Trigonometric Ratios of Some Specific Angles Class 10th

In Trigonometry, For all the trigonometric ratios we will find their values for some specific angles 0°, 30°, 45°, 60°, 90°.

Trigonometric Ratios of Angle 0°

Let △ABC be a right-angled triangle in which ∠B is the right angle and ∠C is an acute angle ϴ. If the angle ϴ becomes 0° then line segment AC(Hypotenuse) will coincide with line segment BC(Base) and line segment AB(Perpendicular) will become 0. it means AB = 0 and AC = BC.

TRIGONOMETRIC RATIOS OF ANGLE 0°

Thus, the values of trigonometric ratios for angle 0°

sin 0° = AB/AC = 0/AC = 0                    cosec 0° = AC/AB = AC/0 = ∞ (infinitive)

cos 0° = BC/AC = BC/BC = 1                 sec 0° = AC/BC = BC/BC = 1

tan 0° = AB/BC = 0/BC = 0                    cot 0° = BC/AB = BC/0 = ∞ (infinitive)

Trigonometric Ratios of Angles 30° and 60°

Let △ABC be an equilateral triangle with each side 2a. Therefore AB=BC=CA=2a and ∠A=∠B=∠C=60°. AD is perpendicular on side BC from vertex A (AD ⊥ BC).

TRIGONOMETRIC RATIOS OF ANGLES 30° AND 60°

In △ABD and △ACD,

∠ADB = ∠ADC = 90°

AB = AC = 2a

AD = AD (Common side)

By RHS criterion, △ABD ≅ △ACD

Therefore, BD = CD = a and ∠BAD = ∠CAD = 30° (by CPCT)                       

In right-angled triangle ABD

AB = 2a, BD = a

AD2 = AB2 – BD2 (by Pythagoras theorem)                                                      

AD2 = (2a)2 – (a)2 = 4a2 – a2

AD = √3a2 

AD = a√3

Trigonometric ratios of angle 30°

sin 30° = BD/AB = a/2a = 1/2              cosec 30° = AB/BD = 2a/a = 2

cos 30° = AD/AB = a√3/2a = √3/2               sec 30° = AB/AD = 2a/a√3 = 2/√3

tan 30° = BD/AD = a/a√3 = 1/√3                 cot 30° = AD/BD = a√3/a = √3

Now Trigonometric ratios of angle 60°

sin 60° = AD/AB = a√3/2a = √3/2                 cosec 60° = AB/AD = 2a/a√3 = 2/√3

cos 60° = BD/AB = a/2a = 1/2                       sec 60° = AB/BD = 2a/a = 2

tan 60° = AD/BD = a√3/a = √3                      cot 60° = BD/AD = a/a√3 = 1/√3 

Trigonometric Ratios of Angle 45°

Let △ABC be a right-angled triangle. In which ∠B is right angle and ∠C = 45°.

TRIGONOMETRIC RATIOS OF ANGLE 45°

Now ∠A + ∠B + ∠C = 180°

∠A + 90° + 45° = 180°

∠A = 180° – 135°

∠A = 45° 

Since ∠A = ∠C

Therefore, AB = BC (opposite sides of equal angles are equal)

Let AB = BC = a

Then AC2 = AB2 + BC2 (by Pythagoras theorem)

AC2 = (a)2 + (a)2 = a2 + a2

AC = √2a2 = a√2

Now   sin 45° = AB/AC = a/a√2 = 1/√2            cosec 45° = AC/AB = a√2/a = √2           

cos 45° = BC/AC = a/a√2 = 1/√2             sec 45° = AC/BC = a√2/a = √2

tan 45° = AB/BC = a/a = 1                       cot 45° = BC/AB = a/a = 1

Trigonometric Ratios of Angle 90°

Let △ABC be a right-angled triangle in which ∠B is the right angle and ∠C is an acute angle ϴ. If the angle ϴ becomes 90° then line segment AC(Hypotenuse) will coincide with line segment AB(Perpendicular) and line segment BC(Base) will become 0. it means BC = 0 and AC = AB.

TRIGONOMETRIC RATIOS OF ANGLE 90°

Thus, the values of trigonometric ratios for angle 90°

sin 90° = AB/AC = AB/AB = 1                cosec 90° = AC/AB = AB/AB = 1

cos 90° = BC/AC = 0/AC = 0                 sec 90° = AC/BC = AC/0 = ∞ (infinitive)

tan 90° = AB/BC = AB/0 = ∞ (infinitive)               cot 90° = BC/AB = 0/AB = 0

From all the above values if we make a table so all these values will be remembered easily.

Trigonometric Ratios∖Angles (ϴ) (in Degree)➡
30°45°60°90°
sin ϴ01/21/√2√3/21
cos ϴ1√3/21/√21/20
tan ϴ01/√31√3
cosec ϴ2√22/√31
sec ϴ12/√3√22
cot ϴ√311/√30

Steps to Remember the Table Quickly –

1) First of all, we have to write numbers 0, 1, 2, 3, and 4 for the angles 0°, 30°, 45°, 60°, and 90° respectively.

2) After that we shall divide each number by 4 as follows

0/4 = 0 1/4 2/4 = 1/2 3/4 4/4 = 1

3) Now we have the numbers 0, 1/4, 1/2, 3/4, and 1. We shall take the square root of these numbers.

√0 = 0 √1/4 = 1/2 √1/2 = 1/√2 √3/4 = √3/2 √1 = 1

These values are for sin ϴ.

4) By writing these values in the reverse order we shall get the values for cos ϴ.

Reverse order → 1 √3/2 1/√2 1/2 0

5) For tan ϴ, we shall divide the values of sin ϴ by values of cos ϴ because tan ϴ = sin ϴ/cos ϴ.

0/1 = 0 1/2 ∕ √3/2 = 1/√3 1/√2 ∕ 1/√2 = 1 √3/2 ∕ 1/2 = √3 1/0 = (infinite)

6) Now we shall take the reciprocal of the values of sin ϴ to get the values of cosec ϴ because cosec ϴ = 1/sin ϴ.

Reciprocal → 1/0 = (infinite) 1 ∕ 1/2 = 2 1 ∕ 1/√2 = √2 1 ∕ √3/2 = 2/√3 1/1 = 1

7) For sec ϴ, we shall write the values of cosec ϴ in the reverse order.

Reverse order → 1 2/√3 √2 2 (infinite)

8) For cot ϴ, we shall take the reciprocal of the values of tan ϴ because cot ϴ = cos ϴ/sin ϴ.

Reciprocal → 1/0 = (infinite) 1 ∕ 1/√3 = √3 1/1 = 1 1/√3 0/1 = 0

Examples –

Example – 1) find the value of sin 60° cos 30° – sin 30° cos 60°.

Solution – sin 60° cos 30° – sin 30° cos 60°

(√3/2)⨯(√3/2) – (1/2)⨯(1/2) = 3/4 – 1/4 = 3-1/4 = 2/4 = 1/2 Ans. 

Example – 2) find the value of tan2 45° + sin2 30° – cos2 60°

Solution – tan2 45° + sin2 30° – cos2 60°

(1)2 + (1/2)2 – (1/2)2 = 1 + 1/4 – 1/4 = 1 Ans.

Trigonometric Ratios of Some Specific Angles Class 10th in Hindi

More About Trigonometric Ratios of Some Specific Angles

Rate this post

Leave a Comment

Your email address will not be published. Required fields are marked *