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. In Maths, trigonometry is one of the branches, where we learn the relationships between angles and sides of a triangle. Trigonometry is derived from Greek words ‘tri’ (means three), ‘gon’ (means sides) and ‘metron’ (means measure). In this chapter, we will learn the basics of trigonometry. Get the complete concept of 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
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
Example: Suppose a right-angled triangle ABC, right-angled at B such that hypotenuse AC = 5cm, base BC = 3cm and perpendicular AB = 4cm. Also, ∠ACB = θ. Find the trigonometric ratios tan θ, sin θ and cos θ.
Solution: Given, in ∆ABC,
Hypotenuse, AC = 5cm
Base, BC = 3cm
Perpendicular, AB = 4cm
Then, by the trigonometric ratios, we have;
tan θ = Perpendicular/Base = 4/3
Sin θ = Perpendicular/Hypotenuse = AB/AC = ⅘
Cos θ = Base/Hypotenuse = BC/AC = ⅗
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.
- sin θ=PQ/OP=PQ/1=PQ
- cos θ=OQ/OP=OQ/1=OQ
- tan θ=PQ/OQ=sin θ/cos θ
- cosec θ=OP/PQ=1/PQ
- sec θ=OP/OQ=1/OQ
- cot θ=OQ/PQ=cos θ/sin θ
Trigonometric Ratios of Specific Angles
The specific angles that are defined for trigonometric ratios are 0°, 30°, 45°, 60° and 90°.
Trigonometric Ratios of 45°
If one of the angles of a right-angled triangle is 45°, then another angle will also be equal to 45°.
Let us say, ABC is a right-angled triangle at B, such that;
∠ A = ∠ C = 45°
Thus, BC = AB = a (say)
Using Pythagoras theorem, we have;
AC2 = AB2 + BC2
= a2 + a2
AC = a√2
Now, from the trigonometric ratios, we have;
- sin 45° = (Opp. side to angle 45°)/Hypotenuse = BC/AC = a/a√2 = 1/√2
- cos 45° = (Adj. side to angle 45°)/Hypotenuse = AB/AC = a/a√2 = 1/√2
- tan 45° = BC/AB = a/a = 1
- cosec 45° = 1/sin 45° = √2
- sec 45° = 1/cos 45° = √2
- cot 45° = 1/tan 45° = 1
Trigonometric Ratios of 30° and 60°
Here, we will consider an equilateral triangle ABC, such that;
AB = BC = AC = 2a
∠A = ∠B = ∠C = 60°
Now, draw a perpendicular AD from vertex A that meets BC at D
According to the congruency of the triangle, we can say;
Δ ABD ≅ Δ ACD
BD = DC
∠ BAD = ∠ CAD (By CPCT)
Now, in triangle ABD, ∠ BAD = 30° and ∠ ABD = 60°
Using Pythagoras theorem,
AD2 = AB2 – BD2
= (2a)2 – (a)2
AD = a√3
So, the trigonometric ratios for 30-degree angle will be;
sin 30° = BD/AB = a/2a = 1/2
cos 30° = AD/AB = a√3/2a = √3/2
tan 30° = BD/AD = a/a√3 = 1/√3
cosec 30° = 1/sin 30 = 2
sec 30° = 1/cos 30 = 2/√3
cot 30° = 1/tan 30 = √3
Similarly, we can derive the values of trigonometric ratios for 60°.
- sin 60° = √3/2
- cos 60° = 1/2
- tan 60° = √3
- cosec 60° = 2/√3
- sec 60° = 2
- cot 60° = 1/√3
Trigonometric Ratios of 0° and 90°
If ABC is a right-angled triangle at B, if ∠A is reduced then side AC will come near to side AB. So, if ∠ A is nearing to 0 degree, then AC becomes almost equal to AB and BC get almost equal to 0.
Hence, Sin A = BC /AC = 0
and cos A = AB/AC = 1
tan A = sin A/cos A = 0/1 = 0
cosec A = 1/sin A = 1/0 = not defined
sec A = 1/cos A = 1/1 = 1
cot A = 1/tan A = 1/0 = not defined
In the same way, we can find the values of trigonometric ratios for a 90-degree angle. Here, angle C is reduced to 0, and the side AB will be nearing side BC such that angle A is almost 90 degrees and AB is almost 0.
Range of Trigonometric Ratios from 0 to 90 degrees
- 0 ≤ sin θ ≤ 1
- 0 ≤ cos θ ≤ 1
- 0 ≤ tan θ < ∞
1 ≤ sec θ < ∞
- 0 ≤ cot θ < ∞
- 1 ≤ cosec θ < ∞
tan θ and sec θ are not defined at 90∘.
cot θ and cosec θ 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
Example: Find the value of sin65°/cos25°.
cos A = sin (90° – A)
cos 25° = sin (90° – 25°)
= sin 65°
Hence, sin65°/sin65° = 1
To know more about Trigonometric Ratios of Complementary Angles, visit here.
The three most important trigonometric identities are:
Example: Prove that sec A (1 – sin A)(sec A + tan A) = 1.
Solution: We will start solving for LHS, to get RHS.
sec A (1 – sin A)(sec A + tan A) = (1/cos A)(1 – sin A)(1/cos A + sin A/cos A)
= [(1 – sin A)(1 + sin A)]/cos2 A
= [1 – sin2A]/cos2A
To know more about Trigonometric Identities, visit here.
- NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry
- Class 10 Maths Chapter 8 Introduction to Trigonometry MCQs
- Important Questions for Class 10 Maths Chapter 8- Introduction to Trigonometry
Trigonometry for Class 10 Solved 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°.
Video Lesson on Trigonometry
Stay tuned with BYJU’S – The Learning App and download the app to learn all Maths-related concepts easily by exploring more videos.