NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1#
Exercise 8.1 introduces trigonometric ratios — sine, cosine, tangent, cosecant, secant, and cotangent — defined with respect to the sides of a right-angled triangle. Students learn to find these ratios from given information.
Trigonometric Ratios in right △ABC (right angle at B), for the acute angle A:
Ratio Definition Reciprocal $\sin A$ $\dfrac{\text{side opposite to } A}{\text{hypotenuse}} = \dfrac{BC}{AC}$ $\cosec A = \dfrac{1}{\sin A}$ $\cos A$ $\dfrac{\text{side adjacent to } A}{\text{hypotenuse}} = \dfrac{AB}{AC}$ $\sec A = \dfrac{1}{\cos A}$ $\tan A$ $\dfrac{\text{side opposite to } A}{\text{side adjacent to } A} = \dfrac{BC}{AB}$ $\cot A = \dfrac{1}{\tan A}$
1. In △ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine: #
Setup
By Pythagoras Theorem:
$$AC^2 = AB^2 + BC^2$$$$\Rightarrow AC^2 = 24^2 + 7^2$$
$$\Rightarrow AC^2 = 625$$
$$\Rightarrow AC = 25$$
So the hypotenuse AC = 25 cm.
(i) sin A, cos A
$$\cos A = \frac{AB}{AC} = \frac{24}{25}$$
∴ $\sin A = \mathbf{\dfrac{7}{25}}$ and $\cos A = \mathbf{\dfrac{24}{25}}$
(ii) sin C, cos C
$$\cos C = \frac{BC}{AC} = \frac{7}{25}$$
∴ $\sin C = \mathbf{\dfrac{24}{25}}$ and $\cos C = \mathbf{\dfrac{7}{25}}$
2. In the figure, find tan P – cot R. #
Answer
From the figure, △PQR is right-angled at Q, with PQ = 12 cm and PR = 13 cm.
By Pythagoras Theorem:
$$QR^2 = PR^2 - PQ^2$$$$\Rightarrow QR^2 = 169 - 144$$
$$\Rightarrow QR^2 = 25$$
$$\Rightarrow QR = 5$$
Now:
$$\tan P = \frac{QR}{PQ} = \frac{5}{12}$$$$\cot R = \frac{QR}{PQ} = \frac{5}{12}$$
We find $\tan P - \cot R$:
$$= \frac{5}{12} - \frac{5}{12}$$$$= 0$$
∴ $\tan P - \cot R = \mathbf{0}$
3. If sin A = $\dfrac{3}{4}$, calculate cos A and tan A. #
Answer
Take a right triangle ABC, right-angled at B.
$\sin A = \dfrac{BC}{AC} = \dfrac{3}{4}$, so let $BC = 3k$ and $AC = 4k$.
By Pythagoras Theorem:
$$AB^2 = AC^2 - BC^2$$$$\Rightarrow AB^2 = 16k^2 - 9k^2$$
$$\Rightarrow AB^2 = 7k^2$$
$$\Rightarrow AB = \sqrt{7}\,k$$
Now:
$$\cos A = \frac{AB}{AC} = \frac{\sqrt{7}\,k}{4k} = \frac{\sqrt{7}}{4}$$$$\tan A = \frac{BC}{AB} = \frac{3k}{\sqrt{7}\,k} = \frac{3}{\sqrt{7}}$$
∴ $\cos A = \mathbf{\dfrac{\sqrt{7}}{4}}$ and $\tan A = \mathbf{\dfrac{3}{\sqrt{7}}}$
4. Given 15 cot A = 8, find sin A and sec A. #
Answer
Take a right triangle ABC, right-angled at B.
$$15 \cot A = 8$$$$\Rightarrow \cot A = \frac{8}{15}$$
$\cot A = \dfrac{AB}{BC} = \dfrac{8}{15}$, so let $AB = 8k$ and $BC = 15k$.
By Pythagoras Theorem:
$$AC^2 = AB^2 + BC^2$$$$\Rightarrow AC^2 = 64k^2 + 225k^2$$
$$\Rightarrow AC^2 = 289k^2$$
$$\Rightarrow AC = 17k$$
Now:
$$\sin A = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17}$$$$\sec A = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8}$$
∴ $\sin A = \mathbf{\dfrac{15}{17}}$ and $\sec A = \mathbf{\dfrac{17}{8}}$
5. Given sec θ = $\dfrac{13}{12}$, calculate all other trigonometric ratios. #
Answer
Take a right triangle ABC, right-angled at B, with $\angle A = \theta$.
