Maths Trigonometry X-A

Introduction to Trignometry

Trigonometric Ratios of Some Specific Angles

Made By-Anirudh Gupta

X-AHoliday HHW

N.E.P.S

• Trigonometry-The Trigonometry is derived from a Greek word, trigonon and metron. The word trigonon means a triangle and the word metron means measure. Therefore, trigonometry means the science of measuring triangles.

Trigonometry Ratios

The most important task of trigonometry is to find the remaining sides and angles of a triangle when some of its sides and angles are given. This problem is solved by using some ratios of the sides of a triangle with respect to its acute angles. These ratios of acute angles are called Trigonometry ratios of angles.

Trigonometric Ratios Of Some Specific Angles

• Trigonometric Ratios of 45• Trigonometric Ratios of • Trigonometric Ratios of

• Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then

the other angle is also 45°, i.e., A = C = 45°∠ ∠ BC = AB (Angles Opp. To equal sides of a Δ are equal)

let BC = AB = a.

By Pythagoras Theorem, AC2 = AB2 + BC2 = a2 + a2 = 2a2,

AC = a 2

As sin 45° = = = =

And cos 45° = = = =

And tan 45° = = = = 1

Also, cosec 45° = 1/sin 45,sec 45 = 1/cos 45 = 2,

cot 45 =1/tan 45 = 1

C

AB

• Trigonometric Ratios of 30° & 60° In equilateral Δ ABC, A = B = C = 60°∠ ∠ ∠ AD perpendicular to BC

∠B = C = 60°∠ AD = AD (common)

∠ADB = ADC (each 90∠ Δ Δ ACD (AAS)

BD = DC and BAD= CAD (CPCT)∠ ∠

let AB = 2a

then BD = ½ BC = a

AD2 = AB2 – BD2 = (2a)2 –a2 = 3a2

AD=a3

Now, sin 30 = BD/AB = a / 2a = ½

cos 30 = AD/AB = a3 / 2a = 3/2

tan 30 = BD/AD = a / a3 = 1/ 3

cosec 30 =1/ sin 30 = 2

sec 30 = 1/cos 30 = 2/ 3

cot 30 = 1/tan 30 = 3

Similarly,

sin 60 = AD/AB = a 3/2a = 3/2

cos 60 = ½

tan 60 = 3

cosec 60 = 2/ 3

sec 60 = 2

cot 60 = 1/ 3

Trigonometric Ratios of 0° & 90°

Let angle XAY = θ be an acute angle and P be

a point on its terminal side AY. Draw

perpendicular PM from P on AX .

IN ∆AMP, we have

Sin θ = PM, cos θ = AM

AP AP

and tan θ = PM

AM

It is evident from ∆AMP that as θ becomes smaller and smaller, line segment PM also becomes smaller and smaller; finally when θ becomes 0 ; the point P will coincide with M. ⁰Consequently , we have PM = 0 and AP=AM Therefore, sin 0 = ⁰ PM = 0 =0 AP AP cos O =⁰ AM =AP = 1 AP AP

Therefore,

sin 90 = ⁰ PM = PM = 1

AP PM

and cos 90 = ⁰ AM = 0 = 0

AP AP

Thus, we have

sin 90 = 1 and cos 90 =0⁰ ⁰Remark:- It is evident from the above

discussion that tan 90 = PM =0 is ⁰not defined . Similarly, sec 90 , ⁰cosec 0 , cot 0 are not defined .⁰ ⁰ and , tan 0 = ⁰ PM = 0 =0

AP AP

Thus, we have

sin 0 = 0, cos 0 = 1 ⁰ ⁰ and tan 0 = 0⁰

•From ∆AMP, it is evident that as θ

increase, line segment AM

becomes smaller and smaller and

finally when θ becomes 90 the ⁰ point M will coincide with A. •Consequently,

we have AM =O, AP =PM

• Trigonometry Ratios of O⁰, 45⁰,30⁰, 60⁰ and 90⁰

Anirudh GuptaX-A

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