Exercise 8.1 solved Archives – Mathemafia https://mathemafia.com/tag/exercise-8-1-solved/ Mathematics is like music - Enjoy it!!! Sat, 08 Mar 2025 07:04:45 +0000 en-US hourly 1 https://wordpress.org/?v=7.1 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png Exercise 8.1 solved Archives – Mathemafia https://mathemafia.com/tag/exercise-8-1-solved/ 32 32 Exercise 8.1 (NCERT) – Introduction to Trigonometry, Class 10 – All Solutions https://mathemafia.com/class-10-trigonometry-exercise-8-1-all-solutions-ncert/ Thu, 06 Feb 2025 11:51:34 +0000 https://mathemafia.com/?p=9799 Complete solution of Exercise 8.1, Chapter 8, Introduction to Trigonometry of NCERT book of Mathematics for Class 10. The solutions are written in pure mathematical way with mobile-friendly view, fully responsive. The free YouTube video tutorial of every question is embedded after the written solution. This educational content has been created with utmost care to make it easily readable and hence understandable by the students of class 10, especially preparing for the final board exams.

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Exercise 8.1 (NCERT) - Trigonometry, Class 10

Chapter 8 Exercise 8.1 Class 10

Below are the Quick links for all questions of Exercise 8.1, Trigonometry, Class 10.

Click a link to view the solution of corresponding question.

Below are the solutions of every question of exercise 8.1 of chapter 8, Introduction to Trigonometry, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.Β Β 

Question 1

In β–³ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :Β (i) sin A, cos A (ii) sin C, cos C

Question 1, Exercise 8.1, Chapter 8, Trigonometry
𝐴𝐡=24π‘π‘š,𝐡𝐢=7π‘π‘š

∡ β–³ ABC is a right angled triangle, let us apply pythagoras theorem

𝐴𝐢2𝐴𝐢=𝐴𝐡2+𝐡𝐢2=242+72=576+49=625=625β€Ύβ€Ύβ€Ύβ€Ύβˆš=25π‘π‘š

Β (i) Let us now find the trigonometric ratios sin A and cos A

When we take A as the angle, the perpendicular will be side opposite to it. So

BC – Perpendicular, AB – Base, AC – Hypotenuse

𝑠𝑖𝑛𝐴=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=725
π‘π‘œπ‘ π΄=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝐴𝐡𝐴𝐢=2425

(ii) Now we have to find the trigonometric ratios sin CΒ and cos C

When we take C as the angle, the perpendicular will be side opposite to it. So

AB – Perpendicular, BC – Base, AC – Hypotenuse

𝑠𝑖𝑛𝐢=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐴𝐡𝐴𝐢=2425
π‘π‘œπ‘ πΆ=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝐡𝐢𝐴𝐢=725
Click the following video to watch it’s solution.

Question 2

In Fig. 8.13, find tan P – cot R.

Question 2, Exercise 8.1, Chapter 8, Trigonometry
𝑃𝑄=12π‘π‘š,𝑃𝑅=13π‘π‘š

∡ β–³ PQR is a right angled triangle, let us apply pythagoras theorem and find QR.

𝑃𝑅2𝑄𝑅2𝑄𝑅=𝑃𝑄2+𝑄𝑅2=𝑃𝑅2–𝑃𝑄2=132–122=169–144=25=25β€Ύβ€Ύβ€Ύβˆš=5π‘π‘š

Let us now find the trigonometric ratios tan P and cot R

When we take P as the angle, the perpendicular will be side opposite to it. So

QR – Perpendicular, PQ – Base, PR – Hypotenuse

π‘‘π‘Žπ‘›π‘ƒ=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝑄𝑅𝑃𝑄=512…(1)

When we take R as the angle, the perpendicular will be side opposite to it. So

PQ – Perpendicular, QR – Base, PR – Hypotenuse

π‘π‘œπ‘‘π‘…=π΅π‘Žπ‘ π‘’π‘ƒπ‘’π‘Ÿπ‘=𝑄𝑅𝑃𝑄=512…(2)

From equations (1) and (2)

π‘‘π‘Žπ‘›π‘ƒβ€“π‘π‘œπ‘‘π‘…=512–512=0
Click the following video to watch it’s solution.