$\sec\theta = \dfrac{AC}{AB} = \dfrac{13}{12}$, so let $AC = 13k$ and $AB = 12k$.
By Pythagoras Theorem:
$$BC^2 = AC^2 - AB^2$$$$\Rightarrow BC^2 = 169k^2 - 144k^2$$
$$\Rightarrow BC^2 = 25k^2$$
$$\Rightarrow BC = 5k$$
Now:
$$\sin\theta = \frac{BC}{AC} = \frac{5}{13}$$$$\cos\theta = \frac{AB}{AC} = \frac{12}{13}$$
$$\tan\theta = \frac{BC}{AB} = \frac{5}{12}$$
$$\cosec\theta = \frac{AC}{BC} = \frac{13}{5}$$
$$\cot\theta = \frac{AB}{BC} = \frac{12}{5}$$
6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B. #
Proof
Take a triangle ABC right-angled at C, so that ∠A and ∠B are its two acute angles.
In this triangle:
$$\cos A = \frac{AC}{AB}$$$$\cos B = \frac{BC}{AB}$$
It is given that $\cos A = \cos B$, so:
$$\frac{AC}{AB} = \frac{BC}{AB}$$$$\Rightarrow AC = BC$$
In △ABC, the angles opposite equal sides are equal, so:
$$\angle B = \angle A$$Hence proved. $\blacksquare$
7. If cot θ = $\dfrac{7}{8}$, evaluate: #
(i) $\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
Take a right triangle ABC, right-angled at B, with $\angle A = \theta$.
$\cot\theta = \dfrac{AB}{BC} = \dfrac{7}{8}$, so let $AB = 7k$ and $BC = 8k$.
By Pythagoras Theorem:
$$AC^2 = AB^2 + BC^2$$$$\Rightarrow AC^2 = 49k^2 + 64k^2$$
$$\Rightarrow AC = \sqrt{113}\,k$$
Now:
$$\sin\theta = \frac{BC}{AC} = \frac{8}{\sqrt{113}}$$$$\cos\theta = \frac{AB}{AC} = \frac{7}{\sqrt{113}}$$
We find $\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$:
$$= \frac{1 - \sin^2\theta}{1 - \cos^2\theta}$$$$= \frac{1 - \frac{64}{113}}{1 - \frac{49}{113}}$$
$$= \frac{\frac{49}{113}}{\frac{64}{113}}$$
$$= \frac{49}{64}$$
∴ The value is $\mathbf{\dfrac{49}{64}}$.
(ii) $\cot^2\theta$
∴ $\cot^2\theta = \mathbf{\dfrac{49}{64}}$
8. If 3 cot A = 4, check whether $\dfrac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A$ or not. #
Answer
Take a right triangle ABC, right-angled at B.
$$3 \cot A = 4$$$$\Rightarrow \cot A = \frac{4}{3}$$
$\cot A = \dfrac{AB}{BC} = \dfrac{4}{3}$, so let $AB = 4k$ and $BC = 3k$.
By Pythagoras Theorem:
$$AC^2 = 16k^2 + 9k^2$$$$\Rightarrow AC = 5k$$
Now:
$$\tan A = \frac{BC}{AB} = \frac{3}{4}$$$$\sin A = \frac{BC}{AC} = \frac{3}{5}$$
$$\cos A = \frac{AB}{AC} = \frac{4}{5}$$
We find the LHS, $\dfrac{1-\tan^2 A}{1+\tan^2 A}$:
$$= \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}}$$$$= \frac{\frac{7}{16}}{\frac{25}{16}}$$
$$= \frac{7}{25}$$
We find the RHS, $\cos^2 A - \sin^2 A$:
$$= \frac{16}{25} - \frac{9}{25}$$$$= \frac{7}{25}$$$$\Rightarrow \frac{7}{25} = \frac{7}{25}$$
∴ LHS = RHS, so yes, $\dfrac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A$.
