Ppt on trignometry by damini

PROJECT ON TRIGONOMETRY

DESIGNED BY :- DAMINI NARANG and

WARQA RAIS 10TH B ROLL NO :-13 and 46

Sin θ

Cot θ

Cos θCosecant θ

∟A

½=θ

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ACKNOWLEDGEMENT

We made this project under the guidance of my mathematics teacher Mrs. Navjeevan Kaur mam

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WELCOME TO THE

WORLD OF TRIGONOMET

RY


INTRODUCTION TO TRIGONOMETRY

The word trigonometry is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides’ ) and ‘metron’ (meaning measure). In fact, Trigonometry is the study of the relationships between the sides and angles of a triangle. Trigonometric ratios of an angle are some ratios of the sides of a right triangle with respect to its acute angles. Trigonometric identities are some trigonometric ratios for some specific angles and some identities involving these ratios.


EXAMPLE

Suppose the students of a school are visiting Eiffel tower . Now, if a student is looking at the top of the tower, a right triangle can be imagined to be made as shown in figure.Can the student find out the height of the tower, without actually measuring it? Yes the student can find the height of the tower with the help of trigonometry.


TRIGONOMETRIC RATIOSLet us take a right angle ABC as shown in figure.Here, ∟CAB or ∟A is an acute angle. Note the position of side BC with respect to ∟A. It faces ∟A. we call it the side opposite to ∟A(perpendicular). AC is hypotenuse of the right angle and the side AB is a part of ∟A. so, we call it the side adjacent to ∟A(base).Next Slide Previous SlideHOME

NAMES OF TRIGONOMETRIC RATIOS

NAMES WRITTEN AS

Sine θ Sin θ

Cosine θ Cos θ

Tangent θ Tan θ

Cosecant θ Cosec θ

Secant θ Sec θ

Cotangent θ Cot θ


The trigonometric ratios of the angle A in the right triangle ABC see in fig.

• Sin of A =side opposite to angle A =BC

hypotenuse AC

• Cosine of A =side adjacent to angle A =AB

hypotenuse AC

• Tangent of A =side opposite to angle A =BC

side adjacent to angle A AB

C

A B

Cosecant of A = 1 = hypotenuse = AC

sin of A side opposite to angle A BC

Secant of A = 1 = hypotenuse = AC

sin of A side adjacent to angle a AB

Cotangent of A= 1 =side adjacent to angle A= AB

tangent of A side opposite to angle A BC

C

A B

These are some easy method to learn these formulas:

• Pandit Badri Prasad Har Har Bhole Sona Chandi Tole

• Pakistan Bhuka Pyasa Hindustan Hara Bhara.S C T

P B P

H H B

INFORMATIONS – Sin θ C – Cos θT – Tan θP – PerpendicularB – Base H – Hypotenuse

RECIPROCALS OF SIN , COS & TAN

Sin θ = reciprocal= Cosec θ

Cos θ = reciprocal = Sec θ

Tan θ = reciprocal = Cot θ

Means :-Sin θ = 1/ Cosec θ (sin θ * cosec θ = 1

)

Cos θ = 1/ Sec θ ( cos θ * sec θ = 1 )

Tan θ = 1/ Cot θ ( tan θ * cot θ =

1 )


QUESTIONS RELATED TO ABOVE TOPICS

1) Calculating the value of other trigonometric ratios, if one is given.

2) Proving type.3) Evaluating by

using the given trigonometric ratio’s value.


TYPE 1 – CALCULATING

VALUE OF OTHER TRIGONOMETRIC RATIOS, IF ONE IS GIVEN.

If Sin A = 3 / 4 , calculate Cos A and Tan A .

Solution - Sin A = P / H = BC / AC = 3 / 4

Let BC = 3K

AND , AC = 4K

THEREFORE, By Pythagoras Theorem,

(AB)² = (AC)² – (BC)²

(AB)² = (4K)² - (3K)²

AB = √7K

Cos A = B / H= AB / AC = √7K / 4K

= √7 / 4

Tan A = P / B = BC / AB = 3K / √7K

= 3 / √7


TYPE 2 – PROVING TYPE

If ∟A and ∟B are acute angles such that Cos A = Cos B, then show that ∟A = ∟B

Solution - Since, Cos A = Cos B

AC / AB = BC / AB

therefore, AC = BC.

