Trigonometry l class 10th l learn trigonometry.
TRIGONOMETRY
}INTRODUCTION
}In this
chapter, we intend to study an important branch of mathematics called “Trigonometry”.
}The word
trigonometry is derived from the greek words, trigonon and metron were ‘trigonon’ mean triangle and
‘metron’ mean measure.
}In
broader sense it is that branch of mathematics which deals with the measurement
of the sides and the angels of a triangle and the problems allied with angles.
History
Aryabhata was the first who uses the idea of sine. “Cosine” and
“tangent” came later in which Cos was
first used by Sir
Jonas Moore an English
Mathematician
TRIGONOMETRIC RATIOS
}The six trigonometric
ratios relates the sides of a right triangle to its angles. Specifically they
are ratios of two sides of a right triangle and a related angle.
}Sin A = Opposite side/hypotenuse
}Cos A = Adjacent side/hypotenuse
}Tan A = Opposite side/Adjacent side
}Cosec A =hypotenuse/Opposite side
}Sec A =hypotenuse/Adjacent side
}Cot A =Adjacent side/Opposite side
Example
In
a ABC, right angled at A, if AB=12,
AC=5 and BC=13, find all the six trigonometric ratios of B.
Solution
}Base= AB,
Perpendicular=AC And, Hypotenuse=BC
Using trigonometric
indents
}Sin B = AC/BC = 5/13
}Cos B =AB/BC=12/13
}Tan B = AC/AB = 5/12
}Cosec B = BC/AC = 13/5
}Sec B =BC/AB =13/12
}Cot B =AB/AC =12/5
TRIGONOMETRIC
RATIO OF
(0,30,45,60,90)
Example
Evaluate the following in the simplest form:
}Sin60oCos30o+Cos60oSin30o
}Solution
}(√3/2)(√3/2)+(1/2)(1/2)
}(√3/2)(√3/2)+(1/2)(1/2)
} (3/4)+(1/4)
}1
TRIGONOMETRIC RATIO OF COMPLEMENTARY ANGLES
}Sin (90o-A) = Cos A
}Tan (90o-A) = Cot A
}Sec (90o-A) = Cosec
A
}Cos (90o-A) = Sin A
}Cot (90o-A) = Tan A
}Cosec (90o-A) = Sec A
Examples
Evaluate the following:
}Sin39 o - Cos51o
}Sin(90o-51o) – Cos51o
}Cos51o-Cos51o
}0
}Cos213 – Sin277
}Cos2 (90o-77o) – Sin277
}Sin277 – Sin277
}0
TRIGONOMETRIC IDENTITIES
}In triangle ABC,
}By Pythagoras thermo,
}AB2 + BC2 = AC2 - 1
}Dividing equation by
AC2
}AB2 + BC2 = AC2
AC2 AC2 AC2
}AB 2 + BC 2 = 1
AC
AC
}Cos
A2 + Sin2 A =
1
}
}Dividing AB in
Equation 1
}AB2 + BC2 = AC2
AB2 AB2 AB2
AB2 AB2 AB2
}1 + BC 2 = AC 2
AB AB
}1 +
Tan2 A = Sec2 A
}Now Dividing BC in Equation 1
}AB2 + BC2 = AC2
BC2 BC2 BC2
}AB 2+ 1 = AC 2
BC
BC
}Cot2 A +1 = Cosec2 A
Example
Evaluate
}Sin25oCos65o+Cos25oSin65o
}Sin(90o-65o)Cos65o + Cos(90o-65o)Sin65o
}Cos65oCos65o + Sin65oSin65o
}Cos265o+Sin265o
Share your problems with us and we try to solve it or you can mail me to on sptrop2016@gmail.com
ReplyDeleteOur team always tray to solve it.
Thanks for our your precious time.