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

Trigonometric Ratio values for selected angles

NCERT solutions class 10 maths ch 8 pt 2

In this series, the topic Introduction to trigonometry is covered.

This series complements Chapter 8 of NCERT Class 10 Maths and consists of 4 parts:

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

This second part deals with, Trigonometric ratios for selected angles.

It has three components:

  • First: concepts explained in an easy to understand way,
  • Second: example problems used for establishing the concepts clearly and,
  • Third: the solutions to the problems in the second exercise in the NCERT Chapter 8.

To Skip the Concept Explanations and Go straight to the Solutions, click here.

The selected angles for which you can evaluate trigonometric ratio values by simple geometry

The selected angles for which trigonometric ratio values are evaluated are only five in number and are—$0^0$, $30^0$, $45^0$, $60^0$ and $90^0$. These ratio values are mostly used and you have to remember these values.

Let us take up evaluating the easiest one—for the angle $45^0$.

Trigonometric ratio values for $45^0$

In the right-angled $\triangle ABC$ right-angled at B, the six trigonometric ratio values will be evaluated on $\angle A$.

height BC is opposite to $\angle A$ and base AB is adjacent to $\angle A$. The third side CA is the hypotenuse. In this special case, $\angle A=45^0$.

ncert-solutions-10-maths-trigo-ch-8-pt-2-1.png

As sum of three angles in a triangle is $180^0$, sum of $\angle A$ and $\angle C$ is $90^0$, $\angle B$ being $90^0$. That is,

$\angle A+\angle C=90^0$.

Again, $\angle A=45^0$, So,

$\angle C=90^0-\angle A=90^0-45^0=45^0=\angle A$.

With two equal angles, the triangles is isosceles, and its two sides adjacent to the equal angles are equal,

$AB=BC$.

By Pythagoras theorem then, the third side hypotenuse,

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

The trigonometric ratios are simply,

$\sin A=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{BC}{CA}=\frac{1}{\sqrt{2}}$.

$\cos A=\displaystyle\frac{\text{base}}{\text{hypotenuse}}=\frac{AB}{CA}=\frac{1}{\sqrt{2}}$.

$\tan A=\displaystyle\frac{\text{height}}{\text{base}}=\frac{BC}{AB}=1$.

$\text{cosec }A=\displaystyle\frac{\text{hypotenuse}}{\text{height}}=\frac{CA}{BC}=\sqrt{2}$.

$\text{sec }A=\displaystyle\frac{\text{hypotenuse}}{\text{base}}=\frac{CA}{AB}=\sqrt{2}$, and

$\text{cot }A=\displaystyle\frac{\text{base}}{\text{height}}=\frac{AB}{BC}=1$.

Next we'll evaluate the ratios for the angles $30^0$ and $60^0$.

Values of trigonometric ratios for $30^0$ and $60^0$

This time $\angle A$ in right-angled $\triangle ABC$ right-angled at B is $30^0$.

As sum of $\angle A$ and $\angle C$ is $90^0$, $\angle C=90^0-30^0=60^0$.

ncert-solutions-10-maths-trigo-ch-8-pt-2-2.png

To evaluate the trigonometric ratios from the special geometric shape of the right-angled triangle by simple geometry,

For two sides of the right-angled triangle, we must be able to express one side as a factor of the second side so that their ratio is a real number. By Pythagoras theorem the third side can also be expressed then as a factor of the second side.

In the previous case for $\angle A=45^0$, you have done precisely that—first you have found that $AB=BC$ with factor coefficient 1, and by Pythagoras theorem evaluated the factor coefficient for the third side as $\sqrt{2}$.

In this case, with $\angle A=30^0$, how would you geometrically relate lengths of two sides? That's the challenge.

The simplest way to achieve the goal is to utilize the angle value of $60^0$ of $\angle C$.

If you drop a line segment from B to the side CA making an angle of $60^0$ with side BC, you would be able to create an equilateral triangle with all three sides equal. And with still more positive results waiting to be used. 

ncert-solutions-10-maths-trigo-ch-8-pt-2-3.png

The line segment BD is drawn making an angle of $60^0$ with BC. In $\triangle BDC$, all three angles being $60^0$, it is an equilateral triangle with all three sides equal.

Let us assume this equal length be, $a$. That is,

$BC=CD=DB=a$.

Turn your attention now to $\triangle ABD$.

As $\angle ABC=90^0$ and $\angle DBC=60^0$,

$\angle ABD=30^0=\angle BAD$.

Result is—$\triangle ABD$ is an isosceles triangle with sides, $AD=DB=a=CD$.

