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NCERT solutions for class 10 maths Chapter 8 Introduction to Trigonometry part 1 Trigonometric ratios

Trigonometric Ratios—properties and concepts

ncert solutions class 10 maths ch 8 pt 1

In this 4 part series, we will cover the topic Introduction to trigonometry.

This series will complement Chapter 8 of NCERT Class 10 Maths and will consist of 4 parts:

  1. Trigonometric ratios,
  2. Trigonometric ratios for selected angles,
  3. Trigonometric ratios for complementary angles, and,
  4. Basic trigonometric identities.

This first part of the series complements Exercise 1 of Chapter 8 of NCERT Class 10 Maths, in short, Ex 8.1 class 10.

In this session, concept explanation and problem solutions on trigonometric ratios will be covered.

This will be made up of three components:

  • First: concepts explained in an easy to understand way,
  • Second: example problems that will be used for establishing the concepts more clearly and,
  • Third: the solutions to the problems in the first exercise in the NCERT Chapter. The problems will be listed first, followed by answers and then solutions.

Starting the conceptual component, let us first know in brief what trigonometry is.

What is trigonometry

Trigonometry is a rich topic favoured in schools as well as in most competitive exams. The reason for this popularity lies in the ways that trigonometry problems can be used for judging pattern identification and problem solving ability of the student.

It is a branch of mathematics that is built primarily on ONLY ONE geometric closed-shape—the right-angled triangle.

This is the first time that through the well-defined trigonometric functions such as $\sin \theta$ or $\cos \theta$, relationships are established between two primary attributes or characteristics of a right-angled triangle—the lengths of a pair of sides and an internal angle.

For example, $\sin \theta$ is the ratio of height and hypotenuse (where height is the side opposite to $\angle \theta$) of a right-angled triangle as shown below,

ncert-10-math-ch8-1-1

Like $\sin \theta$ five other basic trigonometric functions are defined as ratios of combinations of two sides of the right triangle.

These form the foundation of Trigonometry.

In a right-angled triangle, as sum of the two angles other than the right angle is also $90^0$, both these angles are acute, that is less than $90^0$. In such a triangle, it is always possible to express any of the basic trigonometric functions as a positive ratio of two sides of the triangle.

That's why we say, trigonometry starts with trigonometric ratios. Throughout this session, we'll learn about properties of these trigonometric ratios.


Note on maximum and minimum value of trigonometric ratio $\sin \theta$ with change in angle

Numerically the ratio of two lengths must be positive. Also, hypotenuse in denominator of the ratio of $\sin \theta$ is greater than the height except when base is of zero length at $\theta=90^0$ and height becomes equal in length to hypotenuse. With $\theta=90^0$ then, $\sin \theta$ takes the maximum value of 1.

On the other hand, in the limiting case when $\theta=0^0$, the height becomes zero, base becomes equal in length to hypotenuse, and $\sin \theta$ reaches the value 0.

So, in a right-angled triangle, limits of the value of $\sin \theta$ is given by, $0 \leq \sin \theta \leq 1$.

In other words,

The maximum value of ratio $\sin \theta$ is 1 and minimum value is 0 in a right-angled triangle.

These limiting values hold good with $\theta$ in a right-angled triangle.

Being a mathematical entity though, definition of $\sin \theta$ is expanded to include values of $\theta$ more than $90^0$ when the triangle would no longer remain a right-angled triangle. In such a general case, as angle $\theta$ varies from $0^0$ to $360^0$, the trigonometric function $\sin \theta$ varies between its maximum value of $+1$ (at $\theta=90^0$) and minimum value of $-1$ (at $\theta=270^0$).

Naturally, there are other advanced concepts such as "sine inverse theta" or $\sin^{-1} \theta$, or $\theta$ as a compound angle in a topic as rich as Trigonometry.

Those concepts are not difficult and you will learn those in due time.

Right now, we are at the start of our journey to know Trigonometry and concentrate just on Trigonometric ratios, because that's what the trigonometric functions are in a right-angled triangle without any more complication.


Use of Trigonometry in brief

The commonly known use of trigonometry is—in measuring height of a tall object or the distance from observation point to the base of a tall object.

For such measurements, you need to be able to measure, what is termed as, angle of inclination by a separate instrument. Let us explain briefly by the following figure.

ncert-10-math-ch8-1-2

If you can measure the angle of viewing the top of a tall Minar from where you stand, you will know easily the value of a trigonometric ratio function $\tan \theta$ that is the ratio of height and base of the right-angled triangle $\triangle ABC$.

$\tan \theta = \displaystyle\frac{\text{height of minar}}{\text{distance of base of Minar}}$

With value of $\tan \theta$ known, if you know the height, you can easily get the value of distance and vice-versa.

For example, suppose you know the height of the minar to be 73 m, and measured the angle $\theta$ to be such that the value of $\tan \theta$ you find to be $\displaystyle\frac{1}{2}$.

Then you can confidently say that you are standing at a distance of $2\times{73}=146$ m from the base of the minar.


Note: To find the value of $\tan \theta$ for any value of $\theta$ you need to finally look up a table in which the values are pre-calculated for all angles. The method is a little advanced, and won't be taught to you in this class.


