# Trigonometry: Applications and Concepts | Class 10 Mathematics Project

## Introduction

Let us take some examples from our surroundings where right triangles can be imagined to be formed. For instance :

1. Suppose the students of a school are visiting Qutub Minar. Now, if a student is looking at the top of the Minar, a right triangle can be imagined to be made, as shown in Fig 8.1. Can the student find out the height of the Minar, without actually measuring it?
2. Suppose a girl is sitting on the balcony of her house located on the bank of a river. She is looking down at a flower pot placed on a stair of a temple situated nearby on the other bank of the river. A right triangle is imagined to be made in this situation as shown in Fig.8.2. If We know the height at which the person is sitting, can We find the width of the river?

In all the situations given above, the distances or heights can be found by using some mathematical techniques, which come under a branch of mathematics called ‘trigonometry’. The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle

## Trigonometric Ratios

In the above Section, We have seen some right triangles imagined to be formed in different situations. Let us take a right triangle ABC as shown in Fig. 8.4. Here, ∠ CAB (or, in brief, angle A) is an acute angle. Note the position of the side BC with respect to angle A. It faces ∠ A. We call it the side opposite to angle A. AC is the hypotenuse of the right triangle and the side AB is a part of ∠ A. So, we call it the side adjacent to angle A.

Note that the position of the sides changes when We consider angle C in place of A (see Fig. 8.5). We have studied the concept of ‘ratio’ in Wer earlier classes. We now define certain ratios involving the sides of a right triangle and call them trigonometric ratios. The trigonometric ratios of the angle A in right triangle ABC (see Fig. 8.4) are defined as follows :

sine of ∠ A = BC\over AC

cosine of ∠ A = AB\over AC

tangent of ∠ A = BC\over AB

cosecant of ∠ A = AC\over BC

secant of ∠ A = AC\over AB

cotangent of ∠ A = AB\over BC

## Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved. In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities.

In Δ ABC, right-angled at B (see Fig), we have:

AB2 + BC2 = AC2

## Some Example Numerical

Question 1 :

From the figure given below, find the values of:

(i) sin θ

(ii) cos θ

(ii) tan θ

Solution

From right-angled triangle OMP,

By Pythagoras theorem, we get

OP2 = OM2 +MP2

or, MP2 = OP2 + OM2

or, MP2 = (15)2 – (12)2

or, MP2 = 225 – 144

or, MP2 = 81

or, MP2 = 92

or, MP = 9

(i) sin θ = MP\over OP = 9\over 15 = 3\over 5

(ii) cos θ = OM\over OP = 12\over 15 = 4\over 5

(iii) tan θ = MP\over OP = 9\over 12 = 3\over 4

Question 2 :

From the figure given below, find the values of:

(i) sin A

(ii) cos A

(iii) sin2 A + cos2 A

Solution

From the right-angled triangle ABC,

By Pythagoras theorem, we get

AB2 = AC2 + BC2

AB2 = (12)2 + (5)2

AB2 = 144 + 25

AB2 = 169

AB2 = 132

AB = 13

(i) sin A = BC\over AB

= 5\over 13

(ii) cos A = AC\over AB

= 12\over 13

(iii) Sin2 A + cos2 A = {BC\over AB}^2+ {AC\over AB}^2

= {5\over 13}^2 + {12\over 13}^2

= {25\over 169} + {144\over 169}

= {25 + 144}\over 169

= 169\over 169

= 1

Sin2 A + cos2 A = 1

Question 3 :

From the figure given below, find the values of:

(i) tan x

(ii) cos y

(iii) cosec2 y – cot2 y

(iv) 5/sin x + 3/sin y – 3 cot y.

Solution

From right-angled ∆ACD,

By Pythagoras theorem we get

AD= AC2 – CD2  = (13)2 – (5)= 169 – 25

From right-angled ∆ABD,

By Pythagoras angled ∆ABD

By Pythagoras theorem we get

AB= AD2 + BD2 = 400

AB= (20)2

AB = 20

(i) tan x = perpendicular\over Base (in right-angled ∆ACD)

= 5\over 12

(ii) cos y = Base\over Hypotenuse (in right-angled ∆ABD)

= BD\over AB = 20\over 12 = 5\over 3

Cot y = Base\over Perpendicular (in right ∆ABD)

=BD\over AB = 16\over 20 = 4\over 5

(iii) cos y = Hypotenuse\over perpendicular (in right-angled ∆ABD)

= BD\over AB = 20\over 12 = 5\over 3

Cot y = Base\over Perpendicular (in right ∆ABD)

= AB\over AD = 16\over 12 = 4\over3

Cosec2 y – coty = {5\over 3}^2 – {4\over 3}^2

= {25\over 9}{16\over 9} = {25-16\over 9} = {9\over 9} = 1

Hence, cosec2 y – coty = 1

Question 4 :

If θ is an acute angle and tan θ = 8\over 15, find the value of sec θ + cosec θ.