It means hypotenuse CA is bisected at D and its length is $2a$.

Now you have got what you wanted—hypotenuse is double the length of height, that is,

$CA=2a=2BC$.

Apply Pythagoras theorem to get the length of third side AB in terms of BC as,

$AB=\sqrt{CA^2-BC^2}=\sqrt{4a^2-a^2}=\sqrt{3}a$.

With all three sides expressed in terms of common dummy factor $a$, ratio of any two sides will just be a real number and you can easily evaluate all six trigonometric ratios on $\angle A=30^0$ as well as $\angle C=60^0$.

When evaluating the ratios for $\angle C$ you just have to interchange the definition of base and height. That's all.

Values of trigonometric ratios for $\angle 30^0$

In this case, $\text{height}=BC=a$, $\text{base}=AB=\sqrt{3}a$, and $\text{hypotenuse}=CA=2a$.

$\sin A=\sin 30^0=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{a}{2a}=\frac{1}{2}$.

$\cos A=\cos 30^0=\displaystyle\frac{\text{base}}{\text{hypotenuse}}=\frac{\sqrt{3}a}{2a}=\frac{\sqrt{3}}{2}$.

$\tan A=\tan 30^0=\displaystyle\frac{\text{height}}{\text{base}}=\frac{a}{\sqrt{3}a}=\frac{1}{\sqrt{3}}$,

$\text{cosec }A=\text{cosec }30^0=\displaystyle\frac{\text{hypotenuse}}{\text{height}}=\frac{2a}{a}=2$,

$\text{sec }A=\text{sec }30^0=\displaystyle\frac{\text{hypotenuse}}{\text{base}}=\frac{2a}{\sqrt{3}a}=\frac{2}{\sqrt{3}}$, and

$\text{cot }A=\text{cot }30^0=\displaystyle\frac{\text{base}}{\text{height}}=\frac{\sqrt{3}a}{a}=\sqrt{3}$.

Values of trigonometric ratios for $\angle 60^0$

In this case the trigonometric ratios are on $\angle C$ and the $\text{base}=BC=a$, $\text{height}=AB=\sqrt{3}a$, and $\text{hypotenuse}=CA=2a$.

$\sin C=\sin 60^0=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{\sqrt{3}a}{2a}=\frac{\sqrt{3}}{2}$,

$\cos C=\cos 60^0=\displaystyle\frac{\text{base}}{\text{hypotenuse}}=\frac{a}{2a}=\frac{1}{2}$,

$\tan C=\tan 60^0=\displaystyle\frac{\text{height}}{\text{base}}=\frac{\sqrt{3}a}{a}=\sqrt{3}$,

$\text{cosec }C=\text{cosec }60^0=\displaystyle\frac{\text{hypotenuse}}{\text{height}}=\frac{2a}{\sqrt{3}a}=\frac{2}{\sqrt{3}}$,

$\text{sec }C=\text{sec }60^0=\displaystyle\frac{\text{hypotenuse}}{\text{base}}=\frac{2a}{a}=2$, and

$\text{cot }C=\text{cot }60^0=\displaystyle\frac{\text{base}}{\text{height}}=\frac{a}{\sqrt{3}a}=\frac{1}{\sqrt{3}}$.

Values of Trigonometric ratios for $\angle 0^0$

In the figure below, the trigonometric ratios are on $\angle A=\angle \theta$ that decreases as we pull down the vertex C towards the base $AB$. Another important effect of decreasing the $\angle \theta$ is diminishing height BC.

You can see from the four figures starting from left and moving towards right that in the fourth rightmost right triangle, value of $\theta$ is very small as well as the height BC is also very small.

Question is—what happens when $\angle \theta$ reaches the limiting (in this case) value of $0^0$! The hypotenuse merges with the base, that is, $AC=AB$, height becomes 0, that is, $BC=0$ and interestingly the triangle vanishes.

That's what is interesting—even though the triangle physically can't be seen, mathematically it still exists with height 0. And so, even in this extreme case, we can define the trigonometric ratios for which the height does not appear in the denominator—division by 0 is not allowed—it results in undefined state.

ncert-solutions-10-maths-trigo-ch-8-pt-2-4.png

Let us list out then the values of trigonometric ratios for $\angle \theta=0^0$ that is definable.

When $\angle \theta=\angle A=0^0$, height $BC=0$, and $CA=AB$.