We'll devote the whole of next chapter on uses of trigonometry.

Let's now go into the topic of trigonometric ratios in more details.

Trigonometric ratios

We'll base our discussions on a standard form of acute-angled right-angled triangle $\triangle ABC$ with right angle at vertex B as shown below.

ncert-10-math-ch8-1-3

$\text{height}$ is side AB opposite to $\angle \theta$ (on which trigonometric functions or ratios are defined), $\text{base}$ is side BC and $\text{hypotenuse}$ of the right-angled triangle is side CA. The trigonometric functions on $\angle ACB$ or $\angle \theta$ in short will be expressed as trigonometric ratios between combinations of two sides of the right triangle.

There are six trigonometric ratios in all—$\sin \theta$, $\cos \theta$, $\tan \theta$, $\text{cosec } \theta$, $\text{sec } \theta$ and $\text{cot } \theta$.

Just remember:

  1. height is the side opposite to the angle $\angle \theta$ on which the trigonometric functions are defined,
  2. $\sin \theta$ is a trigonometric function value of which is a ratio of lengths of two sides of a right-angled triangle,
  3. As each trigonometric function in a right-angled triangle can be expressed as a positive ratio, at this initial stage, we call the trigonometric functions as trigonometric ratios.

We'll use all through the session, the term trigonometric ratio and not trigonometric function as these two are equivalent in a right-angled triangle.

Each of the six ratios originate as a ratio of two sides of the right triangle $\triangle ABC$.

With three sides, there can be $3$ unique combinations of two sides—(AB, BC), (BC, CA), and (CA, AB).

Again, from each of these pairs of sides, two ratios can be created—direct and inverse. For example, you will get two ratios from the first pair of sides AB and BC as,

$\displaystyle\frac{AB}{BC}$ and $\displaystyle\frac{BC}{AB}$.

That's how taking direct and inverse ratios of three unique combinations of 2 sides, $3\times{2}=6$ ratios in all arise.

Each of the six trigonometric ratios is defined as one of these six ratios of two sides of a right triangle.

Note: The value of a trigonometric ratio varies when the value of angle $\angle \theta$ changes.

We consider the two primary trigonometric ratios as,

$\sin \theta=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{AB}{CA}$, and

$\cos \theta=\displaystyle\frac{\text{base}}{\text{hypotenuse}}=\frac{BC}{CA}$.


Note: Each of these two is complementary to the other—if you know the value of one, you can easily deduce the value of the other. This is because, if you take the square of each and sum up the two, you will get 1 by Pythagoras theorem which states—in a right-angled triangle, square of hypotenuse is equal to the sum of squares of the other two sides. 


The reason why we classify these two trigonometric ratios as primary is because the third basic trigonometric ratio $\tan \theta$ is the ratio of these two,

$\tan \theta=\displaystyle\frac{\text{height}}{\text{base}}=\frac{AB}{BC}=\frac{\sin \theta}{\cos \theta}$.

If you know the values of $\sin \theta$ and $\cos \theta$ you can easily deduce the value of $\tan \theta$ as their ratio.


Patterns to remember relations of trigonometric ratios: In both ratios of $\sin \theta$ and $\cos \theta$, $\text{hypotenuse}$ appears in the denominator and that's why, maximum absolute value of both can reach 1 and no more.

How to Recall the series of these three basic trigonometric ratios easily: The series starts with $\sin \theta$ with its numerator as the height. If you can remember this much, you can form $\cos \theta$ numerator as the base with both denominators as hypotenuse.

$\tan \theta$ is formed by taking the ratio of $\sin \theta$ and $\cos \theta$ where the numerator is $\sin \theta$.

In summary, $\sin \theta$ is MORE primary than its other primary partner $\cos \theta$ (as it appears in the numerator of $\tan \theta$).

Rest of the three trigonometric ratios are formed by inverting $\sin \theta$, $\cos \theta$ and $\tan \theta$.


Furthermore, we classify the three trigonometric ratios $\sin \theta$, $\cos \theta$ and $\tan \theta$ as basic trigonometric ratios because—you'll get the rest of the three trigonometric ratios by just inverting each of the three.

$\text{cosec } \theta=\displaystyle\frac{\text{hypotenuse}}{\text{height}}=\frac{CA}{AB}=\frac{1}{\sin \theta}$,

$\text{sec } \theta=\displaystyle\frac{\text{hypotenuse}}{\text{base}}=\frac{CA}{BC}=\frac{1}{\cos \theta}$, and,

$\text{cot } \theta=\displaystyle\frac{\text{base}}{\text{height}}=\frac{BC}{AB}=\frac{\cos \theta}{\sin \theta}=\frac{1}{\tan \theta}$.

If you analyze the six trigonometric ratios, you will find a basic truth of trigonometry,

If you know the value of any one of the six trigonometric ratios, you'll be able to deduce the values of rest.

This is the most important basic property of the set of trigonometric ratios.

Note: In a general form, we call this result as Trigonometric basic function derivation principle.

Let's verify its truth.

How you can evaluate all other trigonometric ratios from the given value of any of the trigonometric ratios

We'll use an example in which the trigonometric ratios are on acute angle $\angle \theta$ in a right-angled triangle as usual.

Problem example 1.

