Notes on Trigonometric Ratios of Some Standard Angles | Grade 9 > Optional Mathematics > Trigonometry | KULLABS.COM

Notes, Exercises, Videos, Tests and Things to Remember on Trigonometric Ratios of Some Standard Angles

Please scroll down to get to the study materials.

• Note
• Things to remember
• Videos
• Exercise
• Quiz

Different angles have a different value with various trigonometric ratios. We shall consider 0°, 30°, 45° and 90° as the standard angles and we shall learn their values here. In this unit, we shall verify the values of 0°, 30°, 45° and 90° using geometrical proofs.

Trigonometrical Ratio of 45°

Let ABC be a right-angledisosceles trianglewhere $$\angle$$B = 90° and $$\angle$$A = $$\angle$$C = 45°

Also let BC = AC = $$\alpha$$.

We know, in right angled $$\triangle$$ABC

h2 = p2 + b2

or, (AC)2 = (AB)2 + (BC)2

or, (AC)2 = $$\alpha$$2 + $$\alpha$$2

or, (AC)2 = 2 $$\alpha$$2

∴ AC = $$\alpha$$$$\sqrt{2}$$

Taking $$\angle$$C as the reference angle, we get,

sin 45 = $$\frac{AB}{AC}$$= $$\frac{α}{α\sqrt{2}}$$ = $$\frac{1}{\sqrt{2}}$$

cos 45 = $$\frac{BC}{AC}$$= $$\frac{α}{α\sqrt{2}}$$ = $$\frac{1}{\sqrt{2}}$$

tan 45 = $$\frac{AB}{BC}$$= $$\frac{α}{α}$$ = 1

cosec 45 = $$\frac{AC}{AB}$$= $$\frac{α\sqrt{2}}{α}$$ = $$\sqrt{2}$$

sec 45 = $$\frac{AC}{AB}$$= $$\frac{α\sqrt{2}}{α}$$ =$$\sqrt{2}$$

cot 45 = $$\frac{BC}{AB}$$ = $$\frac{α}{α}$$ = 1

Let ABC be an equilateral triangle where$$\angle$$A = $$\angle$$B = $$\angle$$C = 60°

and AB = BC = CA = 2$$\alpha$$

Now, let's draw AD perpendicular to BC so that,

BD = DC = a and $$\angle$$BAD = $$\angle$$DAC = 30°

Now, in right angled $$\triangle$$ADC,

or, (AD)2 = (AC)2 + (DC)2

= (2α)2 - (α)2

= 4$$\alpha$$2 - $$\alpha$$2 = 3$$\alpha$$2

$$\therefore$$ AD = $$\alpha$$$$\sqrt{3}$$

Now, to find the trigonometric ratio for 60°, lets take $$\angle$$C = 60° as the reference angle, we get,

sin 60° = $$\frac{AD}{AC}$$ = $$\frac{\alpha\sqrt{3}}{2α}$$ = $$\frac{\sqrt{3}}{2}$$

cos 60° = $$\frac{DC}{AC}$$= $$\frac{α}{2α}$$ =$$\frac{1}{2}$$

tan 60° = $$\frac{AD}{DC}$$=$$\frac{\alpha\sqrt{3}}{2α}$$ = $$\sqrt{3}$$

cosec 60° = $$\frac{AC}{AD}$$= $$\frac{2α}{\alpha\sqrt{3}}$$ = $$\frac{2}{\sqrt{3}}$$

sec 60° = $$\frac{AC}{DC}$$= $$\frac{2α}{α}$$ = 2

cot 60° = $$\frac{DC}{AD}$$= $$\frac{α}{\alpha\sqrt{3}}$$ = $$\frac{1}{\sqrt{3}}$$

Now, to find out the trigonometric ratio for 30° , let's take $$\angle$$DAC = 30° as the reference angle, we get,

sin 30° =$$\frac{DC}{AC}$$= $$\frac{α}{2α}$$ =$$\frac{1}{2}$$

cos 30° = $$\frac{AD}{AC}$$ = $$\frac{\alpha\sqrt{3}}{2α}$$ = $$\frac{\sqrt{3}}{2}$$

tan 30° =$$\frac{DC}{AD}$$= $$\frac{α}{\alpha\sqrt{3}}$$ = $$\frac{1}{\sqrt{3}}$$

cosec 30° =$$\frac{AC}{DC}$$= $$\frac{2α}{α}$$ = 2

sec 30° = $$\frac{AC}{AD}$$= $$\frac{2α}{\alpha\sqrt{3}}$$ = $$\frac{2}{\sqrt{3}}$$

cot 30° =$$\frac{AD}{DC}$$=$$\frac{\alpha\sqrt{3}}{2α}$$ = $$\sqrt{3}$$

Trigonometrical Ratio of 0°

Let ABC be a right angled triangle where $$\angle$$B = 90° and $$\angle$$C = $$\theta$$.

