INTRODUCTION TO TRIGONOMETRY-Exercise 8.1
Maths - Exercise
\(\tiny\begin{aligned}P&\Rightarrow \text{Perpendicular}\\
B&\Rightarrow \text{Base}\\
H&\Rightarrow \text{Hypotenuse}
\end{aligned}\)
Q1. In \(\triangle ABC\), right-angled at B, AB = 24 cm, BC = 7 cm. Determine
:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
AB = 24, BC = 7
By Pythagoras theorem
Q2. In Fig. 8.13, find tan P – cot R.
Solution:
tan P-cot R
PQ= 12,
PR = 13
QR can be found by using Pythagoras theorem
Q3. If \(\sin A = \frac{3}{4}\), calculate cos A and tan A.
Solution:
$$\begin{aligned}\sin A&=\dfrac{3}{4}\\ \sin A&=\dfrac{P}{H}\\ \dfrac{P}{H}&=\dfrac{3}{4}\\ P&=3k\quad\scriptsize\text{(k is any positive number)}\\ H&=4k\\ B^{2}&=H^{2}-P^{2}\\ B^{2}&=\left( 4k\right) ^{2}-\left( 3k\right) ^{2}\\ B&=\sqrt{16k^{2}-9k^{2}}\\ &=\sqrt{7k^{2}}\\ &=k\sqrt{7}\\\\ \cos A&=\dfrac{B}{H}\\ &=\dfrac{k\sqrt{7}}{4k}\\ &=\dfrac{\sqrt{7}}{4}\\\\ \tan A&=\dfrac{\sin A}{\cos A}\\ &=\dfrac{\frac{3}{4}}{\sqrt{7}/4}\\ &=\dfrac{3}{\sqrt{7}}\end{aligned}$$Q4. Given 15 cot A = 8, find sin A and sec A.
Solution:
$$\begin{aligned}15\times \cot A&=8\\ \cot A&=\dfrac{8}{15}\\ &=\dfrac{B}{P}\\ \dfrac{B}{P}&=\dfrac{8}{15}\\ B&=8k\quad\scriptsize\text{(k is a positive integer)}\\ P&=15k\\ H^{2}&=B^{2}+P^{2}\\ &=\left( 8k\right) ^{2}+\left( 5k\right) ^{2}\\ &=64k^{2}+225k^{2}\\ &=289k^{2}\\ H&=\sqrt{289k^{2}}\\ &=17k\\\\ \sin A&=\dfrac{P}{H}\\ &=\dfrac{15k}{17k}\\ &=\dfrac{15}{17}\\\\ \sec A&=\dfrac{H}{B}\\ &=\dfrac{17k}{8k}\\ &=\dfrac{17}{8}\end{aligned}$$Q5. Given \(\sec \theta = \frac{13}{12}\) , calculate all other trigonometric ratios.
Solution:
$$\begin{aligned}\sec \theta &=\dfrac{13}{12}\\ \sec \theta &=\dfrac{H}{B}\\ \therefore \dfrac{H}{B}&=\dfrac{13}{12}\\ H&=13k\quad\scriptsize\text{(k is a positive number)}\\ B&=12k\end{aligned}$$By pythagoras theorem
$$\begin{aligned}H^{2}&=P^{2}+B^{2}\\ P^{2}&=H^{2}-B^{2}\\ &=13^{2}-12^{2}\\ &=169-144\\ P&=\sqrt{25}\\ &=5\end{aligned}$$Trigonometrical Ratios
$$\begin{aligned}\sin \theta &=\dfrac{P}{H}\\ &=\dfrac{5}{13}\\\\ \cos \theta &=\dfrac{B}{H}\\ &=\dfrac{12}{13}\\\\ \tan \theta &=\dfrac{P}{B}\\ &=\dfrac{5}{12}\\\\ \cot \theta &=\dfrac{B}{P}\\ &=\dfrac{12}{5}\\\\ \text{cosec } \theta &=\dfrac{H}{P}\\ &=\dfrac{13}{5}\end{aligned}$$
Q6. If \(angle A\) and \(\angle B\) are acute angles such that cos A = cos B, then show that \(\angle A = \angle B\).