In right-angled triangle $\triangle ABC$ with right angle at B, if $\tan \theta=\displaystyle\frac{3}{4}$, find the value of rest of the trigonometric ratios.

ncert-10-math-ch8-1-4

Let's solve the problem taking a direct approach.

Solution example 1: Direct approach

Assuming cancelled out HCF for the given ratio value for $\tan \theta=\displaystyle\frac{3}{4}=\frac{AB}{BC}$ to be a positive integer $p$, the actual values of sides AB and BC are expressed as,

$AB=3p$ and $BC=4p$.

Apply Pythagoras theorem on the right-angled triangle to get the actual value of the hypotenuse in terms of $p$,

$CA=\sqrt{9p^2+16p^2}=5p$.

So,

$\sin \theta=\displaystyle\frac{AB}{CA}=\frac{3}{5}$,

$\cos \theta=\displaystyle\frac{BC}{CA}=\frac{4}{5}$,

$\text{cosec } \theta=\displaystyle\frac{1}{\sin \theta}=\frac{5}{3}$,

$\text{sec } \theta=\displaystyle\frac{1}{\cos \theta}=\frac{5}{4}$, and,

$\text{cot } \theta=\displaystyle\frac{1}{\tan \theta}=\frac{4}{3}$.

$p$ cancels out in each ratio.

It would be possible in this way to evaluate the values of all the rest of the trigonometric ratios when value of any other trigonometric ratio is given.

Just find the actual value of the third side in terms of dummy ratio factor $p$ by Pythagoras theorem and express rest of the functions as ratios of two sides—$p$ would cancel out in each case.

Corollary 1: Evaluation of any trigonometric expression in a right-angled triangle if a non-fraction (irrational or integer) value of a single trigonometric ratio is known

It follows from the previous result that,

You can evaluate any trigonometric expression consisting of the six basic trigonometric ratios on an angle $\theta$ of a right-angled triangle, if you know the value of a single trigonometric ratio—that may or may not be a fraction.

We have already seen in problem example 1 that when value of $\tan \theta$ is a rational fraction, you could get the value of all the rest of the trigonometric ratios, and so you could have evaluated any expression consisting of these basic trigonometric ratios as well. It is logical.

Now in second problem example which is a simpler version of the previous example, you'll see that the same result is true if the value of $\tan \theta$ is not a fraction.

Problem example 2.

If $\tan \theta=\sqrt{3}$, find the value of $2\sin \theta \cos \theta$.

Solution example 2.

The following right triangle $\triangle ABC$ with acute angle $\theta$ will be used for solution.

ncert-10-math-ch8-1-5

Given,

$\tan \theta =\sqrt{3}$,

Or, $\tan \theta = \displaystyle\frac{AB}{BC} =\frac{\sqrt{3}}{1}=\frac{\sqrt{3}p}{p}$.

The cancelled out HCF is reintroduced in the ratio terms as factor $p$ that is a positive integer.

So actual values of AB and BC are,

$AB=\sqrt{3}p$, and $BC=p$.

Applying Pythagoras theorem on the right triangle, the hypotenuse CA is,

$CA =\sqrt{3p^2+p^2}=2p$.

It follows,

$\sin \theta = \displaystyle\frac{AB}{CA}=\frac{\sqrt{3}}{2}$, and,

$\cos \theta=\displaystyle\frac{BC}{CA}=\frac{1}{2}$, $p$ cancels out in both cases.

So,

$2\sin \theta \cos \theta =\displaystyle\frac{\sqrt{3}}{2}$.


Note: You will learn in the next session that $\tan \theta =\sqrt{3}$, $\sin \theta=\displaystyle\frac{\sqrt{3}}{2}$, and $\cos \theta=\displaystyle\frac{1}{2}$ occur when $\theta=60^0$.


Corollary 2: Evaluation of any trigonometric expression in a right-angled triangle if actual lengths of any two sides are known

At this stage it easy to conclude that,

You can evaluate any trigonometric expression consisting of the six basic trigonometric ratios on an angle $\theta$ of a right-angled triangle, if you know actual lengths of any two sides of the right triangle.

This also is a simple situation and let's verify the conclusion using an example problem.

Problem example 3.

In a right-angled triangle $\triangle ABC$ with right angle at B, if height AB is 7 cm and hypotenuse CA is 25 cm, find the value of,

  1. $\cos^2 \theta-\sin^2 \theta$
  2. $\cos \theta - \sin \theta$.

Solution example 3.

We'll use the following diagram to explain the solutions.

ncert-10-math-ch8-1-6

With two sides of the right triangle given, the length of the third side by Pythagoras theorem is,

$BC=\sqrt{25^2-7^2}=\sqrt{625-49}=\sqrt{576}=24$ cm.

It follows,

$\sin \theta=\displaystyle\frac{AB}{CA}=\frac{7}{25}$, and

$\cos \theta=\displaystyle\frac{BC}{CA}=\frac{24}{25}$.

So,

$\cos^2 \theta - \sin^2 \theta=\displaystyle\frac{24^2-7^2}{25^2}=\displaystyle\frac{576-49}{625}=\frac{527}{625}$, and

$\cos \theta - \sin \theta=\displaystyle\frac{24}{25}-\displaystyle\frac{7}{25}=\frac{17}{25}$.