Question 3

If sinA = 3/4 calculate cos A and tan A.

Question 3, Exercise 8.1, Chapter 8, Trigonometry
𝑠𝑖𝑛𝐴=34

Now we know that

𝑠𝑖𝑛𝐴=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=34

Note: Now this is an important point to understand. Often students take Perpendicular (BC) as 3 and Hypotenuse (AC) as 4. But 3/4 is ratio of Perpendicular and Hypotenuse. Hence using the above ratio, we must solve it as follows.

𝐡𝐢𝐴𝐢=3π‘˜=4π‘˜

Now, let us apply Pythagoras theorem to find the third side (AB) of the right angled triangle.

𝐴𝐢2𝐴𝐡2𝐴𝐡=𝐴𝐡2+𝐡𝐢2=𝐴𝐢2–𝐡𝐢2=(4π‘˜)2–(3π‘˜)2=16π‘˜2–9π‘˜2=5π‘˜2=(5π‘˜2)β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=5β€Ύβˆšπ‘˜

Let us now calculate cos A and tan A

When we take A as the angle, the perpendicular will be the side opposite to it. Hence

BC – Perpendicular, AB – Base, AC – Hypotenuse

π‘π‘œπ‘ π΄=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝐴𝐡𝐴𝐢=5β€Ύβˆšπ‘˜4π‘˜=5β€Ύβˆš4
π‘‘π‘Žπ‘›π΄=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝐡𝐢𝐴𝐡=3π‘˜5β€Ύβˆšπ‘˜=35β€Ύβˆš
Click the following video to watch it’s solution.

Question 4

Given 15 cot A = 8, find sin A and sec A.

Question 4, Exercise 8.1, Chapter 8, Trigonometry

This question is similar to the previous question that we just solved.Β 

15Γ—π‘π‘œπ‘‘π΄π‘π‘œπ‘‘π΄=8=815

When we take A as the angle, the perpendicular will be side opposite to it. So

BC – Perpendicular, AB – Base, AC – Hypotenuse

Now we know that

π‘π‘œπ‘‘π΄=π΅π‘Žπ‘ π‘’π‘ƒπ‘’π‘Ÿπ‘=𝐴𝐡𝐡𝐢=815

Note: Now this is an important point to understand. Often students take Base (AB) as 8 and Perpendicular (BC) as 15. But 8/15 is ratio of Base and Perpendicular. Hence using the above ratio, we must solve it as follows.

⇒𝐴𝐡𝐡𝐢=8π‘˜=15π‘˜

Now, let us apply Pythagoras theorem to find the third side (AC) of the right angled triangle.

𝐴𝐢2=𝐴𝐡2+𝐡𝐢2=(8π‘˜)2+(15π‘˜)2=64π‘˜2+225π‘˜2=289π‘˜2=289π‘˜2β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=17π‘˜

Let us now calculate sin A and sec A

𝑠𝑖𝑛𝐴=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=15π‘˜17π‘˜=1517
𝑠𝑒𝑐𝐴=π»π‘¦π‘π΅π‘Žπ‘ π‘’=𝐴𝐢𝐴𝐡=17π‘˜8π‘˜=178
Click the following video to watch it’s solution.

Question 5

Given sec ΞΈ = 13/12 calculate all other trigonometric ratios.

