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NCERT exemplar Class 10 Maths solutions chapter 8- Introductions to Trigonometry and Its Equations- NCERT exemplar Class 10 Maths solutions chapter 8 introduces some ratios of sides in a right-angle triangle, which are called Trigonometric ratios. The learnings of this chapter help solve complex problems at later stages in life, whether it be in engineering or architecture. The NCERT exemplar Class 10 Maths chapter 8 solutions follow an in-depth approach and provide solutions at a comprehensive level, enabling the students to study the NCERT Class 10 Maths effectively.
These Class 10 Maths NCERT exemplar chapter 8 solutions, due to their thorough approach, helps the students to build vital concepts of trigonometry and its equations. The NCERT exemplar Class 10 Maths solutions chapter 8 follow the CBSE Syllabus for Class 10 and covers all the topics prescribed by CBSE.
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Question:1
If cos A =4/5 , then the value of tan A is
(b) (c) (d)
Answer:
Answer. [B]
Solution. It is given that cos A = 4/5
value of base = 4
Hypotenuse = 5
Use Pythagoras theorem in ABC
(Hypotenuse)^{2} = (Base)^{2} + (perpendicular)
That is value of perpendicular is 3
Hence option (B) is correct.
Question:2
If sin A =1/2 then the value of cot A is
a) (b) (c) (d) 1
Answer:
Hence option (a) is correct.
Question:3
The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
(A) – 1 (B) 0 (C) 1 (D)
Answer:
Answer. [B]
Solution. Given expression is :
[cosec(75° + q) – sec (15° – q) – tan (55° + q) + cot (35° – q)]
[cosec(90 – (15 – q)) – sec (15° – q) – tan (90 – (35 – q)) + cot (35° – q)]
{ we can write (75 + q) = (90 – (15 – q)) and (55 + q) = (90 – (35 – q))}
{ cosec (90 – q) = sec q and tan (90 – q) = cot q }
[sec(15 – q) – sec (15 – q) – cot (35 – q) + cot (35 – q)]=0
Hence option (B) is correct
Question:4
Given that sinθ =a/b , then cosθ is equal to
(A) (B) (C) (D)
Answer:
Answer. [C]
Solution. It is given that sin = a/b
Hence option (C) is correct.
Question:5
If cos (α + β) = 0, then sin (α – β) can be reduced to
(A) cos β (B) cos 2β (C) sin α (D) sin 2α
Answer:
Hence option B is correct.
Question:6
The value of (tan1° tan2° tan3° ... tan89°) is
(A) 0 (B) 1 (C) 2 (D)1/2
Answer:
Answer. [B]
Solution. Given :-tan1° tan2° tan3° ... tan89°
tan1° tan2° tan3° ... tan89°tan87° tan 88° tan89° …(1)
We can also write equation (1) in the form of
[tan (90^{0} – 89^{0}) . tan (90^{0} – 88^{0}). tan (90^{0} – 87^{0}) …… tan 87° . tan 88° tan 89°]
[ we can write tan 1° in the form of tan (90^{0} – 89^{0}) similarly we can write other values]
[cot 89^{0} . cot 88^{0}. cot 87^{0} …. tan 87^{0} . tan 88^{0} . tan 89^{0}]
[tan (90^{2} – ) = cot ]
Also
Throughout all terms are cancelled by each other and remaining will be tan45
Hence the value is 1
option B is correct.
Question:7
If cos 9α = sinα and 9α < 90° , then the value of tan5α is
(A) (B) (C) 1 (D) 0
Answer:
Answer. [C]
Solution. Given :- cos 9 = sin
cos9 = cos(90 – )
(cos (90 – ) = sin)
9 = 90 –
9+ = 90
10 = 90
Now tan 5 is
Put = 9 we get
tan 5 (9)
tan 45^{0}
= 1
{ from the table of trigonometric ratios of angles we know that tan 45^{0} = 1}
Hence option C is correct.
