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NCERT Solutions for Exercise 8.3 Class 10 Maths Chapter 8 - Introduction to Trigonometry

NCERT Solutions for Exercise 8.3 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Edited By Ramraj Saini | Updated on Nov 27, 2023 09:11 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.3

NCERT Solutions for Exercise 8.3 Class 10 Maths Chapter 8 Introduction to Trigonometry are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 8.3 gives a brief overview of the trigonometric ratios of complementary angles and is explained in trigonometry. If the sum of two angles equals 90°, they are said to be complementary. According to the angle sum property, one angle in a right-angled triangle measures 90°, and the sum of the other two angles also measures 90°. This is the concept from which the formulas for this exercise are derived or the formula of complementary angle. The majority of these questions require proof. As a result, students must carefully study the theory in order to provide detailed answers to these sums.

10th class Maths exercise 8.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Introduction to Trigonometry class 10 chapter 8 Exercise: 8.3

Q1 Evaluate :

(i)\frac{\sin 18^{o}}{\cos 72^{o}}

Answer:

\frac{\sin 18^{o}}{\cos 72^{o}}
We can write the above equation as;
=\frac{\sin (90^0-72^0)}{\cos 72^0}
By using the identity of \sin (90^o-\theta) = \cos \theta
Therefore, \frac{\cos 72^0}{\cos 72^0} = 1

So, the answer is 1.

Q1 Evaluate :

(ii) \frac{\tan 26^{o}}{\cot 64^{o}}

Answer:

\frac{\tan 26^{o}}{\cot 64^{o}}
The above equation can be written as ;

\tan (90^o-64^o)/\cot 64^o .........(i)
It is known that, \tan (90^o-\theta) = \cot \theta
Therefore, equation (i) becomes,
\cot64^o/\cot 64^o = 1

So, the answer is 1.

Q1 Evaluate :

(iii) \cos 48^{o}-\sin 42^{o}

Answer:

\cos 48^{o}-\sin 42^{o}
The above equation can be written as ;
\cos (90^o-42^{o})-\sin 42^{o} ....................(i)
It is known that \cos (90^o-\theta) = \sin \theta
Therefore, equation (i) becomes,
\sin42^{o}-\sin 42^{o} = 0

So, the answer is 0.

Q1 Evaluate :

(iv) cosec \: 31^{o}-\sec 59^{o}

Answer:

cosec \: 31^{o}-\sec 59^{o}

This equation can be written as;
cosec 31^o - \sec(90^o-31^o) .................(i)
We know that \sec(90^o-\theta) = cosec \theta

Therefore, equation (i) becomes;
cosec 31^o - cosec\ 31^o = 0

So, the answer is 0.

Q2 Show that :

(i) \tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1

Answer:

\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1
Taking Left Hand Side (LHS)
= \tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}
\Rightarrow \tan 48^{o}\tan 23^{o}\tan (90^o-48^{o})\tan (90^o-23^{o})
\Rightarrow \tan 48^{o}\tan 23^{o}\cot 48^{o}\cot23^{o} [it is known that \tan (90^0-\theta = \cot\theta) and \cot\theta\times \tan \theta =1
=1

Hence proved.

Q2 Show that :

(ii) \cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0

Answer:

\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0

Taking Left Hand Side (LHS)
= \cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}
= \cos 38^{o}\cos (90^o-38^{o})-\sin 38^{o}\sin (90^o-38^{o})
= \cos 38^{o}\sin38^{o}-\sin 38^{o}\cos 38^{o} [it is known that \sin(90^0-\theta) =\cos \theta and \cos(90^0-\theta) =\sin \theta ]
= 0

Q3 If \tan 2A= \cot (A-18^{o}) , where 2A is an acute angle, find the value of A .