Before taking up solving the exercise problems we'll explain two more basic concepts on trigonometric ratios.

With same equal angle $\theta$ in two triangles, corresponding trigonometric ratios on $\theta$ are equal

The concept is stated as,

If in two right-angled triangles, one angle in one triangle is equal to an angle in the second triangle, corresponding trigonometric ratios on the equal angle for both triangles become equal.

It follows,

The value of a trigonometric ratio in right-angled triangles depends only on the angle on which the ratio is taken (whatever be the size of the triangle).

We'll verify this basic, seemingly obvious, concept using the following diagram.

ncert-10-math-ch8-1-7

In the two triangles, $\triangle ABC$ and $\triangle PQC$, one angle $\angle \theta$ is common, that is, $\angle BCA=\angle QCP$.

To show that all corresponding trigonometric ratios on $\angle \theta$ for the two triangles are equal.

With $\angle \theta$ common and the second right angle also equal, the third angle pair must also be equal, that is,

$\angle CAB = \angle CPQ$.

It follows, $\triangle ABC \sim \triangle PQR$.

So in two similar triangles, ratios of pairs of corresponding sides are equal,

$\displaystyle\frac{AB}{PQ}=\frac{CA}{CP}=\frac{BC}{QC}$.

Taking the first pair of equal ratios,

$\displaystyle\frac{AB}{PQ}=\frac{CA}{CP}$,

Or, $\displaystyle\frac{AB}{CA}=\frac{PQ}{CP}=\sin \theta$ in both triangles.

Taking the second pair of equal ratios,

$\displaystyle\frac{CA}{CP}=\frac{BC}{QC}$,

Or, $\displaystyle\frac{BC}{CA}=\frac{QC}{CP}=\cos \theta$ in both triangles.

Equality of even one trigonometric ratio $\sin \theta$ ensures equality of all the rest of the trigonometric ratios.

Same equal trigonometric ratios signifies similar triangles and equal corresponding angles

This time we'll verify the concept,

In two right-angled triangles, if a trigonometric ratio $\sin \alpha=\sin \beta$, then $\alpha=\beta$.

We'll use the following two right-angled triangles $\triangle ABC$ and $\triangle PQR$ to verify the concept.

ncert-10-math-ch8-1-8

In the two triangles, $\sin \alpha=\sin \beta$.

In ratio form then,

$\sin \alpha=\displaystyle\frac{AB}{CA}$, and,

$\sin \beta=\displaystyle\frac{PQ}{RP}$.

So,

$\displaystyle\frac{AB}{CA}=\frac{PQ}{RP}=m$, say.

Thus,

$AB=mCA$, and by Pythagoras theorem,

$BC=\sqrt{CA^2-AB^2}=(\sqrt{1-m^2})CA$,

Or, $\displaystyle\frac{BC}{CA}=\sqrt{1-m^2}$, and,

$\displaystyle\frac{AB}{BC}=\frac{m}{\sqrt{1-m^2}}$.

Similarly,

$PQ=mRP$,

$QR=\sqrt{RP^2-PQ^2}=(\sqrt{1-m^2})RP$.

So,

$\displaystyle\frac{QR}{RP}=\sqrt{1-m^2}=\frac{BC}{CA}$,

Or, $\displaystyle\frac{BC}{QR}=\frac{CA}{RP}$.

And, $\displaystyle\frac{PQ}{QR}=\frac{m}{\sqrt{1-m^2}}=\frac{AB}{BC}$

So, $\displaystyle\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$.

That is—ratios of pairs of corresponding sides of the two triangles are equal.

This is side-side-side or SSS criterion for similarity of two triangles, stated as,

If in two triangles, three ratios of sides of one triangle to sides of the second triangle (ratio terms of two sides taken from first to second triangle) are equal, corresponding angles of the two triangles are equal and the triangles are similar.

It follows then, the two triangles $\triangle ABC$ and $\triangle PQR$ are similar, and so, the corresponding angles,

$\angle \alpha=\angle \beta$.

Angles opposite to AB and PQ of the first ratio are the corresponding angles $\angle \alpha$ and $\angle \beta$ .

Problems to solve

Convention of naming the angles: In a $\triangle ABC$ or $\triangle PQR$, single letter angles represent angles formed at the corresponding vertices. For example, $\angle A$ represents the angle $\angle BAC$ at vertex A.

Problem 1.

In $\triangle ABC$, right-angled at $B$, $AB=24$ cm and $BC=7$ cm. Find the values of,

  1. $\sin A$, $\cos A$
  2. $\sin C$, $\cos C$.

Problem 2.

In the following diagram find the value of $\tan P-\text{cot }R$.

ncert-10-math-ch8-pt1-qs2

Problem 3.

If $\sin A=\displaystyle\frac{3}{4}$, calculate $\cos A$ and $\tan A$.

Problem 4.

Given $15\text{ cot }A=8$, find $\sin A$ and $\text{sec }A$.

Problem 5.

Given $\text{sec } \theta=\displaystyle\frac{13}{12}$, calculate all other trigonometric ratios.

Problem 6.

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

Problem 7.