Super Tip: We have to find all other trigonometry ratios in this question and for that we need to know the dimensions of all 3 sides of the triangle. But before that, we can directly find the value of cos θ. This way, you will quickly score 20% marks 😊

Question 5, Exercise 8.1, Chapter 8, Trigonometry

cos ΞΈ

π‘π‘œπ‘ ΞΈ=1𝑠𝑒𝑐θ=11312=1213

Let us consider that angle ΞΈ is at vertex A of right angled triangle ABC with right angle at B. This means

BC – Perpendicular, AB – Base, AC – HypotenuseΒ 

𝑠𝑒𝑐θ=π»π‘¦π‘π΅π‘Žπ‘ π‘’=𝐴𝐢𝐴𝐡=1312

Note: Now this is an important point to understand. Often students take Hypotenuse (AC) as 13 and Base (AB) as 12. But 13/12 is ratio of Hypotenuse and Base. Hence using the above ratio, we must solve it as follows.

⇒𝐴𝐢𝐴𝐡=13π‘˜=12π‘˜

Now, let us apply Pythagoras theorem to find the third side (BC) of the right angled triangle.

𝐴𝐢2𝐡𝐢2=𝐴𝐡2+𝐡𝐢2=𝐴𝐢2–𝐴𝐡2=(13π‘˜)2–(12π‘˜)2=169π‘˜2–144π‘˜2=49π‘˜2=(49π‘˜2)β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=7π‘˜

sin ΞΈ

𝑠𝑖𝑛θ=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=5π‘˜13π‘˜=713

cosec ΞΈ

Now cosec ΞΈ can be calculated in 2 ways. (Write only 1 of these methods in exam)

Method 1

π‘π‘œπ‘ π‘’π‘ΞΈ=π»π‘¦π‘π‘ƒπ‘’π‘Ÿπ‘=𝐴𝐢𝐡𝐢=13π‘˜7π‘˜=137

Method 2

π‘π‘œπ‘ π‘’π‘ΞΈ=1𝑠𝑖𝑛θ=1713=137

tan ΞΈ

Now tan ΞΈ can be calculated in 2 ways. (WriteΒ only 1 of these methods in exam)

Method 1

π‘‘π‘Žπ‘›ΞΈ=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝐡𝐢𝐴𝐡=7π‘˜12π‘˜=712

Method 2

π‘‘π‘Žπ‘›ΞΈ=π‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈ=7131213=712

cot ΞΈ

Now cot ΞΈ can be calculated in 2 ways. (WriteΒ only 1 of these methods in exam)

Method 1

π‘π‘œπ‘‘ΞΈ=π΅π‘Žπ‘ π‘’π‘ƒπ‘’π‘Ÿπ‘=𝐴𝐡𝐡𝐢=12π‘˜7π‘˜=127

Method 2

π‘π‘œπ‘‘ΞΈ=π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈ=1213713=127
Click the following video to watch it’s solution.

Question 6

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Copyright - Mathemafia Question 6, Exercise 8.1, Chapter 8, Trigonometry

It is given that ∠A and ∠B are acute angles

β‡’βˆ πΆ=90Β°

Now for angle A

PerpendicularΒ = BC (side opp. to vertex A)

Base = AC (side adjacent to vertex A)

& for angle B

PerpendicularΒ = AC (side opp. to vertex B)

Base = BC (side opposite to vertex B)

Hypotenuse (AB) will remain same as hypotenuse in a right angled triangle is the longest side.Β 

HenceΒ 

π‘π‘œπ‘ π΄π‘π‘œπ‘ π΅=𝐴𝐢𝐴𝐡…(1)=𝐡𝐢𝐴𝐡…(2)

But we know thatΒ 

cosA = cosB

From (1) and (2)

𝐴𝐢𝐴𝐡𝐴𝐢=𝐡𝐢𝐴𝐡=𝐡𝐢

This implies β–³ABC is an isosceles triangle.Β 

β‡’βˆ π΄=∠𝐡
Click the following video to watch it’s solution.

Question 7

If cot ΞΈ = 7/8 evaluate :

(𝑖)  (1+𝑠𝑖𝑛θ)(1–𝑠𝑖𝑛θ)(1+π‘π‘œπ‘ ΞΈ)(1β€“π‘π‘œπ‘ ΞΈ)(𝑖𝑖)  π‘π‘œπ‘‘2ΞΈ

Time saver tip:Β Such questions are very easy to solve. No need to find any other trigonometric ratio to solve this question.

