac{AB}{AC} 12=5AC\frac{1}{2} = \frac{5}{AC} AC=10 cmAC = 10\text{ cm}

Explanation:

By using the trigonometric ratios for 30∘30^\circ, we link the known side (perpendicular) to the unknown sides (base and hypotenuse).

Problem 4:

Evaluate the value of 2tan⁡245∘+cos⁡230∘−sin⁡260∘2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ.

Table of values used in the evaluation

Solution:

Substitute the specific values: tan⁡45∘=1\tan 45^\circ = 1 cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2} sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2} Plugging into the expression: 2(1)2+(32)2−(32)22(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 =2(1)+34−34= 2(1) + \frac{3}{4} - \frac{3}{4} =2= 2

Explanation:

This problem requires substitution of standard trigonometric values and basic arithmetic simplification.