Solution:
$$\begin{aligned}\cos A&=\cos B\\ \cos A&=\dfrac{AC}{AB}\\ \cos B&=\dfrac{BC}{AB}\\ \Rightarrow \dfrac{AC}{AB}&=\dfrac{BC}{AB}\\ AC&=BC\\ \angle B&=\angle A\end{aligned}$$ Angles opposite to equal side of triangleQ7. If \(\cot \theta = \frac{7}{8}\), evaluate :
(i) \(\dfrac{ (1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta)(1 - \cos \theta)}\)
(ii) \(\cot^2 \theta\)
Solution:
$$\begin{aligned}\cot \theta &=\dfrac{7}{8}\\ &=\dfrac{B}{P}\\ \dfrac{B}{P}&=\dfrac{7}{8}\\ B&=7k\quad\scriptsize\text{(k is positive number)}\\ P&=8k\end{aligned}$$By Pythagoras theorem
$$\begin{aligned}H^{2}&=B^{2}+P^{2}\\ H^{2}&=\left( 7k\right) ^{2}+\left( 8k\right) ^{2}\\ &=49k^{2}+64k^{2}\\ &=113k^{2}\\ H&=k\sqrt{113}\end{aligned}$$ $$\begin{aligned}\sin \theta &=\dfrac{P}{H}\\ &=\dfrac{8k}{k\sqrt{113}}\\ &=\dfrac{8}{\sqrt{113}}\\\\ \cos \theta &=\dfrac{B}{H}\\ &=\dfrac{7k}{k\sqrt{113}}\end{aligned}$$(i)
Substituting values
$$\scriptsize\begin{aligned}&\dfrac{\left( 1+\sin \theta \right) \times \left( 1-\sin \theta \right) }{\left( 1+\cos \theta \right) \times \left( 1-\cos \theta \right) }\\ &=\dfrac{\left( 1+\dfrac{8}{\sqrt{113}}\right) \times \left( 1-\dfrac{8}{\sqrt{113}}\right) }{\left( 1+\dfrac{7}{\sqrt{113}}\right) \times \left( 1-\dfrac{7}{\sqrt{113}}\right) }\\ &=\dfrac{1-\left[ \dfrac{8}{\sqrt{113}}\right] ^{2}}{1-\left[ \dfrac{7}{\sqrt{113}}\right] ^{2}}\\ &=\dfrac{113-64}{113-49}\\ &=\dfrac{49}{64}\end{aligned}$$(ii)
$$\begin{aligned}\cot \theta &=\dfrac{7}{8}\\ \cot ^{2}\theta &=\left( \dfrac{7}{8}\right) ^{2}\\ \cot ^{2}\theta &=\dfrac{49}{64}\end{aligned}$$Q8. If 3 cot A = 4, check whether \(\dfrac{1-\tan^2 A}{1+ \tan^2 A}=\cos^2 A-\sin^2 A\) or not.
Solution:
3 cot A = 4
$$\begin{aligned}3\times \cot A&=4\\ &=\dfrac{4}{3}\\ &=\dfrac{B}{P}\\ \dfrac{B}{P}&=\dfrac{4}{3}\\ B&=4k\quad\scriptsize\text{(k is positive number)}\\ P&=3k\end{aligned}$$Using pythagoras theorem we can find H (Hypotenuse of triangle)
$$\begin{aligned}H^{2}&=B^{2}+P^{2}\\ &=\left( 4k\right) ^{2}+\left( 3k\right) ^{2}\\ &=16k^{2}+9k^{2}\\ &=25k^{2}\\ H&=\sqrt{25k^{2}}\\ H&=5k\end{aligned}$$ $$\begin{aligned}\cos A&=\dfrac{B}{H}\\ &=\dfrac{4k}{5k}\\ &=\dfrac{4}{5}\\\\ \cos ^{2}A&=\left( \dfrac{4}{5}\right) ^{2}\\ &=\dfrac{16}{25}\\\\ \sin A&=\dfrac{P}{H}\\ &=\dfrac{3k}{5k}\\ &=\dfrac{3}{5}\\\\ \sin ^{2}A&=\left( \dfrac{3}{5}\right) ^{2}\\ &=\dfrac{9}{25}\end{aligned}$$ $$\begin{aligned}&\cos ^{2}A-\sin ^{2}A\\ &\dfrac{16}{25}-\dfrac{9}{25}\\ &\dfrac{16-9}{25}\\ &=\dfrac{7}{25}\end{aligned}$$ $$\begin{aligned}\cot A&=\dfrac{4}{3}\\ &\dfrac{1}{\cot A}\\ &=\dfrac{3}{4}\\\\ \tan ^{2}A&=\left( \dfrac{3}{4}\right) ^{2}\\ &=\dfrac{9}{16}\end{aligned}$$ $$\begin{aligned}&\dfrac{1-\tan ^{2}A}{1+\tan ^{2}A}\\ &\dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}\\ &\dfrac{\dfrac{16-9}{16}}{\dfrac{16+9}{16}}\\ &\dfrac{7}{25}\\ \therefore \dfrac{1-\tan ^{2}A}{1+\tan ^{2}A}&=\cos ^{2}A-\sin ^{2}A=\dfrac{7}{25}\end{aligned}$$Q9. In triangle ABC, right-angled at B, if tan A = \(\dfrac{1}{\sqrt3}\) find the value
of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