Question:8
If ΔABC is right angled at C, then the value of cos (A+B) is
Answer:
Answer. [A]
Solution. It is given that C = 90°
In ABC
A +B +C = 180 [ sum of interior angles of triangle is 180°]
A + B + 90o = 180 [ C = 90° (given)]
A + B = 180^{0} – 90^{0}
A + B = 90° …(1)
cos(A +B) = cos (90°)
cos(90°) = 0
[ from the table of trigonometric ratios of angles we know that cos 90° = 0]
Hence option A is correct.
Question:9
If sinA + sin^{2}A = 1, then the value of the expression (cos^{2}A+ cos^{4}A) is
(A) 1 (B)1/2 (C) 2 (D) 3
Answer:
Answer. [A]
Solution. It is given that sinA + sin^{2}A = 1 …(*)
sinA = 1 – sin^{2}A
sinA = cos^{2}A …(1) ( 1 – sin^{2}A = cos^{2}A)
Squaring both sides we get
sin^{2} A = cos^{4}A …(2)
Hence cos^{2}A + cos^{4}A =
= sinA + sin^{2}A {using (1) and (2)}
= sinA + sin^{2}A = 1 (given)
Hence option (A) is correct.
Question:10
Given that sinα =1/2 and cosβ =1/2 , then the value of (α + β) is
(A) 0° (B) 30° (C) 60° (D) 90°
Answer:
Answer. [D]
Solution.
Hence option (D) is correct.
Question:11
The value of the expression
(A) 3 (B) 2 (C) 1 (D) 0
Answer:
Answer. [B]
Solution.
= 1+ 1 =2
Hence option (B) is correct.
Question:13
If sinθ – cosθ = 0, then the value of (sin^{4}θ + cos^{4}θ) is
(A) 1 (B)3/4 (C) 1/2 (D) 1/4
Answer:
squaring both sides we get
( (a – b)^{2} = a^{2} + b^{2} – 2ab)
Squaring both sides we get
…(2)
Now squaring both side of equation (1) we get
(Use equation (2))
Hence option (C) is correct.
Question:14
sin (45° + θ) – cos (45° – θ) is equal to
(A) 2cosθ (B) 0 (C) 2sinθ (D) 1
Answer:
Answer. [B]
Solution. Here
:
Sin[90° - (45°- θ)] – cos(45°- θ)
Cos(45°- θ) – cos(45°- θ) [ sin (90 – θ) = cosθ]
= 0
Hence option (B) is correct
Question:15
A pole 6 m high casts a shadow 2 m long on the ground, then the Sun’s elevation is
(A) 60° (B) 45° (C) 30° (D) 90°
Answer:
Answer. [A]
Solution. Given :
height pole = 6 m
Shadow of pole =
Now make figure according to given condition
Let angle of elevation is
a = 60°
Hence the Sun's elevation is 60°.
Question:1
Write ‘True’ or ‘False’ and justify your answer in each of the following:
Answer:
Taking L.H.S.
Hence L.H.S. = R.H.S.
So, the given expression is true.
Question:2
The value of the expression (cos^{2} 23° – sin^{2} 67°) is positive.
Answer:
Answer. [False]
Solution. (cos^{2} 23° – sin^{2} 67°)
=
= Sin^{2}67° - sin^{2}67° (cos(90-θ) = sin θ)
= 0
Hence the value of the expression is neutral
So, the given statement is false.
Question:3
The value of the expression (sin 80° – cos80°) is negative.
Answer:
Answer. [False]
Solution. (sin 80° – cos 80°)
We know that from 0 to 90° sin and cos both are positive i.e.
(always positive)
(always positive)
At 45° both the values of sin and cos are the same but after 45° to 90° value of sin is greater than the value of cos Hence sin 80° > cos 80°.
If we subtract a smaller term from bigger than the result is positive.
Hence (sin 80° – cos80°) > 0
So, the given statement is false
Question:5
If cosA + cos^{2}A = 1, then sin^{2}A + sin^{4}A = 1.