(𝑖𝑖)  π‘π‘œπ‘‘2ΞΈ
π‘π‘œπ‘‘2ΞΈ  π‘Žπ‘π‘‘π‘’π‘Žπ‘™π‘™π‘¦  π‘–𝑠  π‘π‘œπ‘‘ΞΈΓ—π‘π‘œπ‘‘ΞΈ
=(π‘π‘œπ‘‘ΞΈ)2=(78)2=4964
(𝑖)  (1+𝑠𝑖𝑛θ)(1–𝑠𝑖𝑛θ)(1+π‘π‘œπ‘ ΞΈ)(1β€“π‘π‘œπ‘ ΞΈ)
=1–𝑠𝑖𝑛2ΞΈ1β€“π‘π‘œπ‘ 2ΞΈ=π‘π‘œπ‘ 2θ𝑠𝑖𝑛2ΞΈ=π‘π‘œπ‘‘2ΞΈ
=(78)2=4964
Click the following video to watch it’s solution.

Question 8

If 3 cot A = 4, check whether

1β€“π‘‘π‘Žπ‘›2𝐴1+π‘‘π‘Žπ‘›2𝐴=π‘π‘œπ‘ 2𝐴–𝑠𝑖𝑛2𝐴  π‘œπ‘Ÿ  π‘›π‘œπ‘‘.
Question 8, Exercise 8.1, Chapter 8, Trigonometry
3Γ—π‘π‘œπ‘‘π΄π‘π‘œπ‘‘π΄=4=43

When we take A as the angle, then

BC – Perpendicular, AB – Base, AC – Hypotenuse

Now we know that

π‘π‘œπ‘‘π΄=π΅π‘Žπ‘ π‘’π‘ƒπ‘’π‘Ÿπ‘=𝐴𝐡𝐡𝐢=43

Note: Now this is an important point to understand. Often students take Base (AB) as 4 and Perpendicular (BC) as 3. But 4/3 is ratio of Base and Perpendicular. Hence using the above ratio, we must solve it as follows.

⇒𝐴𝐡𝐡𝐢=4π‘˜=3π‘˜

Applying Pythagoras theorem

𝐴𝐢2𝐴𝐢=𝐴𝐡2+𝐡𝐢2=(4π‘˜)2+(3π‘˜)2=16π‘˜2+9π‘˜2=25π‘˜2=(25π‘˜2)β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=5π‘˜

Let’s calculate the values of sin A, cos A & tan A

𝑠𝑖𝑛𝐴=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=3π‘˜5π‘˜=35π‘π‘œπ‘ π΄=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝐴𝐡𝐴𝐢=4π‘˜5π‘˜=45π‘‘π‘Žπ‘›π΄=1π‘π‘œπ‘‘π΄=14/3=34
𝐿𝐻𝑆=1β€“π‘‘π‘Žπ‘›2𝐴1+π‘‘π‘Žπ‘›2𝐴=1–(3/4)21+(3/4)2=1–9/161+9/16=7/1625/16=725
𝑅𝐻𝑆=π‘π‘œπ‘ 2𝐴–𝑠𝑖𝑛2𝐴=(4/5)2–(3/5)2=16/25–9/25=725=𝐿𝐻𝑆

Hence proved.

Click the following video to watch it’s solution.

Question 9

In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Question 9, Exercise 8.1, Chapter 8, Trigonometry
π‘‘π‘Žπ‘›π΄=13β€Ύβˆš

When we take A as the angle, the perpendicular will be side opposite to it. So

BC – Perpendicular, AB – Base, AC – Hypotenuse

Now we know that

π‘‘π‘Žπ‘›π΄=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝐡𝐢𝐴𝐡=13β€Ύβˆšπ»π‘’π‘›π‘π‘’  π΅πΆ=π‘˜,𝐴𝐡=3β€Ύβˆšπ‘˜

If you have any doubt about the above statement, read the first NOTE written on this page.

