TRIGONOMETRY Measurement Of Angles:
Angles can be measured in two units, Degree and Radian. Following is the conversion formula.
1 radian =  degrees  57.2958 degrees
1 degree =  radian  0.0174533 radian.
Example: Convert 135 degree to radian.
 radians
Trigonometric Functions:
Trigonometric ratios are the ratios of the sides of a right-angled triangle. They are:
sin A = opposite / hypotenuse = a/c
cos A = adjacent / hypotenuse = b/c
tan A = opposite / adjacent= a/b
cosec A = 1/sin A= c/a
sec A = 1/cos A= c/b
cot A = 1/tan A= b/a
Example: If the terminal ray of an angle q in standard position passes through (–3, 2), find sin q and cos q .
Solution: Make a sketch as shown. To find the radius r of the circle, use the equation x 2 + y 2 = r 2 with x = –3 and y = 2:
(–3) 2 + 2 2 = 13 = r 2
 = r
Thus : sin q = 
And 
Example: If q is a fourth–quadrant angle and sin q = –  , find cos q .
Solution: Make a sketch of a circle with radius 13 as shown. Since sin q =  and r is always positive, y = –5. To find x, use the circle's equation,  :
x = ± 12
Since q is a fourth–quadrant angle, x = 12. Thus,  .
Law of sine: This law is used to find the unknown parts of a triangle.
In  ABC , 
Example: In a D ABC , if a = 2, b = 3 and sin A =  , find Ð B.
Solution: We have: 
Þ  Þ  Þ sin B = 1 Þ Ð B = 90 °
Measurement Of Angles:
Angles can be measured in two units, Degree and Radian. Following is the conversion formula.
1 radian =  degrees  57.2958 degrees
1 degree =  radian  0.0174533 radian.
Example: Convert 135 degree to radian.
 radians
Trigonometric Functions:
Trigonometric ratios are the ratios of the sides of a right-angled triangle. They are:
sin A = opposite / hypotenuse = a/c
cos A = adjacent / hypotenuse = b/c
tan A = opposite / adjacent= a/b
cosec A = 1/sin A= c/a
sec A = 1/cos A= c/b
cot A = 1/tan A= b/a
Example: If the terminal ray of an angle q in standard position passes through (–3, 2), find sin q and cos q .
Solution: Make a sketch as shown. To find the radius r of the circle, use the equation x 2 + y 2 = r 2 with x = –3 and y = 2:
(–3) 2 + 2 2 = 13 = r 2
 = r
Thus : sin q = 
And 
Example: If q is a fourth–quadrant angle and sin q = –  , find cos q .
Solution: Make a sketch of a circle with radius 13 as shown. Since sin q =  and r is always positive, y = –5. To find x, use the circle's equation,  :
x = ± 12
Since q is a fourth–quadrant angle, x = 12. Thus,  .
Law of sine: This law is used to find the unknown parts of a triangle.
In  ABC , 
Example: In a D ABC , if a = 2, b = 3 and sin A =  , find Ð B.
Solution: We have: 
Þ  Þ  Þ sin B = 1 Þ Ð B = 90 °
Example: If in any triangle the angles be to one another as 1 : 2 : 3, prove that the corresponding sides are 1 :  : 2.
Solution: Let the angles be x , 2 x and 3 x .
Then, x + 2 x + 3 x = 180 ° Þ 6 x = 180 ° Þ x = 30 °
So, the angle are 30 ° , 60 ° and 90 °
Let a , b , c denote the sides apposite to these angles.
Þ a : b : c = sin30 ° : sin60 ° : sin90 °
Þ a : b : c =  Þ a : b : c = 1 :  : 2
Example: The angle of a triangle ABC are in A.P. and it is being given that b : c =  , find Ð A .
Solution: Since Ð A, Ð B, Ð C are in A.P.
\ 2 Ð B = Ð A + Ð C
Þ 3 Ð B = Ð A + Ð B + Ð C
Þ 3 Ð B = 180 ° Þ Ð B = 60 °
Now, 
Þ  Þ  [  b : c =  ]
Þ 
Þ sin C =  Þ Ð C = 45 °
\ Ð A = 180 ° – ( Ð B + Ð C) = 180 ° – (60 ° + 45 ° ) = 75 °
Law of cosine: This law is used to find the unknown parts of a triangle.
In  ABC , 
Example: The lengths of the sides of a triangle are 5, 10, and 12. Solve the triangle.
Solution:
1. 
a » 101.0 °
2. 
b » 24.1 °
3. q » 180 ° – (101.0 ° + 24.1 ° ) = 54.9 °
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