Answer:

We have,
\tan 2A = \cot (A - \18^{0} )
we know that, \\\cot (90^0-\theta) = \tan\theta
\\\Rightarrow \cot (90^0-2A) = \cot (A-18^0)\\\Rightarrow 90^0-2A = A - 18^0\\\Rightarrow 3A = 108\\\Rightarrow A = 108/3 = 36^0

Q4 If \tan A= \cot B , prove that A+B= 90^{o} .

Answer:

We have,
\tan A= \cot B
and we know that \tan (90^0 - \theta)= \cot \theta
therefore,
\tan A= \tan(90^0- B)
A = 90 - B
A + B = 90
Hence proved.

Q5 If \sec 4A= cosec (A-20^{o}) , where 4A is an acute angle, find the value of A .

Answer:

We have,
\sec 4A= cosec (A-20^{o}) , Here 4A is an acute angle
According to question,
We know that cosec(90^0-\theta)= \sec \theta
cosec(90^0-4A)= cosec (A-20^{o})

\\\Rightarrow 90 - 4A = A - 20\\\\ \Rightarrow 5A=110\\\\ \Rightarrow A=\frac{110}{5}\\\:\\ \Rightarrow A=22^o

Q6 If A,B and C are interior angles of a triangle ABC , then show that

\sin (\frac{B+C}{2})= \cos \frac{A}{2}

Answer:

Given that,
A, B and C are interior angles of \Delta ABC
To prove - \sin (\frac{B+C}{2})= \cos \frac{A}{2}

Now,
In triangle \Delta ABC ,
A + B + C = 180^0
\Rightarrow B + C = 180 - A
\Rightarrow B + C/2 = 90^0 - A/2
\sin \frac{B+C}{2}=\sin (90^0-A/2)
\sin \frac{B+C}{2}=\cos A/2
Hence proved.

Q7 Express sin 67^{o}+\cos 75^{o} in terms of trigonometric ratios of angles between 0^{o} and 45^{o} .

Answer:

By using the identity of \sin\theta and \cos\theta
sin 67^{o}+\cos 75^{o}
We know that,
\sin(90-\theta) =\cos \theta and \cos(90-\theta) =\sin \theta
the above equation can be written as;
=\sin (90^0-23^0)+\cos(90^0-15^0)
=\sin (15^0)+\cos(23^0)

More About NCERT Solutions for Class 10 Maths Exercise 8.3

NCERT solutions for Class 10 Maths exercise 8.3- We can find the value of these complementary angles but in the examination, we have very little time. So, we should try to remember these formulas NCERT solutions for Class 10 Maths chapter 8 exercise 8.3 as these can break a large complex trigonometric equation into simpler ones that can be called easily by the other associated terms present in the function.

The formulas are:

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

for all values of angle, A lying between 0° and 90°.

We should always be careful about

tan (0) = 0 = cot (90)

sec (0) = 1 cosec (90) and sec (90), cosec(0), tan(90) and cot (0) are not defined. Students can use these Introduction to Trigonometry Class 10 notes to quick revision of the important concepts discussed in this chapter.

Benefits of NCERT Solutions for Class 10 Maths Exercise 8.3

  • Exercise 8.3 Class 10 Maths, is based on the main concept of Trigonometric Ratios of Complementary Angles.
  • Class 10 Maths chapter 8 exercise 8.3 helps in solving and revising all questions of the previous exercises. It shows us two different aspects of solving the same problem by using complementary angels.
  • Mastering the complementary angle formulas of Class 10 Maths chapter 8 exercise 8.3 can help in simplifying complex trigonometric questions into simpler ones.

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Dear aspirant !

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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.

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Hope this information helps you.

Hello aspirant,

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.

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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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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

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200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

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20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

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K/2\,

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\; K\;

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In the reaction,

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Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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0.02

Option 2)

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1.25 × 10-2

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2.5 × 10-2

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Option 1)

decrease twice

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remain unchanged

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be a function of the molecular mass of the substance.

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Fraction of solute present in water

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twice that in 60 g carbon

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6.023 × 1022

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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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less than 3

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