NCERT Solutions of class 11 maths chapter 3-Trigonometric functions
Here NCERT Solutions of class 11 maths of chapter 3 – Trigonometric functions are solved by an expert teacher of maths. All questions of chapter 3-Trigonometric functions of class 11 are solved by a step-by-step method to help the students in their preparation for the forthcoming exams.
The chapter Trigonometric functions of NCERT textbook of class 11 contains the questions based on the relation between radian and degree measures of angles, the length of the arc of the circle, verification of different kinds of Trigonometric identities
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NCERT Solutions of class 11 maths chapter 3-Trigonometric functions
Exercise 3.1 and 3.2- Trigonometric Functions
Exercise 3.3 – Trigonometric Functions
Miscellaneous Exercise Chapter 3-Trigonometric Functions
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NCERT Solutions of class 11 maths chapter 3 -Trigonometric Functions
The Relation Between Radian and Degree →
A circle makes the angle(θ) of 360° around its center and an arc of the length(l)= 2πr(circumference)
Which shows that 1° = 2Π/360
Exercise 3.1
Q1. Find the radian measure corresponding to the following degree measures.
(i) 25° (ii) –47°30′ (iii) 240° (iv) 520°
Answer.(i) Using the following relationship between rad. and deg. in all questions.
(ii) –47°30′ = –47°+ 30/60(1°= 60′)
Q2.Find the degree measures of the following radian measures.
Answers. Using the following relationship between radian and degree in all questions.
Substituting π = 22/7
(ii)–4 rad.= –4×180/π
Therefore –4 rad.=–229°5’27”
See the video for Q1,Q2,Q3 and Q4
Q3.A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second.
Answer.1 revolution is equivalent to = 360°=2π rad.
In 1 minute the wheel makes = 360 revolution
360 revolution is equivalent to = 360 ×2π = 720π rad.
1 minute = 60”
So, in 1second the wheel will turn=720π/60=12π rad.
Therefore in 1 second, the wheel will turn 12π rad.
Q4.Find the degree measure of the angle subtended at the center of the circle of radius 100 cm by an arc of length 22 cm.
Anwer. Let the given arc subtend an angle of θ.
1 rad. =180/π, where π =22/7
θ= 12°36”
Q5. In a circle diameter 40 cm, the length of a chord is 20 cm, find the length of the minor arc of the chord.
Answer. The diameter of circle = 40 cm, the length of the chord =20 cm,the radius =40/2=20 cm
Drawing a chord AB in a circle of center O and joining its both ends to the center,
in ΔOAB, OA =OB(radii of same circle), OA =OB =OC=20 cm
therefore θ = 60°(ΔOAB is an equilateral Δ)
Q6.In two circles, arcs of the same length subtend angles 60° and 75 ° at the center,find the ratio of thier radii.
Answer. Let r₁ and r₂ are the radii of the circles having the arcs subtending the angles 60°and 75° respectively.
Applying the formula θ = l/r, l=length of the arc and r=radius of the circle
Therefore the ratio between radii of the circles will be 5:4.
Q7.Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length.
(a)10 cm (b) 15 cm (c) 21 cm
(a) Answer. Length of pendulum makes a radius of arcs formed by it
So,r =10 cm, l = 10 cm
So, the angle in radian will be 2/15 rad. by which pendulum swings.
All questions (b) and (c) you can solve in a similar way.
Exercise-3.2
Q1.Find the value of 5 other trigonometric functions if cos x = -1/2 if x lies in third quadrant.
Answer.
The value of sinx is negative at third quadrant, so
Q2. Find the value of other 5 trigonometric functions if ,x lies in the second quadrant.
Answer.
x lies in second quadrant and value of cosx is negative at second quadrant,so
Q3. Find the value of other 5 trigonometric functions,if cotx = ,x lies in third quadrant.
Answer.
Since,x lies in third quadrant and value of secx is negative in third quadrant.
Q4.Find the value of other 5 trigonometric functions,if secx= , x lies in fourth quadrant.
Answer.
x lies in fourth quadrant and value of sinx is negative at 4 th quadrant.
Therefore
Q5. Find the value of other 5 trigonometric function,if tanx =, x lies in the second quadrant.
Answer.
x lies in 2nd quadrant and value of cosecx is + ve in 2nd quadrant.
therefore
Q6. Find the value of the trigonometric function sin765°.
Answer.
As we know sin(2nπ + θ) = sinθ, where n∈ N, means the value of sin repeats after every interval of 2π.
2π = 360°
So dividing 765 by 360 ,we get
765 = 2 ×360 + 45= 2×2π+ 45
sin765 =sin(4π + 45) =sin45
Therefore
Q7.Find the value of the trigometric function cosec(-1410°) .
Answer.
Since ,sin(–θ) = –sinθ
∴cosec(–θ) = –cosecθ
cosec(–1410) = –cosec(1410)
Moreover ,sin(2nπ +θ) = sinθ
On the same way cosec(2nπ +θ) = cosecθ(n∈N) after every interval of 2π, the value of cosecθ also repeats like sinθ.
Therefore dividing 1410 by 360(i.e 2π) ,we get
1410 = 4 ×360 –30 =8π –π/6
Q8. Find the value of the trigonometric function .
Answer.
As we know the value of tan repeats after an interval of π i.e tan so dividing 19π by 3 we get
Q9.Find the value of the trigonometric function .
Answer.As we know
Since the value of sin repeats after an interval of 2π(360°) ,
11π/3 =11×180/3=660° ,dividing it by 360,we get
660=2×360–60=2×2π –π/3= 4π –π/3
As we know sin(–θ)= –sinθ, so
Q10.Find the value of the trigonometric function ..
Answer.
As we know cot(–θ) = –cotθ,therefore
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