Class 11_Chapter 3_Lecture_1.pdf

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Class XI
Mathematics
Chapter- 3
Trigonometric Functions
Lecture - 1
Dr. Pranav Sharma
Maths Learning Centre. Jalandhar.
Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar

youtube.com/@MathematicsOnlineLectures
Measurement of Angles
ANGLE When a ray 𝑶𝑨
⃗⃗⃗⃗⃗⃗ starting from its initial position 𝑶𝑨 rotates about its end point
𝑶 and takes the final position 𝑶𝑩, we say that angle 𝑨𝑶𝑩
(written as ∠𝑨𝑶𝑩) has been formed.
Here, 𝑶𝑨 and 𝑶𝑩 are respectively known as the initial side
and the terminal side of ∠𝑨𝑶𝑩, and the point 𝑶 is called its
vertex.
The amount of rotation from the initial side to the terminal
side is called the measure of the angle.
POSITIVE AND NEGATIVE ANGLES An angle formed by a
rotating ray is said to be positive or negative depending on
whether it moves in an anticlockwise or a clockwise direction
respectively.

MEASURING ANGLES Mainly, there are two systems of measuring angles.
(i) Degree Measure: The angle traced by a moving line about a point from its initial
position to the terminating position in making
𝟏
𝟑𝟔𝟎° the complete revolution of a circle is
said to have a measure of 𝟏 degree, written as 𝟏∘
.
The one‐sixtieth part of a degree, i.e.
𝟏
𝟔𝟎
°
, is called a minute, written as 𝟏′.
The one‐sixtieth part of a minute, i.e.,
𝟏′
𝟔𝟎
, is called a second, written as 𝟏′′
Also, we define 1 right angle = 𝟗𝟎∘
.
An angle measuring 𝟏𝟖𝟎𝐨
is called a straight angle.

(ii) Circular System (Radian Measure): In this system, an angle
is measured in radians.
A radian is an angle subtended at the centre of a circle by an arc
whose length is equal to the radius of the circle.
We denote 1 radian by 𝟏𝒄
.
A radian is a constant angle.
Let us consider a circle with centre 𝑶 and radius 𝒓.
Let 𝑨𝑩 be an arc of length 𝒓. Join 𝑶𝑨 and 𝑶𝑩.
Then, ∠𝑨𝑶𝑩 = 𝟏𝐜
. Produce 𝑨𝑶 to meet the circle at 𝑪.
Then, ∠𝑨𝑶𝑪 = 𝐚 straight angle = 𝟐𝐫𝐭. ∠𝐬. Since the angles at the
centre of a circle are proportional to the lengths of the arcs
subtending them, we have
∠𝑨𝑶𝑩
∠𝑨𝑶𝐂
=
𝐚𝐫𝐜𝑨𝑩
𝐚𝐫𝐜𝑨𝑩𝐂
=
𝒓
𝟏
𝟐
⋅(𝟐𝝅𝒓)
=
𝟏
𝝅
⇒ ∠𝑨𝑶𝑩 =
∠𝑨𝑶𝐂
𝝅
=
𝟐𝐫𝐭∠𝐬
𝝅
= constant ⇒ 𝟏𝐜
=constant. Hence, a radian is a constant angle.

In a circle of radius 𝒓, if an arc of length 𝒍 subtends an angle 𝜽𝐜
at the centre then 𝜽 =
𝒍
𝒓
.
Let us consider a circle with centre 𝑶 and radius 𝒓.
Let ∠𝑨𝑶𝑩 = 𝜽𝐜
and let arc 𝑨𝑩 = 𝒍.
Let 𝑪 be a point on 𝑨𝑩 such that arc 𝑨𝑪 = 𝒓.
Then, ∠𝑨𝑶𝐂 = 𝟏𝐜
.
We know that the angles at the centre of a circle are
proportional to the lengths of the arcs subtending them.
∠𝑨𝑶𝑩
∠𝑨𝑶𝑪
=
𝐚𝐫𝐜𝑨𝑩
𝐚𝐫𝐜𝑨𝐂
=
𝒍
𝒓
⇒
𝜽
𝟏
=
𝒍
𝒓
⇒ 𝜽 =
𝒍
𝒓
.