Now, using Pythagoras theorem

𝐴𝐢2𝐴𝐢=𝐴𝐡2+𝐡𝐢2=π‘˜2+(3β€Ύβˆšπ‘˜)2=π‘˜2+3π‘˜2=4π‘˜2=(4π‘˜2)β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=2π‘˜

(i) sin A cos C + cos A sin C

𝑠𝑖𝑛𝐴=𝐡𝐢𝐴𝐢=π‘˜2π‘˜=12π‘π‘œπ‘ π΄=𝐴𝐡𝐴𝐢=3β€Ύβˆšπ‘˜2π‘˜=3β€Ύβˆš2

When we take C as the angle, the perpendicular will be side opposite to it. So

AB – Perpendicular, BC – Base, AC – Hypotenuse

𝑠𝑖𝑛𝐢=𝐴𝐡𝐴𝐢=3β€Ύβˆšπ‘˜2π‘˜=3β€Ύβˆš2π‘π‘œπ‘ πΆ=𝐡𝐢𝐴𝐢=π‘˜2π‘˜=12π‘ π‘–π‘›π΄π‘π‘œπ‘ πΆ+π‘π‘œπ‘ π΄π‘ π‘–π‘›πΆ
=12Γ—12+3β€Ύβˆš2Γ—3β€Ύβˆš2=14+34=1

(ii) cos A cos C – sin A sin C

π‘π‘œπ‘ π΄π‘π‘œπ‘ πΆβ€“π‘ π‘–π‘›π΄π‘ π‘–π‘›πΆ
=3β€Ύβˆš2Γ—12–12Γ—3β€Ύβˆš2=3β€Ύβˆš4–3β€Ύβˆš4=0
Click the following video to watch it’s solution.

Question 10

In β–³PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Question 10, Exercise 8.1, Chapter 8, Trigonometry
𝐿𝑒𝑑  π‘„𝑅𝑃𝑅+𝑄𝑅⇒𝑃𝑅=π‘₯=25=25–𝑄𝑅=25βˆ’π‘₯

Applying Pythagoras theorem

𝑃𝑅2(25–π‘₯)2β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš625+π‘₯2–50π‘₯625–50π‘₯50π‘₯π‘₯=𝑄𝑅2+𝑃𝑄2=(5)2+(π‘₯)2=25+π‘₯2=25=625–25=600=600/50=12
⇒𝑄𝑅𝑃𝑅=π‘₯=12π‘π‘š=25–𝑄𝑅=25–12=13π‘π‘š

Let’s calculate the values of sin P, cos P & tan P

𝑠𝑖𝑛𝑃=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝑄𝑅𝑃𝑅=1213π‘π‘œπ‘ π‘ƒ=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝑃𝑄𝑃𝑅=513π‘‘π‘Žπ‘›π‘ƒ=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝑄𝑅𝑃𝑄=125
Click the following video to watch it’s solution.

Question 11

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin ΞΈ = 4/3 for some angle ΞΈ

Solution:

(i) The value of tan A is always less than 1 – False

Reason: We know that tan 60° = √3 and √3 > 1, hence this statement is false.

(i) sec A = 12/5 for some value of angle A. – True

Reason: Trigonometric ratio sec is ratio of Hyp/Base and we know that Hypotenuse is always the longest side of a right angled triangle. Hence taking Hyp = 12 and Base = 5 we can get a right angled triangle.Β 

(iii) cos A is the abbreviation used for the cosecant of angle A – False

Reason: cos A is abbreviation of cosine of angle A.

(iv) cot A is the product of cot and A – False

Reason: cot A is single a term and it is cotangent of A.

(v) sin ΞΈ = 4/3 for some angle ΞΈ – False

Reason: Value of sin ΞΈ can never be greater than 1 because it is the ratio of Perpendicular and Hypotenuse and perpendicular can never be longer than hypotenuse.

Click the following video to watch it’s solution.
\(\begin{aligned}Β  \end{aligned}\)

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