If $$\theta$$ tends to 0°

i.e $$\theta$$→0°, AC coincides with BC.

or, AC $$\approx$$BC

so, AC= BC = a (say)

Now, using pythagoras theorem

h2= p2+ b2

or, (AC)2=(AB)2+ (BC)2

or, a2 = (AB)2 + a2

or, (AB)2 = a2 - a2 = 0

$$\therefore$$ AB = 0

Now, taking right angled $$\triangle$$ABC

sin 0° = $$\frac{p}{h}$$ =$$\frac{AB}{AC}$$ =$$\frac{0}{a}$$ = 0

cos 0° =$$\frac{b}{h}$$ = $$\frac{BC}{AC}$$ = $$\frac{a}{a}$$ = 1

tan 0° =$$\frac{p}{b}$$ = $$\frac{AB}{BC}$$ = $$\frac{0}{a}$$ = 0

cosec 0° =$$\frac{h}{p}$$ = $$\frac{AC}{AB}$$ = $$\frac{a}{0}$$ = $$\infty$$

sec 0° =$$\frac{h}{b}$$ = $$\frac{AC}{BC}$$ = $$\frac{a}{a}$$ = 1

cot 0° =$$\frac{b}{p}$$ = $$\frac{BC}{AB}$$ = $$\frac{a}{0}$$ = $$\infty$$

Trigonometric Ratio of 90°

Let ABC be a right angled triangle where $$\angle$$B = 90° and $$\angle$$C = $$\theta$$ be the reference angle.

If $$\theta$$ tends to 90°

i.e $$\theta$$→90°, AC coincides with BC.

or, AC $$\approx$$AB

so, AC= AB = a (say)

Now, using pythagoras theorem

h2= p2+ b2

or, (AC)2=(AB)2+ (BC)2

or, a2 = (BC)2 + a2

or, (BC)2 = a2 - a2 = 0

$$\therefore$$ BC = 0

Now, taking right angled $$\triangle$$ABC

sin 90° = $$\frac{p}{h}$$ =$$\frac{AB}{AC}$$ =$$\frac{a}{a}$$ =1

cos 90° =$$\frac{b}{h}$$ = $$\frac{BC}{AC}$$ = $$\frac{0}{a}$$ = 0

tan 90° =$$\frac{p}{b}$$ = $$\frac{AB}{BC}$$ = $$\frac{a}{0}$$ =$$\infty$$

cosec 90° =$$\frac{h}{p}$$ = $$\frac{AC}{AB}$$ = $$\frac{a}{a}$$ =1

sec 90° =$$\frac{h}{b}$$ = $$\frac{AC}{BC}$$ = $$\frac{a}{0}$$ = $$\infty$$

cot 90° =$$\frac{b}{p}$$ = $$\frac{BC}{AB}$$ = $$\frac{0}{a}$$ =0

#### Complementary Angles

Two angles are Complementary when they add up to 90 degrees.

Let'a see the following example,

What is the angle added to 60° and 90°?

Let the angle be x.

Then, x + 60° = 90°

or, x = 90° - 60° = 30°

Hence, 30° and 60° when added together gives us 90°, So, 30° and 60° are called complements of each other.

The angles are said to be complementary if the sum of the angles is 90°.

Complementary Angles in Trigonometry

Let ABC be a right angledtriangle where $$\angle$$ABC = 90° and $$\angle$$ACB = $$\theta$$

Now, we know,

$$\angle$$ACB +$$\angle$$ABC +$$\angle$$BAC = 180° (sum of angles of a $$\triangle$$)

or, $$\theta$$ + 90° + $$\angle$$BAC = 180°

or, $$\angle$$BAC = 180° - 90° - $$\theta$$ = 90° - $$\theta$$

So, $$\angle$$ACB and$$\angle$$CAB are complementary angles.

Now, taking$$\angle$$A = 90° - $$\theta$$ as the reference angle, we get,

BC = perpendicular (p)

AC = hypotenuse (h)

AB = base (b)

So,

sin(90° - $$\theta$$) = $$\frac{p}{h}$$ =$$\frac{BC}{AC}$$ = cos$$\theta$$ (For reference angle $$\theta$$)