Answer:
Given
cosA + cos^{2}A = 1 …(1)
cos A = 1 – cos^{2}A
cosA = sin^{2}A …(2) ( sin^{2} = 1 – cos^{2})
L.H.S.
(from (2))
cosA + cos^{2}A
= 1 (R.H.S.) (from (1))
Hence sin^{2}A + sin^{4}A = 1
So, the given statement is true.
Question:6
(tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec^{2}θ.
Answer:
(tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec^{2}θ
Taking L.H.S.
(tanθ + 2) (2 tan θ + 1)
tanθ.(2tan θ+1) + 2(2tan θ +1)
We know that
Put the above value in (1)we get
L.H.S. R.H.S.
Hence the given expression is false.
Question:7
Answer:
Answer. [False]
Solution. Let us take 2 cases.
Case 1:
Case 2 :
is the angle when length is small and is the angle when shadow length is increased.
for finding , find tan
In both the case height of the tower is the same but in case 2 length of the shadow is increased and if the length of shadow increased value of decreased.
Hence the given statement is false.
Question:8
Answer:
Answer. [False]
Solution. According to question.
In the figure is the angle of elevation and is the angle of depression of the cloud.
For finding find tanq in ECD
Find tan in ECB for
Here we found that and both are different.
Hence the given statement is false.
Question:9
The value of 2sinθ can be a+1/a , where a is a positive number, and a ≠ 1.
Answer:
Answer. [False]
Solution. We know that
-1≤ sin θ ≤ 1
Multiply by 2.
-2≤ 2 sin θ ≤ 2
Here we found that value of 2 sin is lies from – 2 to 2.
But if we take a > 0 and a 1 then
For example a = 3
3 + 1/3 = 3.33
Hence is always greater than 2 in case of positive number except 1
But value of 2 sin is not greater than 2
Hence the given statement is false
Question:10
where a and b are two distinct numbers such that ab> 0.
Answer:
We know that
We also know that
a^{2} + b^{2} – 2ab > 0
We observe that is always greater than 2ab.
Hence,
Because if we divide a big term by small then the result is always greater than 1.
cos is always less than or equal to 1
Hence the given statement is false.
Question:11
Answer:
Answer. [False]
Solution. According to question
Case: 1
Here BC is the tower.
Let the height of the tower is H and distance AB = a
In ABC
Case :2 When height is doubled
Here ED = a
In
(from (1))
But (If the angle is double)
hence the given statement is false.
Question:12
Answer:
Answer. [True]
Solution. According to question
In case-1. Height is H and observation distance is a.
In case-2, both height and observation distance is increased by 10%.
In case -1
.....(1)
In case -2
from equation (1) and (2) we observe that
Hence the given statement is true.
Question:2
Answer:
Solution.
Taking L.H.S.
Taking L.C.M.
L.H.S. = R.H.S.
Hence proved
Question:3
If tan A = 3/4 , then show that sinAcos A = 12/25 .
Answer:
Given:-
To prove:-
we know that
P=3, B = 4
Using Pythagoras theorem
H = 5
we know that
Hence
Put the value of sinA and cosA in equation (1)
L.H.S. = R.H.S.
Hence proved.
Question:4
Prove the following : (sin α + cos α) (tan α + cot α) = sec α + cosec α
Answer:
Solution.
(sin α + cos α) (tan α + cot α) = sec α + cosec α
Taking L.H.S.
Taking L.C.M.
by separately divide
L.H.S. = R.H.S.
Hence proved.
Question:7
Prove the following: tan θ + tan (90° – θ) = sec θ sec (90° – θ)
Answer:
Solution.
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
Taking L.H.S.
= tan θ + tan (90° – θ)
= tanθ + cotθ ( tan (90 – θ) = cot θ)
Taking L.C.M.
L.H.S. = R.H.S.
Hence proved.
Question:8
Find the angle of elevation of the sun when the shadow of a pole h metres high is h metres long.
