CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
The exercise establishes basic trigonometric foundations by teaching about the six vital ratios, which include sine, cosine, tangent, cosecant, secant and cotangent. The ratios help us to understand angle-per-length relationships through right-angled triangles. We have to master the exercise because it creates essential foundations for higher-level trigonometric subjects and serves as a key part of the Class 10 mathematics syllabus.
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The NCERT Solutions for Class 10 Maths present practical methods for using trigonometric ratios according to the content found in NCERT textbooks. The exercises guide students to evaluate trigonometric ratios when given dimensional triangle measurements and help students determine missing lengths through ratio analysis while confirming trigonometric identity relations. Exercise 8.1 in the NCERT Books helps students strengthen their trigonometry knowledge as it develops their ability to solve complex mathematical problems required for higher educational levels.
Q1 In $\Delta \: ABC$ , right-angled at $B, AB = 24 \: cm$ , $BC = 7 \: cm$ . Determine : $(i)\; \sin A, \cos A$ $(ii)\; \sin C, \cos C$
Answer:
We have,
In $\Delta \: ABC$ , $\angle$ B = 90, and the length of the base (AB) = 24 cm and length of perpendicular (BC) = 7 cm
So, by using Pythagoras theorem,
$\\AC^2 = AB^2 + BC^2\\ AC = \sqrt{AB^2+BC^2}$
Therefore, $AC = \sqrt{576+49}$
$AC = \sqrt{625}$
AC = 25 cm
Now,
(i) $\sin A = P/H = BC/AB = 7/25$
$\cos A = B/H = BA/AC = 24/25$
(ii) For angle C, AB is perpendicular to the base (BC). Here B indicates to Base and P means perpendicular wrt angle $\angle$ C
So, $\sin C = P/H = BA/AC = 24/25$
and $\cos C = B/H = BC/AC = 7/25$
Q2 In Fig. 8.13, find $\tan P - \cot R$ .
Answer:
We have, $\Delta$ PQR is a right-angled triangle, length of PQ and PR are 12 cm and 13 cm respectively.
So, by using Pythagoras theorem,
$QR = \sqrt{13^2-12^2}$
$QR = \sqrt{169-144}$
$QR = \sqrt{25} = 5\ cm$
Now, According to question,
$\tan P -\cot R$ = $\frac{RQ}{QP}-\frac{QR}{PQ}$
= 5/12 - 5/12 = 0
Q3 If $\sin A=\frac{3}{4},$ calculate $\cos A$ and $\tan A$ .
Answer:
Suppose $\Delta$ ABC is a right-angled triangle in which $\angle B = 90^0$ and we have $\sin A=\frac{3}{4},$
So,
Let the length of AB be 4 unit and the length of BC = 3 unit So, by using Pythagoras theorem,
$AB = \sqrt{16-9} = \sqrt{7}$ units
Therefore,
$\cos A = \frac{AB}{AC} = \frac{\sqrt{7}}{4}$ and $\tan A = \frac{BC}{AB} = \frac{3}{\sqrt{7}}$
Q4 Given $15 \: \cot A=8,$ find $\sin A$ and $\sec A$ .
Answer:
We have,
$15 \: \cot A=8,$ $\Rightarrow \cot A =8/15$
It implies that In the triangle ABC in which $\angle B =90^0$ . The length of AB be 8 units and the length of BC = 15 units
Now, by using Pythagoras theorem,
$AC = \sqrt{64 +225} = \sqrt{289}$
$\Rightarrow AC =17$ units
So, $\sin A = \frac{BC}{AC} = \frac{15}{17}$
and $\sec A = \frac{AC}{AB} = \frac{17}{8}$
Q5 Given $\sec \theta =\frac{13}{12},$ calculate all other trigonometric ratios.
Answer:
We have,
$\sec \theta =\frac{13}{12},$
It means the Hypotenuse of the triangle is 13 units and the base is 12 units.
Let ABC is a right-angled triangle in which $\angle$ B is 90 and AB is the base, BC is perpendicular height and AC is the hypotenuse.