RELATION BETWEEN RADIAN AND DEGREE We know that a complete circle
subtends at its centre an angle whose measure is 𝟐𝝅 radians as well as 𝟑𝟔𝟎∘
.
(𝟐𝝅)𝐜
= 𝟑𝟔𝟎∘
. Hence, 𝝅𝐜
= 𝟏𝟖𝟎∘
.
TWO IMPORTANT CONVERSIONS
(i) 𝝅𝐜
= 𝟏𝟖𝟎∘
⇒ 𝟏𝐜
= (
𝟏𝟖𝟎
𝝅
)
∘
= (
𝟏𝟖𝟎×𝟕
𝟐𝟐
)
∘
= 𝟓𝟕∘
𝟏𝟔′
𝟐𝟏′′
(approx.)
(ii) 𝟏𝟖𝟎∘
= 𝝅𝐜
⇒ 𝟏∘
= (
𝝅
𝟏𝟖𝟎
)
𝐜
= (
𝟐𝟐
𝟕
×
𝟏
𝟏𝟖𝟎
)
𝐜
= (𝟎. 𝟎𝟏𝟕𝟒𝟔)𝐜
.
Degrees 𝟑𝟎∘
𝟒𝟓∘
𝟔𝟎∘
𝟗𝟎∘
𝟏𝟖𝟎∘
𝟐𝟕𝟎∘
𝟑𝟔𝟎∘
Radians 𝝅/𝟔 𝝅/𝟒 𝝅/𝟑 𝝅/𝟐 𝝅 𝟑𝝅/𝟐 𝟐𝝅
𝜽𝒄
is represented by the symbol 𝜽 only.

Find the degree measure corresponding to each of the following radian measures:
(i) (
𝟕𝝅
𝟏𝟐
)
𝐜
(ii) (
𝟑
𝟒
)
𝐜
(iii) (−𝟐)𝐜
(i) 𝝅𝐜
= 𝟏𝟖𝟎∘
⇒ 𝟏𝐜
= (
𝟏𝟖𝟎
𝝅
)
∘
⇒ (
𝟕𝝅
𝟏𝟐
)
𝐜
= (
𝟏𝟖𝟎
𝝅
×
𝟕𝝅
𝟏𝟐
)
∘
= 𝟏𝟎𝟓∘
.
(ii) 𝝅𝐜
= 𝟏𝟖𝟎∘
⇒ 𝟏𝐜
= (
𝟏𝟖𝟎
𝝅
)
∘
⇒ (
𝟑
𝟒
)
𝐜
= (
𝟏𝟖𝟎
𝝅
×
𝟑
𝟒
)
∘
= (𝟏𝟖𝟎 ×
𝟕
𝟐𝟐
×
𝟑
𝟒
)
∘
= 𝟒𝟐∘
𝟓𝟕′
𝟏𝟔′′
(iii) 𝝅𝐜
= 𝟏𝟖𝟎∘
⇒ 𝟏𝐜
= (
𝟏𝟖𝟎
𝝅
)
∘
⇒ (−𝟐)𝐜
= {
𝟏𝟖𝟎
𝝅
× (−𝟐)}
∘
= {𝟏𝟖𝟎 ×
𝟕
𝟐𝟐
× (−𝟐)}
∘
= −(𝟏𝟏𝟒∘
𝟑𝟐′
𝟒𝟒′′) .

Find the radian measure corresponding to each of the following degree measures:
(i) 𝟏𝟓∘
(ii) 𝟐𝟒𝟎∘
(iii) −𝟑𝟕∘
𝟑𝟎′
(i) 𝟏𝟖𝟎∘
= 𝝅𝐜
⇒ 𝟏∘
= (
𝝅
𝟏𝟖𝟎
)
𝐜
⇒ 𝟏𝟓∘
= (
𝝅
𝟏𝟖𝟎
× 𝟏𝟓)
𝐜
= (
𝝅
𝟏𝟐
)
𝐜
Hence, 𝟏𝟓∘
= (
𝝅
𝟏𝟐
)
𝐜
(ii) 𝟏𝟖𝟎∘
= 𝝅𝐜
⇒ 𝟏∘
= (
𝝅
𝟏𝟖𝟎
)
𝐜
⇒ 𝟐𝟒𝟎∘
= (
𝝅
𝟏𝟖𝟎
× 𝟐𝟒𝟎)
𝐜
= (
𝟒𝝅
𝟑
)
𝐜
Hence, 𝟐𝟒𝟎∘
= (
𝟒𝝅
𝟑
)
𝐜
(iii) 𝟏𝟖𝟎∘
= 𝝅𝐜
⇒ 𝟏∘
= (
𝝅
𝟏𝟖𝟎
)
𝐜
⇒ −𝟑𝟕∘
𝟑𝟎′
= {
𝝅
𝟏𝟖𝟎
× (
−𝟕𝟓
𝟐
)}
𝐜
= (
−𝟓𝝅
𝟐𝟒
)
𝐜
[−𝟑𝟕∘
𝟑𝟎′
= − (𝟑𝟕
𝟏
𝟐
)
∘
= − (
𝟕𝟓
𝟐
)
∘
]

Find in degrees the angle subtended at the centre of a circle of diameter 50 cm by an arc
of length 11 cm.
Here, 𝒓 = 𝟐𝟓 cm and 𝒍 = 𝟏𝟏 cm.
Let the measure of the required angle be 𝜽𝐜
.
Then, 𝜽𝐜
= (
𝒍
𝒓
)
𝐜
= (
𝟏𝟏
𝟐𝟓
)
𝐜
= (
𝟏𝟏
𝟐𝟓
×
𝟏𝟖𝟎
𝝅
)
∘
[ 𝝅𝐜
= 𝟏𝟖𝟎∘]
= (
𝟏𝟏
𝟐𝟓
×
𝟕
𝟐𝟐
× 𝟏𝟖𝟎)
∘
= (
𝟏𝟐𝟔
𝟓
)
∘
= 𝟐𝟓∘
𝟏𝟐′
. Hence, the required angle is 𝟐𝟓∘
𝟏𝟐′
.

