NCERT Class 10 Maths – Exercise 8.1 Solutions

NCERT Class 10 Maths

Chapter 8 – Introduction to Trigonometry | Exercise 8.1

(Rationalized Syllabus 2025-26)

💡 Trigonometric Ratios

  • sin A = Perpendicular / Hypotenuse (P/H)
  • cos A = Base / Hypotenuse (B/H)
  • tan A = Perpendicular / Base (P/B)
  • cosec A = H/P, sec A = H/B, cot A = B/P
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

By Pythagoras theorem: $AC^2 = AB^2 + BC^2$.
$$AC^2 = (24)^2 + (7)^2 = 576 + 49 = 625$$
$$AC = \sqrt{625} = 25 \text{ cm}$$
(i) For angle A:
Perpendicular (BC) = 7, Base (AB) = 24, Hypotenuse (AC) = 25.
$$\sin A = \frac{BC}{AC} = \frac{7}{25}$$
$$\cos A = \frac{AB}{AC} = \frac{24}{25}$$
✔️ sin A = 7/25, cos A = 24/25
(ii) For angle C:
Perpendicular (AB) = 24, Base (BC) = 7, Hypotenuse (AC) = 25.
$$\sin C = \frac{AB}{AC} = \frac{24}{25}$$
$$\cos C = \frac{BC}{AC} = \frac{7}{25}$$
✔️ sin C = 24/25, cos C = 7/25
Q2

In Fig 8.13, find tan P – cot R.

In $\Delta PQR$, apply Pythagoras theorem:
$$QR = \sqrt{PR^2 – PQ^2} = \sqrt{13^2 – 12^2} = \sqrt{169 – 144} = \sqrt{25} = 5$$
Calculate ratios:
$$\tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ} = \frac{5}{12}$$
$$\cot R = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{QR}{PQ} = \frac{5}{12}$$
$$\tan P – \cot R = \frac{5}{12} – \frac{5}{12} = 0$$
✔️ 0
Q3

If sin A = 3/4, calculate cos A and tan A.

Let side opposite to A be $3k$ and hypotenuse be $4k$.
By Pythagoras theorem:
$$\text{Adjacent}^2 = (4k)^2 – (3k)^2 = 16k^2 – 9k^2 = 7k^2$$
$$\text{Adjacent} = \sqrt{7}k$$
$$\cos A = \frac{\text{Adj}}{\text{Hyp}} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4}$$
$$\tan A = \frac{\text{Opp}}{\text{Adj}} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}}$$
✔️ cos A = $\frac{\sqrt{7}}{4}$, tan A = $\frac{3}{\sqrt{7}}$
Q4

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

$\cot A = \frac{8}{15} = \frac{\text{Base}}{\text{Perpendicular}}$.
Let Base = $8k$, Perpendicular = $15k$.
$$\text{Hypotenuse} = \sqrt{(8k)^2 + (15k)^2} = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$$
$$\sin A = \frac{P}{H} = \frac{15k}{17k} = \frac{15}{17}$$
$$\sec A = \frac{H}{B} = \frac{17k}{8k} = \frac{17}{8}$$
✔️ sin A = 15/17, sec A = 17/8
Q5

Given sec $\theta$ = 13/12, calculate all other trigonometric ratios.

$\sec \theta = \frac{H}{B} = \frac{13}{12}$. Let $H=13k, B=12k$.
Perpendicular $P = \sqrt{(13k)^2 – (12k)^2} = \sqrt{169k^2 – 144k^2} = \sqrt{25k^2} = 5k$.
Ratios:
$$\sin \theta = \frac{P}{H} = \frac{5}{13}, \quad \cos \theta = \frac{B}{H} = \frac{12}{13}$$
$$\tan \theta = \frac{P}{B} = \frac{5}{12}, \quad \cot \theta = \frac{B}{P} = \frac{12}{5}$$
$$\text{cosec } \theta = \frac{H}{P} = \frac{13}{5}$$
Q6

If $\angle A$ and $\angle B$ are acute angles such that cos A = cos B, then show that $\angle A = \angle B$.

