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Introduction to Trigonometry - Trigonometric Ratios of Specific Angles (0°, 30°, 45°, 60°, 90°)

Grade 10CBSE

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

🔑Concepts

Trigonometric ratios are defined based on the relationship between the angles and the sides of a right-angled triangle. For specific angles like 30,4530^\circ, 45^\circ, and 6060^\circ, these ratios are derived from geometric constructions such as equilateral and isosceles right triangles.

For 4545^\circ, we consider an isosceles right triangle where the base and perpendicular are equal (aa). By Pythagoras theorem, the hypotenuse is a2a\sqrt{2}. This visualizes why sin45\sin 45^\circ and cos45\cos 45^\circ are both equal to 12\frac{1}{\sqrt{2}}.

For 3030^\circ and 6060^\circ, we visualize an equilateral triangle with side 2a2a. When an altitude is drawn, it bisects the base and the vertex angle, creating a 30609030^\circ-60^\circ-90^\circ triangle with sides aa, a3a\sqrt{3}, and 2a2a. This explains why sin30=12\sin 30^\circ = \frac{1}{2} and cos30=32\cos 30^\circ = \frac{\sqrt{3}}{2}.

At 00^\circ, we imagine the angle at the base of a right triangle shrinking. As the angle approaches 00^\circ, the length of the perpendicular (opposite side) becomes 00, and the hypotenuse overlaps the base. Thus, sin0=0\sin 0^\circ = 0 and cos0=1\cos 0^\circ = 1.

At 9090^\circ, we imagine the angle at the base increasing. As it approaches 9090^\circ, the base (adjacent side) becomes 00, and the hypotenuse overlaps the perpendicular. Thus, sin90=1\sin 90^\circ = 1 and cos90=0\cos 90^\circ = 0.

The tanθ\tan \theta ratio for these specific angles is always the ratio of sinθ\sin \theta to cosθ\cos \theta. For instance, at 9090^\circ, since cos90=0\cos 90^\circ = 0, the value of tan90\tan 90^\circ is undefined (1/01/0).

The reciprocal ratios (cosec\text{cosec}, sec\sec, cot\cot) are derived by inverting the values of sin\sin, cos\cos, and tan\tan respectively. If a ratio is 00, its reciprocal is undefined; if a ratio is undefined, its reciprocal is 00.

📐Formulae

sin0=0,sin30=12,sin45=12,sin60=32,sin90=1\sin 0^\circ = 0, \sin 30^\circ = \frac{1}{2}, \sin 45^\circ = \frac{1}{\sqrt{2}}, \sin 60^\circ = \frac{\sqrt{3}}{2}, \sin 90^\circ = 1

cos0=1,cos30=32,cos45=12,cos60=12,cos90=0\cos 0^\circ = 1, \cos 30^\circ = \frac{\sqrt{3}}{2}, \cos 45^\circ = \frac{1}{\sqrt{2}}, \cos 60^\circ = \frac{1}{2}, \cos 90^\circ = 0

tan0=0,tan30=13,tan45=1,tan60=3,tan90=Not Defined\tan 0^\circ = 0, \tan 30^\circ = \frac{1}{\sqrt{3}}, \tan 45^\circ = 1, \tan 60^\circ = \sqrt{3}, \tan 90^\circ = \text{Not Defined}

cosec θ=1sinθ\text{cosec } \theta = \frac{1}{\sin \theta}

secθ=1cosθ\sec \theta = \frac{1}{\cos \theta}

cotθ=1tanθ=cosθsinθ\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}

💡Examples

Problem 1:

Evaluate: sin60cos30+sin30cos60\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ

Solution:

Step 1: Substitute the values of trigonometric ratios. sin60=32\sin 60^\circ = \frac{\sqrt{3}}{2} cos30=32\cos 30^\circ = \frac{\sqrt{3}}{2} sin30=12\sin 30^\circ = \frac{1}{2} cos60=12\cos 60^\circ = \frac{1}{2}

Step 2: Place them into the expression: =(32×32)+(12×12)= (\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}) + (\frac{1}{2} \times \frac{1}{2})

Step 3: Simplify the terms: =34+14= \frac{3}{4} + \frac{1}{4}

Step 4: Add the fractions: =3+14=44=1= \frac{3+1}{4} = \frac{4}{4} = 1

Explanation:

This problem uses the standard values of trigonometric ratios for 3030^\circ and 6060^\circ. It also demonstrates the identity sin(A+B)=sinAcosB+cosAsinB\sin(A+B) = \sin A \cos B + \cos A \sin B where A=60A=60^\circ and B=30B=30^\circ.

Problem 2:

In ABC\triangle ABC, right-angled at BB, AB=5AB = 5 cm and ACB=30\angle ACB = 30^\circ. Determine the length of the side BCBC.

Solution:

Step 1: Identify the given information and the required side. Given: Perpendicular (ABAB) = 55 cm, θ=30\theta = 30^\circ. To find: Base (BCBC).

Step 2: Choose the trigonometric ratio that relates Perpendicular and Base. tanθ=PerpendicularBase\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} tan30=ABBC\tan 30^\circ = \frac{AB}{BC}

Step 3: Substitute the known values. 13=5BC\frac{1}{\sqrt{3}} = \frac{5}{BC}

Step 4: Solve for BCBC. BC=53BC = 5\sqrt{3} cm

Explanation:

To find a missing side when an angle and one side are given, identify which trigonometric ratio (sin, cos, or tan) connects the given side and the side to be found. Here, tangent is used because we are dealing with the opposite (perpendicular) and adjacent (base) sides.