If $\text{cot } \theta=\displaystyle\frac{7}{8}$, evaluate,

  1. $\displaystyle\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$,
  2. $\text{cot}^2 \theta$.

Problem 8.

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

Problem 9.

In $\triangle ABC$, right-angled at B, if $\tan A=\displaystyle\frac{1}{\sqrt{3}}$, find the value of,

  1. $\sin A \cos C + \cos A \sin C$
  2. $\cos A \cos C - \sin A \sin C$.

Problem 10.

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

Problem 11.

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

  1. The value of $\tan A$ is always less than 1.
  2. $\text{sec }A=\displaystyle\frac{12}{5}$ for some value of angle $\angle A$.
  3. $\cos A$ is the abbreviation of $\text{cosecant}$ of angle $\angle A$.
  4. $\text{cot }A$ is the product of $\text{cot}$ and $A$.
  5. $\sin \theta =\displaystyle\frac{4}{3}$ for some angle $\theta$.

Answers to the problems

Problem 1.i: $\sin A=\displaystyle\frac{7}{25}$, $\cos A=\displaystyle\frac{24}{25}$; Problem 1.ii: $\sin C=\displaystyle\frac{24}{25}$, $\cos C=\displaystyle\frac{7}{25}$.

Problem 2: 0.

Problem 3: $\cos A=\displaystyle\frac{\sqrt{7}}{4}$, $\tan A=\displaystyle\frac{3}{\sqrt{7}}$.

Problem 4: $\sin A=\displaystyle\frac{15}{17}$, $\text{sec }A=\displaystyle\frac{17}{8}$.

Problem 5: $\sin \theta=\displaystyle\frac{5}{13}$, $\cos \theta=\displaystyle\frac{12}{13}$, $\tan \theta=\displaystyle\frac{5}{12}$, $\text{cosec } \theta=\displaystyle\frac{13}{5}$, $\text{cot }\theta=\displaystyle\frac{12}{5}$.

Problem 7.1: $\displaystyle\frac{49}{64}$; Problem 7.ii: $\displaystyle\frac{49}{64}$.

Problem 8: Yes, the equality relation is correct.

Problem 9.1: 1; Problem 9.ii: 0.

Problem 10: $\sin P=\displaystyle\frac{12}{13}$, $\cos P=\displaystyle\frac{5}{13}$, $\tan P=\displaystyle\frac{12}{5}$.

Problem 11.i: False; Problem 11.ii: True; Problem 11.iii: False; Problem 11.iv: False; Problem 11.v: False.


Solution to the problems

Convention of naming the angles: In a $\triangle ABC$, or $\triangle PQR$ single letter angles represent angles formed at the corresponding vertices. For example, $\angle A$ represents the angle $\angle BAC$ at vertex A.

Problem 1

In $\triangle ABC$, right-angled at $B$, $AB=24$ cm and $BC=7$ cm. Find the values of,

  1. $\sin A$, $\cos A$
  2. $\sin C$, $\cos C$.

Solution 1

The following diagram represents the problem.

ncert-10-math-ch8-pt1-q1

By Pythagoras theorem on the right triangle, hypotenuse,

$CA=\sqrt{AB^2+BC^2}=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25$ cm.

Solution 1.i.

Height is BC (opposite to $\angle A$) and base is AB, when trigonometric ratio is taken on $\angle A$.

$\sin A=\displaystyle\frac{BC}{CA}=\frac{7}{25}$, and,

$\cos A=\displaystyle\frac{AB}{CA}=\frac{24}{25}$.

Solution 1.ii.

Height is AB (opposite to $\angle C$) and base is BC, when trigonometric ratio is taken on $\angle C$.

$\sin C=\displaystyle\frac{AB}{CA}=\frac{24}{25}$, and,

$\cos C=\displaystyle\frac{BC}{CA}=\frac{7}{25}$.

Problem 2.

In the following diagram find the value of $\tan P-\text{cot }R$.

ncert-10-math-ch8-pt1-qs2

Solution 2.

We'll use the diagram in the question to explain the solution.

We must first calculate QR using Pythagoras theorem as,

$QR=\sqrt{RP^2-PQ^2}=\sqrt{13^2-12^2}$

$=\sqrt{169-144}=\sqrt{25}=5$ cm.

When evaluating $\tan P$, the base of the right triangle is PQ, and height is QR (side opposite to $\angle P$).

So, $\tan P=\displaystyle\frac{\text{height}}{\text{base}}=\frac{QR}{PQ}=\frac{5}{12}$.

But at the time of evaluation of $\text{cot }R$, the base is QR and height is PQ (side opposite to $\angle R$). 

So, $\text{cot }R=\displaystyle\frac{1}{\tan R}=\frac{\text{base}}{\text{height}}=\frac{QR}{PQ}=\frac{5}{12}$.

Final answer,

$\tan P-\text{cot }R=\displaystyle\frac{5}{12}-\displaystyle\frac{5}{12}=0$.

Problem 3.

If $\sin A=\displaystyle\frac{3}{4}$, calculate $\cos A$ and $\tan A$.

Solution 3.

The following diagram represents the question and aids solution.

ncert-10-math-ch8-pt1-q3

Introduce the cancelled out HCF in the given ratio as positive integer $p$ so that $\sin A$ can be expressed as a ratio of actual lengths of height and hypotenuse as,

$\sin A=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{3}{4}=\frac{3p}{4p}$.