Find the radius of a circle in which a central angle of 𝟕𝟐∘
intercepts an arc of length 22
cm.(Use 𝝅 =
𝟐𝟐
𝟕
.)
Let the required radius be 𝒓 cm. Then,
𝒍 = 𝟐𝟐 cm, and 𝜽 = 𝟕𝟐°
= (𝟕𝟐 ×
𝝅
𝟏𝟖𝟎
)
𝒄
= (
𝟐𝝅
𝟓
)
𝒄
Now, 𝜽 =
𝒍
𝒓
⇒ 𝒓 =
𝒍
𝜽
= (𝟐𝟐 ×
𝟓
𝟐𝝅
)cm
= (𝟐𝟐 ×
𝟓
𝟐
×
𝟕
𝟐𝟐
) 𝒄𝒎 =
𝟑𝟓
𝟐
𝒄𝒎 = 𝟏𝟕. 𝟐 𝒄𝒎

The minute hand of a watch is 1.4 cm long. How far does its tip move in 45 minutes?
(Use 𝝅 =
𝟐𝟐
𝟕
.)
In 60 minutes, the minutes hand moves through (𝟐𝝅)𝒄
.
(
𝟐𝝅
𝟔𝟎
×
In 45 minutes, the minutes hand moves through
𝟒𝟓)
𝒄
= (
𝟑𝝅
𝟐
)
𝒄
So, 𝒓 = 𝟏. 𝟒 𝒄𝒎 and 𝜽 = (
𝟑𝝅
𝟐
)
𝒄
.
Distance moved by the tip of the minute hand in 45
minutes is given by
𝒍 = 𝒓𝜽 = (𝟏. 𝟒 ×
𝟑𝝅
𝟐
) cm = (𝟏. 𝟒 ×
𝟑
𝟐
×
𝟐𝟐
𝟕
) cm = 𝟔. 𝟔 cm.

If the arcs of the same length in two circles subtend angles of 𝟔𝟎∘
and 𝟕𝟓∘
at their
respective centres, find the ratio of their radii.
Let 𝒓𝟏 and 𝒓𝟐 be the radii of the two circles. Then,
𝜽𝟏 = 𝟔𝟎∘
= (𝟔𝟎 ×
𝝅
𝟏𝟖𝟎
)
𝐜
= (
𝝅
𝟑
)
𝐜
,
and 𝜽𝟐 = 𝟕𝟓∘
= (𝟕𝟓 ×
𝝅
𝟏𝟖𝟎
)
𝐜
= (
𝟓𝝅
𝟏𝟐
)
𝐜
Let the length of each arc be 𝒍 cm. Then,
𝒍 = 𝒓𝟏𝜽𝟏 = 𝒓𝟐𝜽𝟐
⇒ (𝒓𝟏 ×
𝝅
𝟑
) = (𝒓𝟐 ×
𝟓𝝅
𝟏𝟐
) ⇒
𝒓𝟏
𝒓𝟐
= 𝟓/𝟒