Answer:
Answer. [30°]
Solution. According to question
Here BC is the height of the pole i.e. h meters and AB is the length of shadow i.e. .
For finding angle q we have to find tanq in ABC
Hence angle of elevation is 30°.
Question:9
If tan θ = 1, then find the value of sin^{2}θ – cos^{2}θ.
Answer:
Given :
(Because θ = 30^{0})
Taking L.C.M.
Question:10
Answer:
Length of ladder = 15 m
The angle between wall and ladder = 60°
Let the height of wall = H
In ABC C = 60°, B = 90°
We know that
A + B + C = 180° (Sum of interior angles of a triangle is 180)
A + 90 + 60 = 180°
A = 30°
In ABC
Hence the height of the wall is 7.5 m
Question:12
If 2sin^{2}θ – cos^{2}θ = 2, then find the value of θ.
Answer:
2sin^{2}θ – cos^{2}θ = 2
Hence value of is 90°
Question:14
Answer:
Answer. [45°]
Solution. According to the question.
In EDC EC = 20.5 m
DC = 20.5 m
To find angle in EDC we need to find tan.
Hence the angle of elevation is 45°.
Question:15
Show that tan^{4}θ + tan^{2}θ = sec^{4}θ – sec^{2}θ.
Answer:
Taking L.H.S.
tan^{4}θ + tan^{2}θ
(tan^{2}θ) + tan^{2}θ…(1)
We know that sec^{2}θ – tan^{2}θ = 1
Put
in (1)
LHS = RHS
Hence proved
Question:1
If cosecθ + cotθ = p, then prove that
.
Answer:
Given: cosecθ + cotθ = p …(1)
Taking right hand side.
Put value of p from equation (1) we get
[by taking LCM]
which is equal to the eft-hand side
Hence proved.
Question:3
Answer:
Answer. [27.322 m]
Solution.
The angle of elevation of the top of a tower AB from certain point C is 30°
Let observer moves from C to D that is CD = 20m
Now angle of elevation increased by 15° that is 45° on point D
In ABD
BD = AB …(1)
In ABC
By cross multiplication we get
Now put the value of BD from equation (1) we have
AB = 27.322
Hence the height of the tower is 27.322 m
Question:4
If 1 + sin^{2}θ = 3sinθ cosθ, then prove that tanθ = 1 or1/2
Answer:
Solution. Given : 1 + sin^{2}θ = 3sinθ cosθ
To Prove - tanθ = 1 or 1/2
Dividing both side by sinθ we get
We know that
Hence proved.
Question:5
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
Answer:
Solution. Given:- sinθ + 2cosθ = 1
squaring both sides we have
To prove :
Taking the left-hand side
On squaring equation (2) we get
[use equation (1)]
= 4
So here we get the value of (2sin – cos)^{2} is 4
Hence proved
Question:6
Answer:
Solution. According to question
Let the height of tower = h
the distance of the first point from its foot = s
the distance of the second point from its foot = t
Multiply equation (1) and (2) we get
Hence proved.
Question:7
Answer:
Solution. According to question
Let the height of tower = h
[by cross multiplication]
[from equation (1)]
Question:8
Answer:
According to question
Here h is the height of flagstaff AD.
Let is the height of the tower
and be the angle of elevation of the bottom and the top of the flagstaff.
In BDC
In ABC
equate equation (1) and (2) we get
[by cross multiplication]
Hence Proved
Question:9
If tanθ + secθ =, then prove that
Answer:
Solution. Given : tanθ + secθ =l
There fore
(R.H.S)
Hence proved
Question:10
If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p^{2} – 1) = 2p.
Answer:
Solution. Given :-sinθ + cosθ = p
and secθ + cosecθ = q
To prove :-q (p^{2} – 1) = 2p
Taking left hand side
q.(p^{2}– 1) =
Put value of q and p we get
=
2p (R.H.S)
Hence proved.
Question:11
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ =
Answer:
Solution. Given:- asinθ + b cosθ = c
squaring both side we get
To prove : acosθ – b sinθ =
Taking left hand side : a cosθ – b sinθ and square it we get
Hence
Hence proved.