9. In triangle ABC, right-angled at B, if tan A = $\dfrac{1}{\sqrt{3}}$, find the value of: #
Setup
$\tan A = \dfrac{BC}{AB} = \dfrac{1}{\sqrt{3}}$, so let $BC = k$ and $AB = \sqrt{3}\,k$.
By Pythagoras Theorem:
$$AC^2 = AB^2 + BC^2$$$$\Rightarrow AC^2 = 3k^2 + k^2$$
$$\Rightarrow AC^2 = 4k^2$$
$$\Rightarrow AC = 2k$$
Now:
$$\sin A = \frac{BC}{AC} = \frac{1}{2}$$$$\cos A = \frac{AB}{AC} = \frac{\sqrt{3}}{2}$$
$$\sin C = \frac{AB}{AC} = \frac{\sqrt{3}}{2}$$
$$\cos C = \frac{BC}{AC} = \frac{1}{2}$$
(i) sin A cos C + cos A sin C
Using the values from the setup, we find $\sin A \cos C + \cos A \sin C$:
$$= \frac{1}{2} \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$$$$= \frac{1}{4} + \frac{3}{4}$$
$$= 1$$
∴ The value is 1.
(ii) cos A cos C – sin A sin C
Using the values from the setup, we find $\cos A \cos C - \sin A \sin C$:
$$= \frac{\sqrt{3}}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{\sqrt{3}}{2}$$$$= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}$$
$$= 0$$
∴ The value is 0.
10. In △PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. #
Answer
Let QR = $x$ cm.
PR + QR = 25 cm, so:
$$PR = 25 - x$$By Pythagoras Theorem:
$$PR^2 = PQ^2 + QR^2$$$$\Rightarrow (25 - x)^2 = 5^2 + x^2$$
$$\Rightarrow 625 - 50x + x^2 = 25 + x^2$$
$$\Rightarrow 50x = 600$$
$$\Rightarrow x = 12$$
So QR = 12 cm and PR = 25 − 12 = 13 cm.
Now:
$$\sin P = \frac{QR}{PR} = \frac{12}{13}$$$$\cos P = \frac{PQ}{PR} = \frac{5}{13}$$
$$\tan P = \frac{QR}{PQ} = \frac{12}{5}$$
∴ $\sin P = \mathbf{\dfrac{12}{13}}$, $\cos P = \mathbf{\dfrac{5}{13}}$ and $\tan P = \mathbf{\dfrac{12}{5}}$
11. State whether the following are true or false. Justify your answer. #
(i) The value of tan A is always less than 1.
False.
$\tan A = \dfrac{\text{side opposite to } A}{\text{side adjacent to } A}$, and the opposite side can be longer than the adjacent side.
For example, in a right triangle with opposite side 4 and adjacent side 3:
$$\tan A = \frac{4}{3}$$$$\Rightarrow \frac{4}{3} > 1$$∴ tan A is not always less than 1.
(ii) sec A = $\dfrac{12}{5}$ for some value of angle A.
True.
$\sec A = \dfrac{\text{hypotenuse}}{\text{side adjacent to } A}$. Take a right triangle with hypotenuse 12 and adjacent side 5. The third side is:
$$\sqrt{12^2 - 5^2} = \sqrt{119}$$Such a triangle exists, since the hypotenuse is longer than the adjacent side:
$$\Rightarrow 12 > 5$$∴ sec A = $\dfrac{12}{5}$ for the angle A of this triangle.
(iii) cos A is the abbreviation used for the cosecant of angle A.
False.
cos A is the abbreviation for the cosine of angle A. The cosecant of angle A is written as cosec A.
(iv) cot A is the product of cot and A.
False.
cot A is one symbol, meaning the cotangent of angle A. “cot” on its own has no meaning, so cot A is not a product.
(v) sin θ = $\dfrac{4}{3}$ for some angle θ.
False.
$\sin\theta = \dfrac{\text{side opposite to } \theta}{\text{hypotenuse}}$, and the hypotenuse is the longest side of a right triangle. So:
$$\Rightarrow \sin\theta < 1$$$$\Rightarrow \frac{4}{3} > 1$$∴ sin θ can never be $\dfrac{4}{3}$.