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Trigonometric ratios of some Specific Angles
  • Last Updated : 23 Mar, 2021

Trigonometry is all about triangles or to more precise about the relation between the angles and sides of a right-angled triangle. In this article we will be discussing about the ratio of sides of a right-angled triangle respect to its acute angle called trigonometric ratios of the angle and find the trigonometric ratios of specific angles: 0°, 30°, 45°, 60°, and 90°.

Consider the following triangle:

The side BA is opposite to angle ∠BCA so we call BA the opposite side to ∠C and AC is the hypotenuse, the other side BC is the adjacent side to ∠C.

Trigonometric Ratios of angle C

Sine: Sine of ∠C is the ratio between BA and AC that is the ratio between the side opposite to C and the hypotenuse.



sin\, C = \frac{BA}{AC}
 
Cosine: Cosine of ∠C is the ratio between BC and AC that is the ratio between the side adjacent to C and the hypotenuse.

cos\, C = \frac{BC}{AC}
 
Tangent: Tangent of ∠C is the ratio between BA and BC that is the ratio between the side opposite and adjacent to C

tan\, C = \frac{BA}{BC}
 
Cosecant: Cosecant of ∠C is the reciprocal of sin C that is the ratio between the hypotenuse and the side opposite to C.

csc\, C = \frac{BA}{AC}
 
Secant: Secant of ∠C is the reciprocal of cos C that is the ratio between the hypotenuse and the side adjacent to C.
sec\, C = \frac{BA}{AC}
 
Cotangent: Cotangent of ∠C is the reciprocal of tan C that is the ratio between the side adjacent to C and side opposite to C.
cot\, C = \frac{BA}{AC}

Finding trigonometric ratios for angle 0°, 30°, 45°, 60°, 90° 

A. For angles 0° and 90°

If an angle A = 0° then the length of the opposite side would be zero and hypotenuse = adjacent side and if A = 90° then the hypotenuse = opposite side. So by using the above formulas for the trigonometric ratios and if the length of the hypotenuse is a.

if A = 0°
 
 \\ sin A = \frac{BC}{AC} = \frac{0}{a} = 0 \\\quad\\ cos A = \frac{AB}{AC} = \frac{a}{a} =1 \\\quad\\ tan A = \frac{BC}{AB} = \frac{a}{a} = 0 \\\quad\\ csc A = \frac{AC}{BC} = not\, defined \\\quad\\ sec A = \frac{AC}{AB} = 1 \\\quad\\ cot A = \frac{AB}{BC} = not\, defined \\\quad\\

if A = 90°
 
 \\ sin A = \frac{AB}{AC} = \frac{0}{a} = 1 \\\quad\\ cos A = \frac{BC}{AC} = \frac{a}{a} = 0 \\\quad\\ tan A = \frac{AB}{BC} = \frac{a}{a} = not\, defined \\\quad\\ csc A = \frac{AC}{AB} = 1 \\\quad\\ sec A = \frac{AC}{BC} = not\, defined \\\quad\\ cot A = \frac{BC}{AB} = 0

Here csc 0, cot 0, tan 90 and sec 90 are not defined as at the particular angle it is divided by 0 which is undefined.



B. For angles 30° and 60°

Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore,

∠A = ∠B = ∠C = 60°.

∆ABD is a right triangle, right-angled at D with ∠BAD = 30° and ∠ABD = 60°, 

Here ∆ADB and ∆ADC are similar as they are Corresponding parts of Congruent triangles(CPCT).

 In\, \Delta ABD\;,AB=a\,,BD=\frac{a}{2} \\and\,AB^2=BD^2+AD^2\\ \quad\\\implies AD^2=AB^2-BD^2 \\ \quad\\\implies AD^2=a^2-(\frac{a}{2})^2\\ \quad\\ \implies AD^2=a^2-\frac{a^2}{4} \\ \quad\\ \implies AD^2=\frac{3a^2}{4} \\\quad\\ \implies AD= \frac{\sqrt{3} a}{2}

Now we know the values of AB, BD, and AD, So the trigonometric ratios for angle 30 are

sin\ 30=\frac{BD}{AB}=\frac{1}{2} \\ \quad\\cos\ 30=\frac{AD}{AB}=\frac{\sqrt{3}}{2} \\ \quad\\tan\ 30=\frac{BD}{AD}=\frac{1}{\sqrt{3}} \\\quad\\sec\ 30=\frac{AB}{BD}=2 \\\quad\\csc\ 30=\frac{AB}{AD}=\frac{2}{\sqrt{3}} \\\quad\\cot\ 30=\frac{AD}{BD}=\sqrt{3}

For angle 60°

sin\ 60=\frac{AD}{AB}=\frac{\sqrt{3}}{2} \\ \quad\\cos\ 60=\frac{BD}{AB}=\frac{1}{2} \\\quad\\tan\ 60=\frac{AD}{BD}=\sqrt{3} \\\quad\\sec\ 60=\frac{AB}{AD}=\frac{2}{\sqrt{3}} \\\quad\\csc\ 60=\frac{AB}{BD}=2 \\\quad\\cot\ 60=\frac{BD}{AD}=\frac{1}{\sqrt{3}}

C. For angle 45°

In a right-angled triangle if one angle is 45° then the other angle is also 45° thus making it an isosceles right angle triangle

If the length of side BC=a then length of AB=a and Length of AC(hypotenuse) is a√2, then

sin\ A = \frac{BC}{AC} = \frac{a}{a√2} = \frac{1}{√2}\\ \quad\\ cos\ A = \frac{AB}{AC} = \frac{a}{a√2} = \frac{1}{√2}\\ \quad\\ tan\ A = \frac{BC}{AB} = \frac{a}{a} = 1\\ \quad\\ csc\ A = \frac{1}{sin\ A} = √2\\ \quad\\ sec\ A = \frac{1}{cos\ A} = √2\\ \quad\\ cot\ A = \frac{1}{tan\ A} = 1\\

Result:

        ∠A         0°        30°        45°        60°        90°
        sin A                  0        1/2        1/√2        √3/2        1
        cos A        1        √3/2                1/√2                1/2                0
        tan A        0        1/√3        1        √3Not defined
        sec A Not defined        2        √2        2/√3        1
        csc A        1        2/√3        √2        2Not defined
        cot ANot defined        √3        1        1/√3        0
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