By using Pythagoras theorem,
$BC = \sqrt{169-144}=\sqrt{25}$
BC = 5 unit
Therefore,
$\sin \theta = \frac{BC}{AC}=\frac{5}{13}$
$\cos \theta = \frac{BA}{AC}=\frac{12}{13}$
$\tan \theta = \frac{BC}{AB}=\frac{5}{12}$
$\cot \theta = \frac{BA}{BC}=\frac{12}{5}$
$\sec \theta = \frac{AC}{AB}=\frac{13}{12}$
$\csc \theta = \frac{AC}{BC}=\frac{13}{5}$
Answer:
We have, A and B are two acute angles of triangle ABC and $\cos A =\cos B$
According to question, In triangle ABC,
$\cos A =\cos B$
$\frac{AC}{AB}=\frac{BC}{AB}$
$\Rightarrow AC = AB$
Therefore, $\angle$ A = $\angle$ B [angle opposite to equal sides are equal]
Q7 If $\cot \theta =\frac{7}{8},$ evaluate: $(i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ $(ii)\; \cot ^{2}\theta$
Answer:
Given that,
$\cot \theta =\frac{7}{8}$
$\therefore$ perpendicular (AB) = 8 units and Base (AB) = 7 units
Draw a right-angled triangle ABC in which $\angle B =90^0$
Now, By using Pythagoras theorem,
$AC^2 = AB^2+BC^2$
$AC = \sqrt{64 +49} =\sqrt{113}$
So, $\sin \theta = \frac{BC}{AC} = \frac{8}{\sqrt{113}}$
and $\cos \theta = \frac{AB}{AC} = \frac{7}{\sqrt{113}}$
$\Rightarrow \cot \theta =\frac{\cos \theta}{\sin \theta} = \frac{7}{8}$
$(i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$\Rightarrow \frac{(1-\sin^2\theta)}{(1-\cos^2\theta)} = \frac{\cos^2\theta}{\sin^2\theta} = \cot ^2\theta$
$=(\frac{7}{8})^2 = \frac{49}{64}$
$(ii)\; \cot ^{2}\theta$
$=(\frac{7}{8})^2 = \frac{49}{64}$
Q8 If $3\cot A=4,$ check wether $\frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A$ or not.
Answer:
Given that,
$3\cot A=4,$
$\Rightarrow \cot = \frac{4}{3} = \frac{base}{perp.}$
ABC is a right-angled triangle in which $\angle B =90^0$ and the length of the base AB is 4 units and length of perpendicular is 3 units
By using Pythagoras theorem, In triangle ABC,
$\\AC^2=AB^2+BC^2\\ AC = \sqrt{16+9}\\ AC = \sqrt{25}$
AC = 5 units
So,
$\tan A = \frac{BC}{AB} = \frac{3}{4}$
$\cos A = \frac{AB}{AC} = \frac{4}{5}$
$\sin A = \frac{BC}{AC} = \frac{3}{5}$
$\frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A$
Put the values of above trigonometric ratios, we get;
$\Rightarrow \frac{1-9/4}{1+9/4} = \frac{16}{25}-\frac{9}{25}$
$\Rightarrow -\frac{5}{13} \neq \frac{7}{25}$
LHS $\neq$ RHS
Q9 In triangle $ABC$ , right-angled at $B$ , if $\tan A =\frac{1}{\sqrt{3}},$ find the value of:
$(i) \sin A\: \cos C + \cos A\: \sin C$
$(ii) \cos A\: \cos C + \sin A\: \sin C$
Answer:
Given a triangle ABC, right-angled at B and $\tan A =\frac{1}{\sqrt{3}}$ $\Rightarrow A=30^0$
According to question, $\tan A =\frac{1}{\sqrt{3}} = \frac{BC}{AB}$
By using Pythagoras theorem,
$\\AC^2 = AB^2+BC^2\\ AC = \sqrt{1+3} =\sqrt{4}$
AC = 2
Now,
$\\\sin A = \frac{BC}{AC} = \frac{1}{2}\\ \sin C =\frac{AB}{AC} = \frac{\sqrt{3}}{2}\\ \cos A = \frac{AB}{AC} = \frac{\sqrt{3}}{2}\\ \cos C = \frac{BC}{AC} = \frac{1}{2}$
Therefore,
$(i) \sin A\: \cos C + \cos A\: \sin C$
$\\\Rightarrow \frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\\ \Rightarrow1/4 +3/4\\ \Rightarrow4/4 = 1$
$(ii) \cos A\: \cos C + \sin A\: \sin C$
$\\\Rightarrow \frac{\sqrt{3}}{2}\times \frac{1}{2}+\frac{1}{2}\times\frac{\sqrt{3}}{2}\\ \Rightarrow \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\\ \Rightarrow \frac{\sqrt{3}}{2}$
Answer:
We have, PR + QR = 25 cm.............(i)
PQ = 5 cm
and $\angle Q =90^0$
According to question,
In triangle $\Delta$ PQR,
By using Pythagoras theorem,
$\\PR^2 = PQ^2+QR^2\\ PQ^2 =PR^2-QR^2 \\ 5^2= (PR-QR)(PR+QR)\\ 25 = 25(PR-QR) \\$
PR - QR = 1........(ii)
From equation(i) and equation(ii), we get;
PR = 13 cm and QR = 12 cm.