$$\begin{aligned}\tan A&=\dfrac{1}{\sqrt{3}}\\ \tan A&=\dfrac{P}{B}\\ \dfrac{P}{B}&=\dfrac{1}{\sqrt{3}}\\ P&=1k\\ B&=\sqrt{3}k\end{aligned}$$Using pythagoras theorem we can find Hypotenuse
$$\begin{aligned}H^{2}&=P^{2}+B^{2}\\ &=k^{2}+\left( \sqrt{3}k\right) ^{2}\\ &=k^{2}+3k^{2}\\ &=4k^{2}\\ H&=\sqrt{4k^{2}}\\ H&=2k\end{aligned}$$ $$\begin{aligned}\sin A&=\dfrac{P}{H}\\ &=\dfrac{k}{2k}\\ &=\dfrac{1}{2}\\\\ \sin C&=\dfrac{P}{H}\\ &=\dfrac{\sqrt{3}k}{2k}\\ &=\dfrac{\sqrt{3}}{2}\\\\ \cos A&=\dfrac{B}{H}\\ \cos A=\dfrac{\sqrt{3}k}{2k}\\ &\\=\dfrac{\sqrt{3}}{2}\\ \cos C&=\dfrac{B}{H}\\ &=\dfrac{k}{2k}\\ &=\dfrac{1}{2}\end{aligned}$$(i)
Sin A Cos C+ cos A Sin C
substituting values in equation
(ii)
$$\begin{aligned}&\cos A\times \cos C-\sin A\times \sin C\\ &\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}\\ &=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}\\ &=0\end{aligned}$$
Q10. In \(\triangle PQR\), right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
PR + QR = 25
PQ = 5
Let QR = x
PR = 25-x
using Pythagoras theorem
Calculating Ratios sin P, cos P, tan P
$$\begin{aligned}\sin P&=\dfrac{QR}{PR}\\ &=\dfrac{12}{13}\\\\ \cos P&=\dfrac{PQ}{QR}\\ &=\dfrac{5}{13}\\\\ \tan P&=\dfrac{QR}{PQ}\\ &=\dfrac{12}{5}\end{aligned}$$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 \(\theta\).
Solution:
(i) False.\(\tan A=\frac{\text{Perpendicular}}{\text{Base}}\) can be less than, equal to, or greater than 1 depending on the triangle; for example, when the perpendicular is greater than the base, \(\tan A > 1\)
(ii) True.
\(\sec A=\frac{\text{Hypotenuse}}{\text{Adjacent side}}\) ; since the hypotenuse is the largest side, this ratio can be \(\frac{12}{5}\gt 1\) for a suitable right triangle (e.g. sides proportional to 5, 12, 13).
(iii) False.
“cos A” is the standard abbreviation for cosine of angle A, while cosecant is written as “cosec A” or “csc A”, so cos A does not stand for cosecant.
(iv) False.
“cot A” means the cotangent of angle A, a single trigonometric ratio, not a product of some quantity “cot” and the number A.
(v) False.
\(\sin \theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\),and in a right triangle the hypotenuse is the longest side, so this ratio is always less than or equal to 1; it can never be \(\frac{4}{3}\gt 1\)
Frequently Asked Questions
Trigonometry is the branch of mathematics that studies the relationship between the sides and angles of a right-angled triangle usin g trigonometric ratios such as sin e, cos in e, and tan gent.
Trigonometric ratios are ratios of the lengths of the sides of a right triangle with respect to one of its acute angles. They include sin , cos , tan , cos ec, sec , and cot .
The six ratios are: sin \(\theta\), cos \(\theta\), tan \(\theta\), cos ec\(\ \theta\), sec \(\theta\), and cot \(\theta\).
sin \(\theta\) = Opposite side ÷ Hypotenuse.
cos \(\theta\) = Adjacent side ÷ Hypotenuse.
tan \(\theta\) = Opposite side ÷ Adjacent side.
tan \(\theta\) = sin \(\theta\) ÷ cos \(\theta\).
cosec\(\ \theta\) = 1 ÷ sin \(\theta\) = Hypotenuse ÷ Opposite side.
sec \(\theta\) = 1 ÷ cos \(\theta\) = Hypotenuse ÷ Adjacent side.
cot \(\theta\) = 1 ÷ tan \(\theta\) = Adjacent side ÷ Opposite side.
