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Trigonometry is a branch of Mathematics that deals in establishing relationships between the angles and sides of a triangle. It deals entirely with right angle triangles and it basically sets up the connection between the lengths of sides and angles in it.
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As we now know that the right angled triangle is what we are going to deal with, it is important to know the three basic terms that belong to the right angled triangle. These terms are the base, height and the hypotenuse.
Trigonometry today finds its application in a number of fields like that of oceanography and sound production and hence, knowing this chapter is very important. Moreover, it carries a good percentage of marks with it for class 10th board and being well versed with it’s concepts will help students score better. Here, in this article, students will find all the necessary study material for this chapter that will help them do a quick revision before the exam.
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Trigonometric Ratios
Trigonometric Ratios are the ratios of the sides and the acute angle in a right angled triangle.
In the figure given above, we have angle B= 90° and angle A is an acute angle. Considering these, we will have :-
- AB as the base as it is adjacent to the acute angle,
- BC as the perpendicular as it is opposite to the acute angle and,
- AC as the hypotenuse as this side is opposite to the right angle.
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In the table given below, we have all the trigonometric Ratios with respect to angle A. Knowing these formulas will help students solve the numericals, so students should learn them thoroughly.
| Ratio Formula Short form Value | |
|---|---|
| Sin A Perpendicular/ hypotenuse P/H BC/AC | Cos A Base/hypotenuse B/H AB/AC |
| Tan A Perpendicular/Base P/B BC/AB | Cosec A Hypotenuse/ Perpendicular H/P AC/BC |
| Sec A Hypotenuse/Base H/B AC/AB | Cot A Base/ Perpendicular B/P AB/BC |
Trigonometric Ratios of Complementary Angles
Before knowing the Trigonometric Ratios of Complementary angles, one should know the definition of Complementary angles. When the sum of any two angles of a triangle comes out to be 90°, they are known as complementary angles. According to the angle sum property of a triangle, we know that the sum of three angles of a triangle is 180°. In a Right angled triangle, one of the angles for sure will be 90°. So, the sum of remaining two angles will be 90° and hence they will also be complementary angles and will also prove the angle sum property.
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The Trigonometric Ratios of Complementary angles is as follows:-
sin (90° – A) = cos A,
cos (90° – A) = sin A,
tan (90° – A) = cot A,
cot (90° – A) = tan A,
sec (90° – A) = cosec A,
cosec (90° – A) = sec A
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Trigonometric Identities
An equation is said to be a trigonometric identity when it contains the Trigonometric ratio of an angle and also satisfies it’s value.
In the right angled triangle given above, Q= 90°
According to Pythagoras Theorem,
PQ2 + QR2= PR2. (eq. 1)
Dividing eq. 1 with PR2
PQ2/PR2 + QR2/PR2 = PR2/PR2
Solving this, we get
(Sin R)2 + (Cos R)2 = 1
Sin2R + Cos2R = 1
All the trigonometric identities are:-
- Sin2R + Cos2R = 1
- Cot2R + 1 = Cosec2R
- 1+ tan2 R = Sec2R
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Application of Trigonometric Identities
Trigonometric Identities are used to answer questions related to Line of sight, Angle of elevation and Angle of depression.
Line of sight:- The angle of sight is formed when an imaginary horizontal line is drawn from the observer's eye to the point where an object is kept. This imaginary line that is drawn is called the line of sight.
Angle of elevation:- It is formed when the object that the observer is looking at is kept above the line of sight.
Angle of depression:- It is opposite to the angle of elevation. In this, the object is kept below the line of sight.
Students should be clear about these definitions as many of the questions are based on this.
When a question is given and students draw all the imaginary lines, the resulting triangle is a right angled triangle. Students can apply trigonometric identities to solve such questions.
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Reciprocal Relations between the Trigonometric Ratios
The quantities cosec A, Sec A and cot A are related to cos A, sin A and tan A. The relationship between these quantities is mentioned below.
Sin A = 1/Cosec A
Cos A= 1/Sec A
Tan A= 1/Cot A Cosec A = 1/Sin A
Sec A= 1/Cos A Cot A= 1/Tan A
Sin A cosec A= 1 Cos A Sec A= 1
Cot A Tan A= 1 Tan A= Sin A/ Cos A
Cot A= Cos A/ Sin A
Read More: Trigonometry Table
Trigonometric Ratios of some angles are given in the table below :-
| 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|
| Sin A | 0 | 1/2 | 1/√2 | √3/2 1 |
| Cos A | 1 | √3/2 | 1/√2 | 1/2 0 |
| Tan A | 0 | 1/√3 | 1 √3 | Not known |
| Cosec A | Not known | Not known | 2 √2 | 2/√3 1 |
| Sec A | 1 | 2/√3 | √2 2 | Not known |
| Cot A | Not known | Not known | √3 1 | 1/√3 0 |
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Points to Remember
- If angle C becomes the acute angle, then BC will become the base and AB will become the perpendicular, so the ratios should be taken accordingly.