Question:12
Answer:
Solution To prove :-
Taking left hand side
Multiply nominator and denominator by (1 – sin )
Hence proved
Question:13
Answer:
Solution. According to the question
Here 30 m is the length of tower AB.
Let h is the height of tower DC
Let the distance between them is x
In ABC
In BDC
(using (1))
h = 10m
Hence the height of the second tower is 10
Distance between them
.
Question:14
Answer:
According to question
Let x and y are two objects and and use the angles of depression of two objects.
In AOX
In AOY
(from equation (1))
Hence the distance between two objects is
.
Question:15
Answer:
Solution. According to the question:-
Here a and b be the angles of indication when the ladder at rest and when it pulled away from the wall
In AOB
Similarly In DOC
Now subtract equation (1) from (3) we get
OC – OB = DC cos – AB cos
Here OC – OB = P
and DC = AB because length of ladder remains P = AB cos
– AB cos
P = AB (cos – cos) …(5)
Subtract equation (4) from (2) we get
AO – OD = AB sina – DC sin
Here AO – OD = q
and AB = DC because length of ladder remains same
q = AB sin – AB sin
q = AB (sin – sin ) …(6)
on dividing equation (5) and (6) we get
Hence proved
Question:16
Answer:
Solution. According to question
Let h is the height of the tower
In ABE
In EDC
from equation (1) and (2)
Hence the height of the tower
Question:17
Answer:
Solution. According to the question :
Let the height of the other house is X.
In DCF.
In DAB
.
from equation (1) and (2)
separately divide
Hence proved
Question:18
Answer:
Solution
Let y be the height of the balloon from the second window
In AOB
In OCD
Equating equation (1) & (2) we get
y = 2
Height of balloon = 2 + 4 + y
= 2 + 4 + 2
= 8m
These Class 10 Maths NCERT exemplar chapter 8 solutions provide an introduction to the trigonometric ratios and identities. These trigonometric ratios are defined for the acute angle of the right-angle triangle. Sine X is the ratio of perpendicular side and hypotenuse. Cos X is the ratio of base side and hypotenuse. Tan X is the ratio of perpendicular side and base. A variety of practice problems provided in the exemplar can brush up the skills of the student and enhance the skills related to Trigonometry and its equations based practice problems.
The NCERT exemplar Class 10 Maths Chapter 8 solutions Introduction to Trigonometry and Its Equations consists of a plethora of questions along with detailed solutions and is sufficient to prepare the student to attempt other books such as NCERT Class 10 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths, A textbook of Mathematics by Monica Kapoor et cetera.
NCERT exemplar Class 10 Maths solutions chapter 8 pdf download is a unique feature for the students to provide them with an uninterrupted learning experience as it offers the pdf version of the solutions which can be used while attempting NCERT exemplar Class 10 Maths chapter 8.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
We know that sine is the ratio of perpendicular and hypotenuse.
We know that cos is the ratio of base and hypotenuse.
Therefore, we can say that for complementary angles sine and cosine will give the same values.
We know that sine is the ratio of perpendicular and hypotenuse. Hypotenuses cannot have a smaller length on the perpendicular side; hence, the maximum possible value can be one.
The chapter Introduction to Trigonometry & Its Equations is quite important for Board exams as it carries around 8-10% weightage of the whole paper.
Generally, MCQs, Very short, Short, and Long answers type of questions are asked in the board examinations and NCERT exemplar Class 10 Maths solutions chapter 8 are adequate to score well in this chapter.
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Hope you get it !
Thanking you
Hello aspirant,
The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.
To get the complete datesheet, you can visit our website by clicking on the link given below.
https://school.careers360.com/boards/cbse/cbse-date-sheet
Thank you
Hope this information helps you.
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The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.
You don't need to worry. The class 7th paper will be simple and made by your own school teachers.
Thank you
Hope it helps you.
The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.
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