Angles of a triangle are in 𝑨𝑷 and the ratio of the number of degrees in the least to the
number of radians in the greatest is 𝟔𝟎: 𝝅. Find angles in degrees and radians.
Let the angles of the triangle be (𝒂 − 𝒅)∘
, 𝒂∘
and (𝒂 + 𝒅)∘
. Then,
(𝒂 − 𝒅) + 𝒂 + (𝒂 + 𝒅) = 𝟏𝟖𝟎 ⇒ 𝟑𝒂 = 𝟏𝟖𝟎 ⇒ 𝒂 = 𝟔𝟎.
Thus, the angles are (𝟔𝟎 − 𝒅)∘
, 𝟔𝟎∘
and (𝟔𝟎 + 𝒅)∘
.
Number of degrees in the least angle = (𝟔𝟎 − 𝒅) .
Number of radians in the greatest angle = {(𝟔𝟎 + 𝒅) ×
𝝅
𝟏𝟖𝟎
}.
(𝟔𝟎 − 𝒅)
(𝟔𝟎 + 𝒅) ×
𝝅
𝟏𝟖𝟎
=
𝟔𝟎
𝝅
⇔ (𝟔𝟎 − 𝒅) × 𝝅 = 𝟔𝟎 × (𝟔𝟎 + 𝒅) ×
𝝅
𝟏𝟖𝟎
⇔ 𝟑(𝟔𝟎 − 𝒅) = (𝟔𝟎 + 𝒅) ⇔ 𝟒𝒅 = 𝟏𝟐𝟎 ⇔ 𝒅 = 𝟑𝟎.
the required angles are (𝟔𝟎— 𝟑𝟎)∘
, 𝟔𝟎∘
and (𝟔𝟎 + 𝟑𝟎)∘
, i.e. 𝟑𝟎∘
, 𝟔𝟎∘
and 𝟗𝟎∘
.
These angles in radians are(𝟑𝟎 ×
𝝅
𝟏𝟖𝟎
)
𝐜
, (𝟔𝟎 ×
𝝅
𝟏𝟖𝟎
)
𝐜
, (𝟗𝟎 ×
𝝅
𝟏𝟖𝟎
)
𝐜
, i.e., (
𝝅
𝟔
)
𝐜
, (
𝝅
𝟑
)
𝐜
, (
𝝅
𝟐
)
𝐜

In a right‐angled triangle, the difference between the two acute angles is (
𝝅
𝟏𝟓
)
𝐜
. Find the
angles in degrees.
Clearly, the sum of the two acute angles of a right triangle is 𝟗𝟎∘
.
Difference between the acute angles = (
𝝅
𝟏𝟓
)
𝐜
= (
𝝅
𝟏𝟓
×
𝟏𝟖𝟎
𝝅
)
∘
= 𝟏𝟐∘
.
Let the two acute angles be 𝒙∘
and 𝒚∘
. Then,
𝒙 + 𝒚 = 𝟗𝟎 (i) 𝒙 − 𝒚 = 𝟏𝟐 (ii)
Solving (i) and (ii), we get 𝒙 = 𝟓𝟏 and 𝒚 = 𝟑𝟗.
Hence, the angles of the triangle 𝐚𝐫𝐞𝟓𝟏∘
, 𝟑𝟗∘
and 𝟗𝟎∘
.

A wheel makes 360 revolutions in 1 minute. Through how many radians does it turn in
𝟏 second?
Number of revolutions made in 60 seconds = 𝟑𝟔𝟎.
Number of revolutions made in 1 second =
𝟑𝟔𝟎
𝟔𝟎
= 𝟔.
Angle moved in 1 revolution = (𝟐𝝅)𝐜
.
Angle moved in 6 revolutions = (𝟐𝝅 × 𝟔)𝐜
= (𝟏𝟐𝝅)𝐜
.

Find the angle between the minute hand and the hour hand of a clock when the time is
7.20.
Angle traced by the hour hand in 12 hours = 𝟑𝟔𝟎∘
.
Angle traced by it in 𝟕𝐡𝟐𝟎 𝐦𝐢𝐧 , i.e., in
𝟐𝟐
𝟑
hours = (
𝟑𝟔𝟎
𝟏𝟐
×
𝟐𝟐
𝟑
)
∘
= 𝟐𝟐𝟎∘
.
Angle traced by the minute hand in 60 𝐦𝐢𝐧 = 𝟑𝟔𝟎∘
.
Angle traced by the minute hand in 20 𝐦𝐢𝐧 (
𝟑𝟔𝟎
𝟔𝟎
× 𝟐𝟎)
∘
= 𝟏𝟐𝟎∘
.
Hence, the required angle between the two hands= (𝟐𝟐𝟎∘
− 𝟏𝟐𝟎∘) = 𝟏𝟎𝟎∘
.

A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping
the rope tight, and describes 88 metres when it traces 𝟕𝟐∘
at the centre, find the length of
the rope.
Let us denote the post by 𝑷 and let 𝑷𝑨 be the length of the rope in the tightest position.
Suppose that the horse moves along the arc 𝑨𝑩 so that ∠𝑨𝑷𝑩 = 𝟕𝟐∘
and arc 𝑨𝑩 = 𝟖𝟖𝐦.
Let the length of the rope 𝑷𝑨 be 𝒓 metres.
Then, 𝜽 = 𝟕𝟐∘
= (𝟕𝟐 ×
𝝅
𝟏𝟖𝟎
)
𝐜
= (
𝟐𝝅
𝟓
)
𝐜
and 𝒍 = 𝟖𝟖𝐦.
𝒓 =
𝒍
𝜽
=
𝟖𝟖
(𝟐𝝅/𝟓)
𝐦 = (𝟖𝟖 ×
𝟓
𝟐
×
𝟕
𝟐𝟐
) 𝐦 = 𝟕𝟎𝐦.
Hence, the length of the rope is 70 𝐦.

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