Applying Pythagoras theorem, evaluate actual length of base AB in terms of $p$ as,

$AB=\sqrt{CA^2-BC^2}=(\sqrt{16-9})p=\sqrt{7}p$.

Then,

$\cos A=\displaystyle\frac{\text{base}}{\text{hypotenuse}}=\frac{\sqrt{7}}{4}$, and,

$\tan A=\displaystyle\frac{\text{height}}{\text{base}}=\frac{3}{\sqrt{7}}$, $p$ cancels out in both ratios.

Problem 4.

Given $15\text{ cot }A=8$, find $\sin A$ and $\text{sec }A$.

Solution 4.

Given $15\text{ cot }A=8$,

Or, $\text{ cot }A=\displaystyle\frac{8}{15}=\frac{\text{base}}{\text{height}}$.

Assuming cancelled put HCF of the ratio to be positive integer $p$, actual values of base and height are,

$\text{base}=8p$, and $\text{height}=15p$.

By Pythagoras theorem, the hypotenuse is then,

$\text{hypotenuse}=\sqrt{\text{height}^2+\text{base}^2}=(\sqrt{15^2+8^2})p$

$=\sqrt{289}p=17p$.

So, $\sin A=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{15}{17}$, $p$ cancels out.

And inverse of $\cos A$,

$\text{sec }A=\displaystyle\frac{\text{hypotenuse}}{\text{base}}=\frac{17}{8}$, again $p$ cancels out.

Problem 5.

Given $\text{sec } \theta=\displaystyle\frac{13}{12}$, calculate all other trigonometric ratios.

Solution 5.

$\text{sec }\theta$ is inverse of $\cos \theta$, and so,

$\text{sec }\theta=\displaystyle\frac{\text{hypotenuse}}{\text{base}}=\frac{13p}{12p}$.

$p$ is the cancelled out HCF reintroduced in the ratio as factors to the two ratio terms to get actual values of the ratio terms,

$\text{hypotenuse}=13p$, and,

$\text{base}=12p$.

Apply Pythagoras theorem on the right-angled triangle to get the length of the 3rd side $\text{height}$ in terms of $p$ as,

$\text{height}=\sqrt{\text{hypotenuse}^2-\text{base}^2}$

$=(\sqrt{13^2-12^2})p=(\sqrt{169-144})p$

$=(\sqrt{25})p=5p$

With values of all three sides obtained in terms of $p$, the rest of the five trigonometric ratios are evaluated as,

$\sin \theta=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{5p}{13p}=\frac{5}{13}$,

$\cos \theta=\displaystyle\frac{1}{\text{sec }\theta}=\frac{12}{13}$,

$\tan \theta=\displaystyle\frac{\sin \theta}{\cos \theta}=\frac{5}{12}$,

$\text{cosec }\theta=\displaystyle\frac{1}{\sin \theta}=\frac{13}{5}$, and,

$\text{cot }\theta=\displaystyle\frac{1}{\tan \theta}=\frac{12}{5}$.

Problem 6.

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

Solution 6.

The following diagram will be used for explaining the solution.

ncert-10-math-ch8-pt1-q6

In right-angled triangles, both the other two angles must be acute.

In $\triangle PQA$,

$\cos A=\displaystyle\frac{QA}{AP}$, and,

In $\triangle MNB$,

$\cos B=\displaystyle\frac{NB}{BM}$.

With two cosine ratios equal you have,

$\displaystyle\frac{QA}{AP}=\frac{NB}{BM}$....................(6.1),

Or, $\displaystyle\frac{QA}{NB}=\frac{AP}{BM}$..............(6.2).

Square up the two sides in equation (6.1) and subtract each from 1 (not 1 from each).

Result would be,

$\displaystyle\frac{AP^2-QA^2}{AP^2}=\frac{BM^2-NB^2}{BM^2}$,

Or, $\displaystyle\frac{PQ^2}{AP^2}=\frac{MN^2}{BM^2}$, applying Pythagoras theorem,

Or, $\displaystyle\frac{PQ}{AP}=\frac{MN}{BM}$..............(6.3)

Dividing equation (6.3) by equation (6.1),

$\displaystyle\frac{PQ}{QA}=\frac{MN}{NB}$,

Or, $\displaystyle\frac{PQ}{MN}=\frac{QA}{NB}=\frac{AP}{BM}$, using equation (6.2).

Ratios of all three pairs of corresponding sides being equal, by side-side-side or SSS property the two triangles are similar,

$\triangle PQA \sim \triangle MNB$.

It follows, corresponding angles, $\angle A=\angle B$, being opposite to the corresponding sides, PQ and MN of the first ratio.

Problem 7.

If $\text{cot } \theta=\displaystyle\frac{7}{8}$, evaluate,

  1. $\displaystyle\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$,
  2. $\text{cot}^2 \theta$.

Solution 7.i

Without going into the labour of evaluating $\sin \theta$ and $\cos \theta$ individually, we'll use the algebraic pattern of the relation of $(a+b)(a-b)=a^2-b^2$ to simplify the target equation to,

$\displaystyle\frac{1-\sin^2 \theta}{1-\cos^2 \theta}$.

