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.
Trigonometry
Trigonometry is a scientific study of measurement of triangle and its angle as well as sides.
Ray: A ray having one end point and end with and where.
Line segment: A line segment having two end point are called as line segment.
Note: In trigonometry studying only about right angle triangle.
Angle of elevation
+𝜗
−𝜗
Angle of depression
a) Angle of Elevation(+𝝑) : If a ray rotating in anticlockwise direction then formed angle (+𝜃) are
called as angle of elevation.
b) Angle of Depression(−𝜽) : If a ray rotating in clockwise direction then formed angle (−𝜃) are
called as angle of depression.
B
90°
𝜃
C
A
𝜃
𝜃
Then ∆𝐴𝐵𝐶 are called right angled triangle.
By using angle sum property
∠𝐴 + ∠𝐵 + ∠𝐶 = 180°
1
1
1
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∠𝐴 = 90° + 𝜃 + 180°
∠𝐴 = 180° − 90° − 𝜃
∠𝐴 = 90° − 𝜃 = 𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑎𝑛𝑔𝑙𝑒
Note: Acute angle less than 90°, Obtuse angle more than 90°.
Complementary angle: If sum of two acute angles is 90° are called as complementary angle.
∠𝐴 + ∠𝐶 = 90°\
(90° − 𝜃) + 𝜃
90° − 𝜃 + 𝜃
90°, ℎ𝑒𝑛𝑐𝑒 𝑝𝑟𝑜𝑣𝑒𝑑
Then (90° − 𝜃)and angle 𝜃 are complementary angle to each other.
Supplementary Angle:
If sum of two angles is 180° are called as supplementary angles.
Then (180° − 𝜃)and 𝜃are supplementary to each other.
Pythagoras Theorem (PGT)
Sum of squares of two sides is equal to square of third side are called s Pythagoras theorem (PGT).
2
2
2
ADDRESS: INFRONT OF SHRAVAN KANTA PALACE PLOT NO. 10,ABOVE AXIS BANK,
AYODHYA BYPASS ROAD, BHOPAL ,Pin code 462022,
Cont.No. 07552625412, 7697542831
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A
H
P
90°
𝜃
B
C
(AC)2=(AB)2 + (BC)2
H2 = P2 +B2
NOTE: where using PGT
1. PGT always using in right angled triangle.
2. Q. Why using in right angled triangle.
A. For finding third side if two sides are given
A
H =?
𝜃
𝜃
B
B
C
Given P = 3 cm
B = 4 cm
H =?
By using PGT;
H2 = P2 + B2
(AC)S = (AB)2 + (BC)2
3
3
3
ADDRESS: INFRONT OF SHRAVAN KANTA PALACE PLOT NO. 10,ABOVE AXIS BANK,
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Cont.No. 07552625412, 7697542831
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H2 = (3)2 + (4)2
H2 = 9 + 16 = H2 = √25 = 5𝐶𝑀
a) Side opposite 90° (right angle) Hypotenuse.
b) Side opposite angle 𝜃 always perpendicular.
c) Side adjacent to angle 𝜃 always Base.
Trigonometric Ratios:
A
P
H
𝜃
B
𝑠𝑖𝑛 𝜃
B
C
𝑐𝑜𝑠 𝜃
𝑡𝑎𝑛 𝜃
P
B
P
H
H
B
𝑠𝑒𝑐 𝜃
𝑐𝑜𝑡 𝜃
𝑐𝑠𝑐 𝜃
PANDIT
BADRI
PRASAD
HAR
HAR
BOLE
A right angle triangle is shown Figure. ∠𝐵 𝑖𝑠 𝑜𝑓 90° 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 ∠𝐵 𝑏𝑒 𝑐𝑎𝑙𝑙𝑒𝑑 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒.
4
4
4
ADDRESS: INFRONT OF SHRAVAN KANTA PALACE PLOT NO. 10,ABOVE AXIS BANK,
AYODHYA BYPASS ROAD, BHOPAL ,Pin code 462022,
Cont.No. 07552625412, 7697542831
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There are two other angle i.e. ∠𝐴 𝑎𝑛𝑑 ∠𝐶. It we consider ∠𝐶 as 𝜃,then opposite side to this angle is
called perpendicular and side adjacent to 𝜃is called base.