- The hypotenuse of a right angled triangle is always the longest side. So, the value of sin A and cos A can never exceed 1 and is always equal to or less than 1.
- Similarly, the value of sec A and Cosec A can never be less than 1 and is always greater than or equal to 1.
- The trigonometric Ratios are based on the angle and not on the sides. So, if the angles are the same, the ratios will also remain the same irrespective of the length of the sides.
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Important Questions from Introduction to Trigonometry
Question 1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A, (ii) sin C, cos C
Answer: In a given triangle ABC, right-angled at B = ∠B = 90°
Given: AB = 24 cm and BC = 7 cm
That means, AC = Hypotenuse
According to the Pythagoras Theorem,
In a right-angled triangle, the squares of the hypotenuse side are equal to the sum of the squares of the other two sides.
By applying Pythagoras theorem, we get
AC2 = AB2 + BC2
AC2 = (24)2 + 72
AC2 = (576 + 49)
AC2 = 625 cm2
Therefore, AC = 25 cm
(i) We need to find Sin A and Cos A.
As we know, sine of the angle is equal to the ratio of the length of the opposite side and hypotenuse side. Therefore,
Sin A = BC/AC = 7/25
Again, the cosine of an angle is equal to the ratio of the adjacent side and hypotenuse side. Therefore,
cos A = AB/AC = 24/25
(ii) We need to find Sin C and Cos C.
Sin C = AB/AC = 24/25
Cos C = BC/AC = 7/25
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Question 2. If sin2 A = 12tan2 45°, where A is an acute angle, then find the value of A.
Answer: sin2A = ½ tan2 45°
⇒ sin2A = ½ (1)2 [\(\because\) tan 45° = 1]
= sin2 A = ½
⇒ sin A = 1/√2
So angle A =45°
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Questions 3. If 3 cot A = 4, check whether (1 – tan2A)/(1 + tan2A) = cos2 A – sin2A or not.
Answer: Let us consider a triangle ABC, right-angled at B.
Given,
3 cot A = 4
cot A = 4/3
Since, tan A = 1/cot A
tan A = 1/(4/3) = 3/4
BC/AB = 3/4
Let BC = 3k and AB = 4k
By using Pythagoras theorem, we get;
Hypotenuse2 = Perpendicular2 + Base2
AC2 = AB2 + BC2
AC2 = (4k)2 + (3k)2
AC2 = 16k2 + 9k2
AC = √25k2 = 5k
sin A = Opposite side/Hypotenuse
= BC/AC
=3k/5k
=3/5
In the same way,
cos A = Adjacent side/hypotenuse
= AB/AC
= 4k/5k
= 4/5
To check: (1-tan2A)/(1+tan2A) = cos2 A – sin2 A or not
Let us take L.H.S. first;
(1-tan2A)/(1+tan2A) = [1 – (3/4)2]/ [1 + (3/4)2]
= [1 – (9/16)]/[1 + (9/16)] = 7/25
R.H.S. = cos2 A – sin2 A = (4/5)2 – (3/5)2
= (16/25) – (9/25) = 7/25
Since,
L.H.S. = R.H.S.
Hence, proved.
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Question 4. Evaluate 2 tan2 45° + cos2 30° – sin2 60°.
Answer: Since we know,
tan 45° = 1
cos 30° = √3/2
sin 60° = √3/2
Therefore, putting these values in the given equation:
2(1)2 + (√3/2)2 – (√3/2)2
= 2 + 0
= 2
Read More: Sin2x Formula
Question 5. If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.
Answer: As we know, for any given triangle, the sum of all its interior angles is equals to 180°.
Thus,
A + B + C = 180° ….(1)
Now we can write the above equation as;
⇒ B + C = 180° – A
Dividing by 2 on both the sides;
⇒ (B + C)/2 = (180° – A)/2
⇒ (B + C)/2 = 90° – A/2
Now, put sin function on both sides.
⇒ sin (B + C)/2 = sin (90° – A/2)
Since,
sin (90° – A/2) = cos A/2
Therefore,
sin (B + C)/2 = cos A/2
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