This project on trigonometry was designed by two 10th grade students to introduce various topics in trigonometry. It includes sections on the introduction and definition of trigonometry, trigonometric ratios and their names in a right triangle, examples of applying ratios to find unknown sides, reciprocal identities of ratios, types of problems involving calculating ratios and evaluating expressions, value tables for common angles, formulas relating ratios, and main trigonometric identities. The project was created under the guidance of the students' mathematics teacher.

1. PROJECT ON
TRIGONOMETRY
DESIGNED BY :-
DAMINI NARANG and
WARQA RAIS
10TH B
ROLL NO :-13 and 46
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2. ACKNOWLEDGEMENT
We made this project under the
guidance of my mathematics
teacher Mrs. Navjeevan Kaur
mam
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3. WELCOME TO
THE WORLD OF
TRIGONOMETRY
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4. 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.
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5. 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.
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6. TRIGONOMETRIC RATIOS
Let 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).
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7. 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
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8. 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

9. 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

10. 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
INFORMATION
S – Sin θ
C – Cos θ
T – Tan θ
P – Perpendicular
B – Base
H – Hypotenuse

11. 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 )
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12. 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.
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13. 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
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14. 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
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15. 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 7
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16. 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 NOT
DEFINED
Cosec
θ
NOT
DEFINED 2 √2 2/√3 1
Sec θ 1 2/√3 √2 2 NOT
DEFINED
Cot θ NOT
DEFINED √3 1 1/√3 0Next Slide Previous SlideHOME

17. EXAMPLES ON VALUES OF
TRIGONOMETRIC RATIOS
1)Evaluation
2)Finding values of A and B.
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18. TYPE 1 - EVALUATION
Sin 60° * cos 30° + sin 30° * cos60°
=√3 / 2 * √3 / 2 + 1 / 2 * 1 / 2
= 3 / 4 + 1 / 4
= 4 / 4
= 1
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19. 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 °
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20. FORMULAS
Sin ( 90° – θ ) = Cos θ
Cos ( 90° – θ ) = Sin θ
Tan ( 90° – θ ) = Cot θ
Cot ( 90° – θ ) = Tan θ
Cosec ( 90° – θ ) = Sec
θ
Sec ( 90° – θ ) = CosecNext Slide Previous SlideHOME

21. EXAMPLE ON FORMULAS
oEvaluate : -
(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 °
= 0
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22. MAIN IDENTITIES
Sin²θ + Cos² θ = 1
1 + Tan² θ = Sec² θ
1 + Cot² θ = Cosec² θ
Sinθ / Cos θ = Tan θ
Cosθ / Sin θ = Cot θ
Sin² θ / Cos² θ = Tan² θ
Cos² θ / Sin² θ = Cot² θ
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23. 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.
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