Now apply the trigonometric relation $\sin^2 \theta+cos^2 \theta=1$ to further simplify,

$\displaystyle\frac{1-\sin^2 \theta}{1-\cos^2 \theta}=\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$

$=\text{cot }^2 \theta=\displaystyle\frac{49}{64}$.

Solution 7.ii.

Just square up both sides of the given equation,

$\text{cot }^2 \theta=\displaystyle\frac{49}{64}$.

Why $\sin^2 \theta+\cos^2 \theta=1$:

$\sin^2 \theta=\displaystyle\frac{\text{height}^2}{\text{hypotenuse}^2}$, and,

$\cos^2 \theta=\displaystyle\frac{\text{base}^2}{\text{hypotenuse}^2}$.

Sum up the two,

$\sin^2 \theta+\cos^2 \theta=\displaystyle\frac{\text{height}^2+\text{base}^2}{\text{hyptenuse}^2}=1$, as, sum of squares of base and height equals the square of hypotenuse by Pythagoras theorem.

Problem 8.

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

Solution 8.

Given,

$3\cot A=4$,

Or, $\cot A=\displaystyle\frac{4}{3}=\frac{4p}{3p}=\frac{\text{base}}{\text{height}}$, introducing cancelled out HCF in the ratio terms as positive integer factor $p$.

So, $\text{base}=4p$, and $\text{height}=3p$.

Applying Pythagoras theorem,

$\text{hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

$=(\sqrt{4^2+3^2})p=5p$.

With $\tan A=\displaystyle\frac{1}{\text{cot }A}=\frac{3}{4}$, the LHS of the target equation is simplified to,

$\displaystyle\frac{1-\displaystyle\frac{9}{16}}{1+\displaystyle\frac{9}{16}}=\frac{7}{25}$.

From actual lengths of three sides,

$\sin A=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{3}{5}$, and,

$\cos A=\displaystyle\frac{\text{base}}{\text{hypotenuse}}=\frac{4}{5}$, $p$ cancels out in both ratios.

It follows that the RHS of the target equation,

$RHS=\displaystyle\frac{4^2}{5^2}-\displaystyle\frac{3^2}{5^2}=\displaystyle\frac{7}{25}=LHS$.

Quicker solution 8 without evaluation of length of hypotenuse

$LHS=\displaystyle\frac{1-\displaystyle\frac{\sin^2 A}{\cos^2 A}}{1+\displaystyle\frac{\sin^2 A}{\cos^2 A}}$

$=\displaystyle\frac{\cos^2 A-\sin^2 A}{\cos^2 A+\sin^2 A}$

$=\cos^2 A-\sin^2 A$, as $\sin^2 A+\cos^2 A=1$,

$=RHS$.

Problem 9.

In $\triangle ABC$, right-angled at B, if $\tan A=\displaystyle\frac{1}{\sqrt{3}}$, find the value of,

  1. $\sin A \cos C + \cos A \sin C$
  2. $\cos A \cos C - \sin A \sin C$.

Solution 9.

We'll use the following diagram to explain the solutions.

ncert-10-math-ch8-pt1-q9

Introducing the cancelled out HCF positive integer $p$ as factors of both ratio terms,

$\tan A=\displaystyle\frac{BC}{AB}=\frac{1}{\sqrt{3}}=\frac{p}{\sqrt{3}p}$.

So actual values are, $BC=p$, and $AB=\sqrt{3}p$.

Applying Pythagoras theorem,

$CA=\sqrt{AB^2+BC^2}=2p$.

$\sin A=\displaystyle\frac{BC}{CA}=\frac{1}{2}$,

$\cos A=\displaystyle\frac{AB}{CA}=\frac{\sqrt{3}}{2}$,

$\sin C=\displaystyle\frac{AB}{CA}=\frac{\sqrt{3}}{2}$, and

$\cos C=\displaystyle\frac{BC}{CA}=\frac{1}{2}$.

Solution 9.i.

$\sin A \cos C + \cos A \sin C=\displaystyle\frac{1}{4}+\displaystyle\frac{3}{4}=1$.

Solution 9.ii.

$\cos A \cos C - \sin A \sin C=\displaystyle\frac{\sqrt{3}}{4}-\displaystyle\frac{\sqrt{3}}{4}=0$.

Quicker solution 9 without evalation of trigonometric ratios

From the above figure,

$\sin A=\displaystyle\frac{BC}{CA}$,

$\cos A=\displaystyle\frac{AB}{CA}$,

$\sin C=\displaystyle\frac{AB}{CA}$,

$\cos C=\displaystyle\frac{BC}{CA}$.

Solution 9.i.

$\sin A \cos C + \cos A \sin C=\displaystyle\frac{BC^2}{CA^2}+\displaystyle\frac{AB^2}{CA^2}$

$=\displaystyle\frac{AB^2+BC^2}{CA^2}$

$=\displaystyle\frac{CA^2}{CA^2}$, applying Pythagoras theorem,

$=1$.

Solution 9.ii.

$\cos A \cos C - \sin A \sin C=\displaystyle\frac{AB.BC}{CA^2}-\displaystyle\frac{BC.AB}{CA^2}=0$.