cos (90° - $$\theta$$) = $$\frac{b}{h}$$ =$$\frac{AB}{AC}$$ = sin$$\theta$$

tan(90° - $$\theta$$) =$$\frac{p}{h}$$ =$$\frac{BC}{AB}$$ = cot$$\theta$$

cot(90° - $$\theta$$) =$$\frac{b}{p}$$ =$$\frac{AB}{BC}$$ = tan$$\theta$$

sec(90° - $$\theta$$) =$$\frac{h}{b}$$ =$$\frac{AC}{AB}$$ = cosec$$\theta$$

cosec(90° - $$\theta$$) =$$\frac{h}{p}$$ =$$\frac{AB}{BC}$$ = sec$$\theta$$

Hence,

sin(90° - $$\theta$$) = cos$$\theta$$

cos(90° - $$\theta$$) = sin$$\theta$$

tan(90° - $$\theta$$) =cot$$\theta$$

cot(90° - $$\theta$$) =tan$$\theta$$

sec(90° - $$\theta$$) = cosec$$\theta$$

cosec(90° - $$\theta$$) =sec$$\theta$$

• h2 = p2 + b2
• Different angles have a different value with various trigonometric ratios. We shall consider 0°, 30°, 45° and 90° as the standard angles and we shall learn their values here. In this unit, we shall verify the values of 0°, 30°, 45° and 90° using geometrical proofs.
.

#### Click on the questions below to reveal the answers

Given that A = 30°

Then, LHS = sin3A

= sin3 × 30°

= sin90°

= 1

RHS = 3sinA - 4sin3A

= 3sin30° - 4(sin30°)3

= 3$$\frac{1}{2}$$ - 4($$\frac{1}{2}$$)3

= $$\frac{3}{2}$$ - 4 $$\frac{1}{8}$$

= $$\frac{3}{2}$$ - $$\frac{1}{2}$$

= $$\frac{2}{2}$$

= 1

$$\therefore$$ LHS = RHS verified.

Solution

LHS = cos 60

= $$\frac{1}{2}$$

RHS = cos230° - sin230°

($$\frac{\sqrt(3)}{2}$$)2 - ($$\frac{1}{2}$$)2

$$\frac{3}{4}$$ - $$\frac{1}{4}$$

$$\frac{2}{4}$$

$$\frac{1}{2}$$

$$\therefore$$ LHS = RHS proved.

Solution

sec2$$\frac{π}{4}$$ sec2$$\frac{π}{3}$$(cosec$$\frac{π}{6}$$ - cosec$$\frac{π}{2}$$)

= sec2 $$\frac{180°}{4}$$ × sec2$$\frac{180°}{3}$$ (cosec$$\frac{180°}{6}$$ - cosec$$\frac{180°}{2}$$)

= sec245° × sec260° (cosec30° - cosec90°)

= ($$\sqrt(2)$$)2 × (2)2 (2-1)

= 2×4×1

= 8

0%

Same
Opposite
Different

p2 + b2
p + b
p2 - b2
b2 - p2
• ### sin 45° = ______

(frac{1}{sqrt 2})
(frac{2}{1})
(frac{sqrt{2}}{sqrt{3}})
(frac{sqrt{3}}{5})
• ### cos 45° = ______

(frac{sqrt{2}}{sqrt{3}})
(frac{2}{1})
(frac{sqrt{3}}{5})
(frac{1}{sqrt 2})

2
0
(frac{1}{2})
1
• ### cosec 45° = ______

(sqrt 2)
(frac{1}{2})
(sqrt{3})
(frac{sqrt{2}}{sqrt{3}})
• ### sec 45° = ______

(sqrt{3})
(frac{2}{3})
(sqrt 2)
(frac{sqrt{3}}{2})
• ### cot 45° = ______

(sqrt{3})
1
(frac{sqrt{3}}{2})
0
• ### sin 60° = ______

1
(sqrt{2})
(frac{sqrt {3}}{2})
(frac{1}{2})

1
(frac{1}{2})
(frac{3}{2})

(frac{1}{2})
(frac{3}{2})
(sqrt{2})
(sqrt3)
• ### cosec 60° = ______

(frac{2}{sqrt {3}})
1
(frac{1}{2})
(frac{3}{2})

2
3
0
1
• ### cot 60° = ______

(frac{2}{3})
(frac{1}{2})
(frac{sqrt{3}}{2})
(frac{1}{sqrt 3})
• ### sin 30° = ______

(frac{sqrt{3}}{2})
(frac{1}{sqrt 2})
(frac{2}{3})
(frac{1}{2})
• ### sec 30° = ______

(frac{2}{sqrt 3})
(sqrt{3})
(frac{sqrt{3}}{2})
(frac{1}{sqrt{3}})
• ### tan 30° = ______

(sqrt{2})
(frac{1}{2})
(frac{2}{sqrt{3}})
(frac{1}{sqrt 3})
• ### cos 0° = _______

(frac{2}{sqrt{3}})
(sqrt{2})
0
1

2
(infty)
0
1

(infty)
1
0
(frac{1}{2})

(infty)
2
0
1
• ### tan 90° = ________

(frac{1}{sqrt{3}})
(frac{1}{2})
(infty)
(frac{2}{sqrt{3}})
• ### sec 90° = ________

(infty)
1
(frac{2}{sqrt{3}})
0