therefore,
$\\\sin P= \frac{QR}{PR}= 12/13\\ \cos P = \frac{PQ}{RP} = 5/13\\ \therefore \tan P = \frac{\sin P}{\cos P} = 12/5$
Q11 State whether the following are true or false. Justify your answer.
(i) The value of $\tan A$ is always less than 1.
(ii) $\sec A=\frac{12}{5}$ 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) $\sin \Theta =\frac{4}{3}$ for some angle $\Theta .$
Answer:
(i) False,
because $\tan 60 = \sqrt{3}$ , which is greater than 1
(ii) True,
because $\sec A \geq 1$
(iii) False,
Because $\cos A$ abbreviation is used for cosine A.
(iv) False,
because the term $\cot A$ is a single term, not a product.
(v) False,
because $\sin \theta$ lies between (-1 to +1) [ $-1\leq \sin \theta\leq 1$ ]
1. Calculating Trigonometric Ratios: Individuals can determine the values of trigonometric ratios when given the lengths of right-angled triangles.
2. Determining Side Lengths: Calculating the unknown side lengths of right-angled triangles becomes possible when applying known trigonometric ratios.
3. Verifying Trigonometric Identities: By using both calculation and reasoning, students verify fundamental trigonometric identities.
4. Real-Life Applications: Applying trigonometric principles helps solve real-life situations where trigonometry measures sides and angles in practical problems.
5. Definition of Trigonometric Ratios: Understanding sine, cosine, tangent, cosecant, secant, and cotangent in the context of right-angled triangles.
Trigonometric Ratios | Formula |
sin θ | $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$ |
cos θ | $\frac{\text{Base}}{\text{Hypotenuse}}$ |
tan θ | $\frac{\text{Perpendicular}}{\text{Base}}$ |
cosec θ | $\frac{\text{Hypotenuse}}{\text{Perpendicular}}$ |
sec θ | $\frac{\text{Hypotenuse}}{\text{Base}}$ |
cot θ | $\frac{\text{Base}}{\text{Perpendicular}}$ |
Also, see-
Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.
Students must check the NCERT Exemplar solutions for class 10 of Mathematics and Science Subjects.
Frequently Asked Questions (FAQs)
The three basic trigonometric ratios are sine, cosine and tangent.
The ratio of the opposite side to the hypotenuse of the right angle triangle is known as the sine.
Sin θ=opposite side/hypotenuse
Cos means Cosine is the ratio of Adjacent Side and Hypotenuse
tan θ=sin θ/cos θ
The multiplicative inverse of sine is known as the cosecant
The six trigonometric ratios are
Sine
Cosine
Tangent
Cotangent
Cosecant
Secant.
NCERT solutions for Class 10 Maths chapter 8 exercise 8. 1 consists of 11 Questions in which 7 are short answers, 3 of them are long answers and the remaining one is a short answer with reasoning and all the questions are based on trigonometric ratios.
On Question asked by student community
Hello,
You can find the Class 10 Half-Yearly Exam Question Papers for all subjects on the Careers360 website. It provides PDFs of all subject-wise question papers along with answer keys. It also gives you a detailed idea of the exam overview and is very useful for your preparation.
Follow the Link:
Hello,
The CBSE exam fee for Class 10 students is as follows:
For up to 5 subjects: Rs. 1,600 per student
For each additional subject: Rs. 320
Late fee (after the due date): Rs. 2,000
These fees are applicable for students studying in India as per the latest CBSE notification.
The school fee depends upon the particular school.
Hope it helps !
Hello aspirant,
The Sample Question Paper (SQP) and marking guidelines have been made available by the Central Board of Secondary Education (CBSE). Although the board does not formally provide distinct half-yearly sample papers, many of the final CBSE sample papers' questions address subjects that are covered in the exams.
To get the sample papers, you can visit our site through following link:
https://school.careers360.com/boards/cbse/cbse-class-12-half-yearly-sample-papers-2025-26
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