 Six Trigonometry Ratio are:
𝑃
𝐴𝐵
𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝜃
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 90°
𝐵
𝐵𝐶
𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝜃
𝑐𝑜𝑠 𝜃 = 𝐶𝑜𝑠𝑖𝑛𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝜃 = 𝐻 = 𝐴𝐶 = 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 90°
𝑃
𝐴𝐵
𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝜃
𝑡𝑎𝑛 𝜃 = 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝜃 = 𝐵 = 𝐵𝐶 = 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 90°
𝐻
𝐴𝐶
𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 90°
𝑐𝑠𝑐 𝜃 = 𝑐𝑜𝑠𝑒𝑐 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝜃 = =
=
𝑃
𝐴𝐵
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝜃
𝐻
𝐴𝐶
𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 90°
𝑠𝑒𝑐 𝜃 = 𝑠𝑒𝑐𝑎𝑛𝑡 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝜃 = =
=
𝐵
𝐵𝐶
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝜃
𝐵
𝐵𝐶
𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 90°
𝑐𝑜𝑡 𝜃 = 𝑐𝑜𝑡𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝜃 = 𝑃 = 𝐴𝐶 = 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝜃
1. 𝑠𝑖𝑛 𝜃 = Sine of angle 𝜃 = 𝐻 = 𝐴𝐶 =
2.
3.
4.
5.
6.
 Inter relationship is Basic Trigonometric Ratio :
1
𝑡𝑎𝑛 𝜃 = 𝑐𝑜𝑡 𝜃
1
𝑐𝑜𝑡 𝜃 = 𝑐𝑜𝑡𝜃
𝑐𝑜𝑠 𝜃 =
1
𝑠𝑒𝑐 𝜃
𝑠𝑒𝑐 𝜃 =
1
𝑐𝑜𝑠 𝜃
𝑠𝑖𝑛 𝜃 =
1
𝑐𝑜𝑠𝑒𝑐 𝜃
𝑐𝑠𝑐 𝜃 =
1
𝑠𝑖𝑛 𝜃
Trigonometric Table:
5
5
5
ADDRESS: INFRONT OF SHRAVAN KANTA PALACE PLOT NO. 10,ABOVE AXIS BANK,
AYODHYA BYPASS ROAD, BHOPAL ,Pin code 462022,
Cont.No. 07552625412, 7697542831
SUNSHIELD CLASSES
All progress, change, and Success is based on a foundation at convenience
Trigonometric ratios of complementary angles:
6
6
6
ADDRESS: INFRONT OF SHRAVAN KANTA PALACE PLOT NO. 10,ABOVE AXIS BANK,
AYODHYA BYPASS ROAD, BHOPAL ,Pin code 462022,
Cont.No. 07552625412, 7697542831
SUNSHIELD CLASSES
All progress, change, and Success is based on a foundation at convenience
Illustra tion :
Ex .1 In the given triangle AB =5cm. find all trigonometric ratios.
Sol. Using Pathygoras theorem
𝐴𝐶 2 = 𝐴𝐵2 + 𝐵𝐶 2
A
52 = 32 + P2
16 = P2 ⇒
P = cm
B
H
Here P = 4cm, B = 3cm, H = 5cm
∴
𝑃
𝐻
𝑠𝑖𝑛 𝜃 =
4
=5
𝑃
𝐻
𝑃
=
𝐵
4
𝐶𝑜𝑠 𝜃 =
=5
𝑡𝑎𝑛 𝜃
=
C
B
P
4
3
𝐵
3
𝐻
5
𝐻
𝑃
=4
𝑐𝑜𝑡 𝜃 = 𝑃 = 4
𝑠𝑒𝑐 𝜃 = 𝐵 = 3
𝑐𝑠𝑐 𝜃 =
5
𝑚
𝑛
Ex.2 If 𝑡𝑎𝑛 𝜃 = ,then find 𝑠𝑖𝑛 𝜃.
Sol. Let P = 𝑚 ∝ 𝑎𝑛𝑑 𝛽 = 𝑛 ∝
∴ 𝑡𝑎𝑛 𝜃 =
A
𝑃 𝑚
=
𝐵 𝑛
H2 = P2 + B2
H2 = 𝑚2 ∝2 + 𝑛2 ∝2
𝑚∝
H
𝐻 =∝ √𝑚2 + 𝑛2
∴
𝑡𝑎𝑛 𝜃 =
𝑠𝑖𝑛 𝜃 =
𝑃
𝐻
=
𝑚𝑎
∝√𝑚2 +𝑛2
B
𝑛∝
C
𝑚
√𝑚2 +𝑛2
7
7
7
ADDRESS: INFRONT OF SHRAVAN KANTA PALACE PLOT NO. 10,ABOVE AXIS BANK,
AYODHYA BYPASS ROAD, BHOPAL ,Pin code 462022,
Cont.No. 07552625412, 7697542831