Introduction To Trigonometry Class 10 Notes
CBSE Class 10 Maths Trigonometry Notes:-Download PDF Here
Class 10 Maths Chapter 8 Introduction to Trigonometry Notes
The notes for trigonometry class 10 Maths is provided here. Get the complete concept on trigonometry which is covered in Class 10 Maths. Also, get the various trigonometric ratios for specific angles, the relationship between trigonometric functions, trigonometry tables, various identities given here.
Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:
- Introduction to Trigonometry Short Notes
- Introduction to Trigonometry MCQ Practice Questions
- Introduction to Trigonometry MCQ Practice Solutions
Opposite & Adjacent Sides in a Right Angled Triangle
In the ΔABC right-angled at B, BC is the side opposite to ∠A, AC is the hypotenuse and AB is the side adjacent to ∠A.
For the right ΔABC, right-angled at ∠B, the trigonometric ratios of the ∠A are as follows:
- sin A=opposite side/hypotenuse=BC/AC
- cos A=adjacent side/hypotenuse=AB/AC
- tan A=opposite side/adjacent side=BC/AB
- cosec A=hypotenuse/opposite side=AC/BC
- sec A=hypotenuse/adjacent side=AC/AB
- cot A=adjacent side/opposite side=AB/BC
To know more about Trigonometric Ratios, visit here.
Visualization of Trigonometric Ratios Using a Unit Circle
Draw a circle of the unit radius with the origin as the centre. Consider a line segment OP joining a point P on the circle to the centre which makes an angle θ with the x-axis. Draw a perpendicular from P to the x-axis to cut it at Q.
Relation between Trigonometric Ratios
- cosec θ =1/sin θ
- sec θ = 1/cos θ
- tan θ = sin θ/cos θ
- cot θ = cos θ/sin θ=1/tan θ
Trigonometric Ratios of Specific Angles
Range of Trigonometric Ratios from 0 to 90 degrees
tanθ and secθ are not defined at 90∘.
cotθ and cosecθ are not defined at 0∘.
Variation of trigonometric ratios from 0 to 90 degrees
As θ increases from 0∘ to 90∘
- sin θ increases from 0 to 1
- cos θ decreases from 1 to 0
- tan θ increases from 0 to ∞
- cosec θ decreases from ∞ to 1
- sec θ increases from 1 to ∞
- cot θ decreases from ∞ to 0
Standard values of Trigonometric ratios
|tan A||0||1/√3||1||√3||not defined|
|cosec A||not defined||2||√2||2/√3||1|
|sec A||1||2/√3||√2||2||not defined|
|cot A||not defined||√3||1||1/√3||0|
To know more about Trigonometric Ratios of Standard Angles, visit here.
Trigonometric Ratios of Complementary Angles
Complementary Trigonometric ratios
If θ is an acute angle, its complementary angle is 90∘−θ. The following relations hold true for trigonometric ratios of complementary angles.
- 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) = cosec θ
To know more about Trigonometric Ratios of Complementary Angles, visit here.
To know more about Trigonometric Identities, visit here.
Trigonometry for Class 10 Problems
Find Sin A and Sec A, if 15 cot A = 8.
Given that 15 cot A = 8
Therefore, cot A = 8/15.
We know that tan A = 1/ cot A
Hence, tan A = 1/(8/15) = 15/8.
Thus, Side opposite to ∠A/Side Adjacent to ∠A = 15/8
Let BC be the side opposite to ∠A and AB be the side adjacent to ∠A and AC be the hypotenuse of the right triangle ABC respectively.
Hence, BC = 15x and AB = 8x.
Hence, to find the hypotenuse side, we have to use the Pythagoras theorem.
(i.e) AC2 = AB2 + BC2
AC2 = (8x)2+(15x)2
AC2 = 64x2+225x2
AC2 = 289x2
AC = 17x.
Therefore, the hypotenuse AC = 17x.
Finding Sin A:
We know Sin A = Side Opposite to ∠A / Hypotenuse
Sin A = 15x/17x
Sin A = 15/17.
Finding Sec A:
To find Sec A, find cos A first.
Thus, cos A = Side adjacent to ∠A / Hypotenuse
Cos A = 8x/17x
We know that sec A = 1/cos A.
So, Sec A = 1/(8x/17x)
Sec A = 17x/8x
Sec A = 17/8.
Therefore, Sin A = 15/17 and sec A = 17/8.
If tan (A+ B) =√3, tan (A-B) = 1/√3, then find A and B. [Given that 0° <A+B ≤ 90°; A>B ]
Tan (A+B) = √3.
We know that tan 60 = √3.
Thus, tan (A+B) = tan 60° = √3.
Hence A+B= 60° …(1)
Similarly, given that,
Tan (A-B) = 1/√3.
We know that tan 30° = 1/√3.
Thus, tan (A-B) = tan 30° = 1/√3.
Hence, A-B = 30° …(2)
Now, adding the equations (1) and (2), we get
A+B+A-B = 60° + 30°
2A = 90°
A = 45°.
Now, substitute A = 45° in equation (1), we get
45° +B = 60°
B = 60°- 45°
B = 15°
Hence, A = 45 and B = 15°.
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