Note: For value of $\tan A=\displaystyle\frac{1}{\sqrt{3}}$, $\angle A=30^0$, and $\angle C=60^0$.

Also, in problem 9.i,

$\sin A \cos C + \cos A \sin C=\sin (A+C)=\sin 90^0=1$.

And and in problem 9.ii,

$\cos A \cos C - \sin A \sin C=\cos (A+C)=\cos 90^0=0$.

The values of trigonometric ratios for specific angles you'll learn in next part of this four part series, but the relations for $\sin (A+C)$, and $\cos (A+C)$, the concepts of Compound angles, you'll learn not before the next class (unless of course, you learn it by yourself earlier than that).


Problem 10.

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

Solution 10.

Unless you find out individual values of the three sides, it won't be possible to find the three required values the trigonometric ratios.

To achieve this goal we'll use simple algebra and Pythagoras relation.

First square up $PR+QR$ and add it with $PQ^2$. The result would be,

$PR^2+2PR.QR+QR^2+PQ^2=625+25$,

Or, $2PR^2+2PR.QR=650$, as by Pythagoras theorem, $PR^2=PQ^2+QR^2$,

Or, $PR(PR+QR)=325$, taking PR common factor,

Or, $25PR=325$, substituting given value of $PR+QR=25$,

So in right-angled triangle $\triangle PQR$ right-angled at Q, $PR=13$,

$QR=25-13=12$, and,

$PQ=5$.

Thus,

$\sin P=\displaystyle\frac{QR}{PR}=\frac{12}{13}$,

$\cos P=\displaystyle\frac{PQ}{PR}=\frac{5}{13}$, and,

$\tan P=\displaystyle\frac{PQ}{QR}=\frac{12}{5}$.

The solution figure is shown below.

ncert-10-math-ch8-pt1-q10

Problem 11.

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

  1. The value of $\tan A$ is always less than 1.
  2. $\text{sec }A=\displaystyle\frac{12}{5}$ for some value of angle $\angle A$.
  3. $\cos A$ is the abbreviation of $\text{cosecant}$ of angle $\angle A$.
  4. $\text{cot }A$ is the product of $\text{cot}$ and $A$.
  5. $\sin \theta =\displaystyle\frac{4}{3}$ for some angle $\theta$.

Solution 11.i.

The trigonometric ratio, $\tan A=\displaystyle\frac{\text{height}}{\text{base}}$.

In a right-angled triangle, it is true that both the base and height are always less than the hypotenuse, but there is no such constraint on relative values of height and base. For example, it may very well be,

$\text{height}=4$ cm,

$\text{base}=3$ cm, and

$\text{hypotenuse}=5$ cm.

In this case, $\tan A=\displaystyle\frac{4}{3} \gt 1$.

Answer: False.

Solution 11.ii.

When $\text{sec }A=\displaystyle\frac{12}{5}$,

$\displaystyle\frac{\text{hypotenuse}}{\text{base}}=\frac{12}{5}=\frac{12p}{5p}$, introducing positive integer $p$ as the cancelled out HCF as factor to both the ratio terms.

So hypotenuse is 12p units, base is 5p units and,

$\text{height}=(\sqrt{12^2-5^2})p=(\sqrt{119})p$ units.

These three sides would then form a valid right-angled triangle with a specific value for $\angle A$ without any constraint.

Answer: True.

Solution 11.iii.

Answer: False. 

$\cos A$ is the abbreviation for $\text{cosine}$ of $\text{A}$.

Solution 11.iv.

Answer: False.

$\text{cot }A$ is abbreviation for $\text{cotangent}$ of angle $\text{A}$. 

It is a trigonometric ratio function on angle A, not a product of cot and A.

Solution 11.v.

If $\sin \theta=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{4}{3}=\frac{4p}{3p}$, actual length of height will be more than the hypotenuse, which makes the right-angled triangle invalid.

Truth is,

In every right-angled triangle, hypotenuse is longer than both the base and height.

Answer: False.


NCERT Solutions for Class 10 Maths

Chapter 1: Real Numbers

NCERT Solutions for Class 10 Maths on Real numbers part 3, HCF and LCM by factorization and problem solutions

NCERT Solutions for Class 10 Maths on Real numbers part 2, Euclid’s division algorithms, HCF and problem solutions

NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions

Chapter 2: Polynomials

NCERT Solutions for Class 10 Maths Chaper 2 Polynomials 1 Geometrical Meaning of Zeroes of Polynomials

Chapter 3: Linear Equations

NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection

NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form

NCERT solutions for class 10 maths Chapter 3 Linear equations 5 Algebraic solution by Cross Multiplication

NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination

NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution

NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions

NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.

Chapter 4: Quadratic equations

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 3 Solution by Completing the square

NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 4 Nature of roots of a quadratic equation

Chapter 6: Triangles

NCERT solutions for class 10 maths chapter 6 Triangles 1 Similarity of Triangles and Polygons

Solutions to Exercise 2 Chapter 6 NCERT X Maths, Characteristics of Similar triangles

Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 2 Ratio values for selected angles

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 3 Complementary angle relations

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities

Chapter 8: Introduction to Trigonometry, only solutions to selected problems

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2

NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1


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