Solution

Given tan θ = 8\over 15

θ is an acute angle

In the figure triangle, OMP is a right-angled triangle,

∠M = 90o and ∠Q = θ

Tanθ = PM\over OL = 8\over 15

Therefore, PM = 8, OM = 15

But OP2 = OM2 + PM2 using Pythagoras theorem,

= 152 + 82 = 225 + 64 = 289 = 172

Therefore, OP = 17

Sec θ = OP\over OM = 17\over 15

Cosec θ = OP\over PM = 17\over 8

Now, Sec θ + cosec θ = (17\over 15) + (17\over 8)

= (136 + 255)\over 120 = 391\over 120

Question 5 :

Prove that

(i) (sec A + tan A) (1 – sin A) = cos A

(ii) (1 + tan2 A) (1 – sin A) (1 + sin A) = 1

Solution

(i) (sec A + tan A) (1 – sin A)

= ({1\over cos\ A}+{sin\ A\over cos\ A})(1-sin\ A)

= ({1+sin\ A\over cos\ A})(1-sin\ A)

= {(1+sin\ A)(1-sin\ A)\over cos\ A}

= {1-sin^2\ A\over cos\ A}

= {cos^2 A\over cos A} = cos A

(ii) (1 + tan2 A) (1 – sin A) (1 + sin A)

= (1 + tan^2 A)(1-sin^2 A)

= sec^2\ A \times cos^2\ A

= {1\over cos^2A}\times cos^2\ A =1

Question 6 :

An electric pole is 10 metres high. If its shadow is 10√3 metres in length, find the elevation of the sun.

Solution

Consider AB as the pole and OB as its shadow.

It is given that

AB = 10 m, OB = 10√3 m and θ is the angle of elevation of the sun.

We know that

tan θ = AB\over OB

Substituting the values

tan θ = 10\over 10√3 = 1\over √3

So we get

tan 30º = 1\over √3

θ = 30º

∴ Elevation of the sun = 30º

Question 7 :

What is the angle of elevation of the sun when the length of the shadow of a vertical pole is equal to its height?

Solution

Consider AB as the pole and CB as its shadow

θ is the angle of elevation of the sun

Take AB = x m and BC = x m

We know that

tan θ = AB\ove CB = x\over x = 1

So we get

tan 45º = 1

θ = 45º

∴ Elevation of the sun = 45º

Question 8 :

From point P on level ground, the angle of elevation of the top of a tower is 300. If the tower is 100 m high, how far is P from the foot of the tower?

Solution

Consider AB as the tower and P is at a distance of x m from B, which is the foot of the tower.

Height of the tower = 100 m

Angle of elevation = 300

We know that

tan θ = AB\over PB

Substituting the values

tan 300 = 100\over x

So we get

1\over √3 = 100\over x

By cross multiplication

x = 100√3

x = 100 (1.732) = 173.2 m

Hence, the distance of P from the foot of the tower is 173.2 m.

Question 9 :

A bridge across a river makes an angle of 450 with the river bank. If the length of the bridge across the river is 200 metres, what is the breadth of the river.

Solution

Consider AB as the width of river = x m

Length of bridge AC = 200 m

Angle with the river bank = 450

We know that

sin θ = AB/AC

Substituting the values

sin 450 = x/200

So we get

1/√2 = x/200

By cross multiplication

x = 200/√2

Multiplying and dividing by √2

x = 200/√2 × √2/√2

By further calculation

x = 200(1.414)/2

x = 100 (1.414)

x = 141.4 m

Hence, the breadth of the river is 141.4 m.

Question 10 :

In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x0 where tan x = 2/5 and AF = 200 m. The elevation of T from B, where AB = 80 m, is y0. Calculate:

(i) the height of the tower TF.

(ii) the angle y, correct to the nearest degree.

Solution

Consider the height of the tower TF = x

It is given that

tan x = 2\over 5, AF = 200 m, AB = 80 m

(i) In right triangle ATF

tan x0 = TF\overAF

Substituting the values

2\over5 = x\over 200

So we get

x = (2 × 200)\over 5

x = 400\over 5

x = 80 m

Hence, the height of the tower is 80 m.

(ii) In right triangle TBF

tan y = TF\over BF

Substituting the values

tan y = 80\over (200 – 80)

tan y = 80\over120

tan y = 2\over 3 = 0.6667

So we get

y = 330 41' = 340

## Acknowledgement

I would like to express my heartfelt gratitude to my Mathematics teacher, Mr./Ms. _______[Teacher's Name], and our esteemed principal, Mr./Ms. _______ [Principal's Name], for generously providing me with the golden opportunity to undertake this fascinating project on the topic of _______ [Project Topic]. Their unwavering support and guidance have been invaluable throughout this journey.

I extend my sincere thanks to my parents and friends whose constant encouragement and assistance played a crucial role in finalizing this project within the stipulated time frame. Lastly, I am grateful to all my supporters who have motivated me to complete this project before the deadline, pushing me to fulfil their expectations.

## Bibliography

To successfully complete my project file. I have taken help from the following websites and books

1. Flash Education – A helpful website with educational content. (Website: FlashEducation.online)
2. Wikipedia – An online encyclopedia where I found useful facts. (Website: Wikipedia)
3. NCERT Mathematics Book
4. Ganit Prakash Mathematics Book
5. ML Aggarwal Mathematics Book
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