∟B = ∟A (angles opposite to

equal sides )

Therefore , ∟A = ∟B


TYPE 3 – EVALUATING BY PUTTING THE GIVEN TRIGONOMETRIC RATIO’S VALUE

If Sec A = 5 / 4 , evaluate 1 – Tan A .

1 + Tan A

Solution – Sec A = H / B =AC / AB = 5 / 4

Let AC / AB = 5K / 4K.

By Pythagoras Theorem ,

(BC)² = (AC ) ² – (AB) ²

Therefore, BC = 3K

So, Tan A = P / B = BC / AB = 3K / 4K = 3 / 4

1 – Tan A = 1 – 3 / 4 = 1 / 4 = 1

1 + Tan A 1 + 3 / 4 7 / 4 7Next Slide Previous SlideHOME

VALUES OF TRIGONOMETRIC RATIOS

∟θ 0° 30° 45° 60° 90°

Sin θ 0 1/2 1/√2 √3/2 1

Cos θ 1 √3/2 1/√2 1/2 0

Tan θ 0 1/√3 1 √3 NOTDEFINED

Cosec θ

NOTDEFINED 2 √2 2/√3 1

Sec θ 1 2/√3 √2 2 NOT DEFINED

Cot θ NOT DEFINED √3 1 1/√3 0


EXAMPLES ON VALUES OF TRIGONOMETRIC RATIOS

1)Evaluation2)Finding values of A and B.


TYPE 1 - EVALUATIONSin 60° * cos 30° + sin 30° * cos60° =√3 / 2 * √3 / 2 + 1 / 2 * 1 / 2 = 3 / 4 + 1 / 4 = 4 / 4 = 1


TYPE 2 – FINDING VALUES OF A AND B

If Tan (A+B) = √3 and tan ( A – B) = 1/ √3 ; 0° < A + B ≤ 90° ; A> B , find A and B. Solution – tan (A + B ) = √3 tan (A+ B ) = tan 60°

A+ B = 60° - ( 1)

tan (A- B) = 1 / √3 tan (A- B) = tan 30°

A – B = 30° - ( 2 )

From ( 1 ) & ( 2)

A = 45 °

B = 15 °


FORMULASSin ( 90° – θ ) = Cos θCos ( 90° – θ ) = Sin θ

Tan ( 90° – θ ) = Cot θ Cot ( 90° – θ ) = Tan θ

Cosec ( 90° – θ ) = Sec θ

Sec ( 90° – θ ) = Cosec θ


EXAMPLE ON FORMULASoEvaluate : -(1) Sin 18 ° / Cos 72 ° = Sin (90 – 72 ) ° / Cos 72 ° = Cos 72 ° / Cos 72 ° = 1( 2) Cos 48 ° – Sin 42 ° = Cos ( 90 – 42 ) ° – Sin 42 ° = Sin 42 ° – Sin 42 ° = 0Next Slide Previous SlideHOME

MAIN IDENTITIES

Sin²θ + Cos² θ = 11 + Tan² θ = Sec² θ1 + Cot² θ = Cosec² θSinθ / Cos θ = Tan θCosθ / Sin θ = Cot θSin² θ / Cos² θ = Tan² θCos² θ / Sin² θ = Cot² θ


STEPS OF PROVING THE IDENTITIES

1)Solve the left hand side or right hand side of the identity.

2)Use an identity if required.3)Use formulas if required.4)Convert the terms in the form of sinθ

or cos θ according to the question.5)Divide or multiply the L.H.S. by sin θ

or cos θ if required.6)Then solve the R.H.S. if required.7)Lastly , verify that if L.H.S. = R.H.S.


THE END

PROJECT MADE BY –

DaMiNi NaRaNg

AND WaRqA rAiSPrevious SlideHOME

Ppt on trignometry by damini

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