Let’s consider two right triangles $\Delta APQ$ and $\Delta BRS$.
Given $\cos A = \cos B$.
$$\frac{AP}{AQ} = \frac{BR}{BS} = k \quad (\text{say})$$
Then $AP = k(BR)$ and $AQ = k(BS)$.
By Pythagoras: $PQ = \sqrt{AQ^2 – AP^2}$ and $RS = \sqrt{BS^2 – BR^2}$.
$$\frac{PQ}{RS} = \frac{\sqrt{k^2 BS^2 – k^2 BR^2}}{\sqrt{BS^2 – BR^2}} = \frac{k \sqrt{BS^2 – BR^2}}{\sqrt{BS^2 – BR^2}} = k$$
Since all sides are proportional ($\frac{AP}{BR} = \frac{AQ}{BS} = \frac{PQ}{RS}$), $\Delta APQ \sim \Delta BRS$.
Therefore, $\angle A = \angle B$.
✔️ Hence Proved
Q7

If cot $\theta$ = 7/8, evaluate: (i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ (ii) cot$^2 \theta$

Given $\cot \theta = 7/8$.
(i) Evaluate Expression:
$$\frac{1 – \sin^2 \theta}{1 – \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \cot^2 \theta$$
$$= \left(\frac{7}{8}\right)^2 = \frac{49}{64}$$
(ii) Evaluate $\cot^2 \theta$:
$$= \left(\frac{7}{8}\right)^2 = \frac{49}{64}$$
✔️ (i) 49/64, (ii) 49/64
Q8

If 3 cot A = 4, check whether $\frac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A – \sin^2 A$ or not.

$\cot A = 4/3 \Rightarrow \tan A = 3/4$.
Let Base=4k, Perp=3k. Hypotenuse = 5k.
So, $\sin A = 3/5, \cos A = 4/5$.
LHS:
$$\frac{1 – (3/4)^2}{1 + (3/4)^2} = \frac{1 – 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$$
RHS:
$$\left(\frac{4}{5}\right)^2 – \left(\frac{3}{5}\right)^2 = \frac{16}{25} – \frac{9}{25} = \frac{7}{25}$$
✔️ Yes, LHS = RHS
Q9

In $\Delta ABC$, right-angled at B, if tan A = $1/\sqrt{3}$, find the value of:

$\tan A = \frac{1}{\sqrt{3}} \Rightarrow P=1k, B=\sqrt{3}k$.
$$H = \sqrt{(1k)^2 + (\sqrt{3}k)^2} = \sqrt{1+3}k = 2k$$
For A: $\sin A = 1/2, \cos A = \sqrt{3}/2$.
For C (Base becomes BC=1k, Perp becomes AB=$\sqrt{3}k$):
$\sin C = \sqrt{3}/2, \cos C = 1/2$.
(i) $\sin A \cos C + \cos A \sin C$:
$$= (\frac{1}{2})(\frac{1}{2}) + (\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) = \frac{1}{4} + \frac{3}{4} = 1$$
(ii) $\cos A \cos C – \sin A \sin C$:
$$= (\frac{\sqrt{3}}{2})(\frac{1}{2}) – (\frac{1}{2})(\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4} – \frac{\sqrt{3}}{4} = 0$$
✔️ (i) 1, (ii) 0
Q10

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

Given $PR + QR = 25 \Rightarrow QR = 25 – PR$.
By Pythagoras: $PR^2 = PQ^2 + QR^2$.
$$PR^2 = 5^2 + (25 – PR)^2$$
$$PR^2 = 25 + 625 + PR^2 – 50PR$$
$$50PR = 650 \Rightarrow PR = 13 \text{ cm}$$
So, $QR = 25 – 13 = 12$ cm.
$$\sin P = \frac{QR}{PR} = \frac{12}{13}, \quad \cos P = \frac{PQ}{PR} = \frac{5}{13}, \quad \tan P = \frac{QR}{PQ} = \frac{12}{5}$$
✔️ sin P = 12/13, cos P = 5/13, tan 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.
False. In a right triangle, the perpendicular can be greater than the base. Example: if P=12, B=5, tan A = 2.4.
(ii) sec A = 12/5 for some value of angle A.
True. Sec A is ratio H/B. Since H > B is always true, 12/5 > 1 is a valid ratio.
(iii) cos A is the abbreviation used for the cosecant of angle A.
False. Cos A is cosine. Cosecant is cosec A.
(iv) cot A is the product of cot and A.
False. cot A is a single term representing a ratio for angle A. Separated, they have no meaning.
(v) sin $\theta$ = 4/3 for some angle $\theta$.
False. Sin $\theta$ = P/H. The hypotenuse H must be the longest side. Here P > H (4 > 3), which is impossible.
🎉 Exercise 8.1 Completed
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