Values include: sin 0\(^\circ\)=0, sin 30\(^\circ\)=1/2, sin 45\(^\circ\)=v2/2, sin 60\(^\circ\)=v3/2, sin 90\(^\circ\)=1 (others similarly defined).
They help solve real-life problems involving heights, distan ces, angles of elevation/depression, navigation, physics, engineering, and architecture.
The angle formed between the horizontal line and the line of sight when the observer looks upward at an object.
The angle formed between the horizontal line and the line of sight when the observer looks downward from a higher point.
sin ²\(\ \theta\) + cos ²\(\ \theta\) = 1.
1 + tan ²\(\ \theta\) = sec ²\(\ \theta\) and 1 + cot ²\(\ \theta\) = cos ec²\(\ \theta\).
tan \(\theta\) × cot \(\theta\) = 1.
sin \(\theta\) × cosec\(\ \theta\) = 1.
cos \(\theta\) × sec \(\theta\) = 1.
45\(^\circ\), because sin 45\(^\circ\) = cos 45\(^\circ\) = v2/2.
Only acute angles (0\(^\circ\) < \(\theta\) < 90\(^\circ\)) are considered in this chapter.
No, negative angles and circular trigonometric functions are not introduced at this level.
Trigonometric ratios help determine unknown heights, widths, and distan ces by relating them to measured angles.
tan \(\theta\) = sin \(\theta\)/cos \(\theta\).
Use the mnemonic: SOH-CAH-TOA.or Pundit Badri Prasad Har Har Bole
Because it lies opposite the right angle, based on the Pythagorean theorem.
tan 90\(^\circ\) is undefined because cos 90\(^\circ\)=0.
sec 90\(^\circ\) = 1/cos 90\(^\circ\) = undefined.
Because tan 0\(^\circ\)=0, so cot 0\(^\circ\) = 1/0, which is undefined.
Two angles whose sum is 90\(^\circ\).
sin (90\(^\circ\)-\(\ \theta\))=cos \(\theta\), cos (90\(^\circ\)-\(\ \theta\))=sin \(\theta\), tan (90\(^\circ\)-\(\ \theta\))=cot \(\theta\), etc.
Ratios depend only on the angle, not on the actual size of the triangle.
No. For a given angle, the trigonometric ratios remain constan t.
\(\sin \theta\) = tan \(\theta\sqrt{(1+tan ²\ \theta}\).
\(\cos \theta = \dfrac{1}{\sqrt{(1+tan ² \theta}}\).
\(\tan \theta = \dfrac{\sin \theta}{\sqrt{(1-sin^2\ \theta}}\).
\(\tan \theta= \dfrac{\sqrt{(1-cos^2\ \theta)}}{\cos \theta}\).
Problems involving complementary angles and stan dard value tables are frequently tested.
Measuring mountain heights, building construction, aviation, satellite tracking, GPS, and navigation.
Class 10 NCERT curriculum covers only degree measure for introductory learning.
Yes, but such applications require advanced formulas (sin e rule, cos in e rule) taught in higher classes.
Theodolites, sextan ts, clinometers, laser rangefinders, and surveying instruments.
A surveying instrument used to measure horizontal and vertical angles for calculating heights and distan ces.
It is used in designing structures, mechanical components, electrical waves, circuits, and digital systems.
It creates equal opposite and adjacent sides, leading to simple trigonometric values.
Three core identities: sin ²\(\ \theta\) + cos ²\(\ \theta\) = 1; 1 + tan ²\(\ \theta\) = sec ²\(\ \theta\); 1 + cot ²\(\ \theta\) = cosec²\(\ \theta\).
They help students quickly recall stan dard values essential for solving exam problems.
Usin g calculators, ignoring diagrams, and misidentifying opposite/adjacent sides.
Visualizing the triangle reduces mistakes and clarifies angle–side relationships.
Understan d the geometric meaning rather than memorizing formulas blindly.
Identify the angle first; the side directly across it is opposite, and the side touching it (except hypotenuse) is adjacent.
Yes, because all angles considered (0\(^\circ\)–90\(^\circ\)) lie in the first quadrant.
It leads to understan ding functions, calculus, vectors, coordinate geometry, and physics waveforms.
sin 30\(^\circ\) = 1/2, derived from the geometry of a 30\(^\circ\)-60\(^\circ\) right triangle.
Yes, cartography uses trigonometric principles to estimate distan ces and directions.
To introduce students to trigonometric ratios, identities, and basic applications in a right triangle.
Stan dard values, complementary angles, identities, and basic height-distan ce problems.
An angle greater than 0\(^\circ\) and less than 90\(^\circ\).
Finding heights usin g the angle of elevation and a known distan ce.
It teaches ratio relationships, spatial interpretation, and analytical problem-solving.