$\sin A=\sin 0^0=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=\frac{0}{CA}=0$,

$\cos A=\cos 0^0=\displaystyle\frac{\text{base}}{\text{hypotenuse}}=\frac{\text{base}}{\text{base}}=1$, as $\text{base}=\text{hypotenuse}$

$\tan A=\tan 0^0=\displaystyle\frac{\text{height}}{\text{base}}=\frac{0}{AB}=0$,

$\text{cosec }A=\text{cosec }0^0=\displaystyle\frac{\text{hypotenuse}}{\text{height}}$ is undefined as 0 height is in denominator,

$\text{sec }A=\text{sec }0^0=\displaystyle\frac{\text{hypotenuse}}{\text{base}}=1$, and

$\text{cot }A=\text{cot }0^0=\displaystyle\frac{\text{base}}{\text{height}}$ is also undefined.

Values of Trigonometric ratios for $\theta=90^0$

In this case, instead of the hypotenuse merging with the base, the base length gets reduced gradually as the value of $\angle \theta$ increases from $0^0$ towards $90^0$.

As the height approaches the hypotenuse, base AB approaches zero length and $\angle \theta$ approaches $90^0$—the triangle gets thinner and taller.

When finally the base length reaches value 0, and $\angle \theta$ reaches value $90^0$ the height becomes equal to hypotenuse and the triangle physically vanishes but exists mathematically.

Just like the previous case the values of the trigonometric ratios for which base appears in the denominator, value becomes undefined.

The gradual merging of height with hypotenuse is shown in the figure below.

ncert-solutions-10-maths-trigo-ch-8-pt-2-5.png

Let us list out the value of the trigonometric ratios for $\angle A=\angle \theta=90^0$, when base $AB=0$, and $BC=CA$.

$\sin A=\sin 90^0=\displaystyle\frac{\text{height}}{\text{hypotenuse}}=1$, as height equals hypotenuse,

$\cos A=\cos 90^0=\displaystyle\frac{\text{base}}{\text{hypotenuse}}=0$, as base is of zero length,

$\tan A=\tan 90^0=\displaystyle\frac{\text{height}}{\text{base}}$ is undefined as division by zero length base is not allowed,

$\text{cosec }A=\text{cosec }90^0=\displaystyle\frac{\text{hypotenuse}}{\text{height}}=1$,

$\text{sec }A=\text{sec }90^0=\displaystyle\frac{\text{hypotenuse}}{\text{base}}$ is undefined, and

$\text{cot }A=\text{cot }90^0=\displaystyle\frac{\text{base}}{\text{height}}=0$.

The following is table of values of trigonometric ratios for the selected mostly used angles.

ncert-solutions-10-maths-trigo-ch-8-pt-2-6-ratio-value-table.png

The pink-coloured four values of $\sin \theta$ and $\cos \theta$ are the most important ones to remember.

This is easy if you just remember that value of $\sin \theta$ increases with increase in value of $\theta$ and $\displaystyle\frac{1}{2} \lt \displaystyle\frac{\sqrt{3}}{2}$.

These two pointers fix the value of $\sin 30^0=\displaystyle\frac{1}{2}$ and $\sin 60^0=\displaystyle\frac{\sqrt{3}}{2}$.

Value of $\cos \theta$ decreases from 1 as $\theta$ value increases, and take values complementary to $\sin \theta$.

When $\sin 0^0=0$, $\cos 0^0=1$, the maximum.

When $\sin 30^0=\displaystyle\frac{1}{2}$, $\cos 30^0=\displaystyle\frac{\sqrt{3}}{2}$.

And when $\sin 60^0=\displaystyle\frac{\sqrt{3}}{2}$, $\cos 60^0=\displaystyle\frac{1}{2}$.

A more comprehensive set of simple rules for remembering the trigonometric ratios are given below.

You may skip this section if you decide so, but do have a look at the figure showing variations of $\sin \theta$ and $\cos \theta$ as $\angle \theta$ changes.

Remember, it is easier to recall a picture showing values than recalling the values directly.


How to remember the values of trigonometric ratios for selected angles easily

If you remember the picture below, it would never be any problem to recall the values of the trigonometric ratios for selected angles.

ncert-solutions-10-maths-trigo-ch-8-pt-2-7-sine-cosine-curves.png

The green and red-coloured curves show how values of $\sin \theta$ and $\cos \theta$ change as $\angle \theta$ changes.

Concentrate on the region bounded by blue-coloured lines. This is the region in which $\theta$ varies from $0^0$ to $90^0$. That is your interest area now and the values of $\sin \theta$, $\cos \theta$ and other trigonometric ratios tabulated above, all belong to this region.

Observe the variation of $\sin \theta$ with $\angle \theta$.

First rule: $\sin \theta$ has value 0 at $\theta=0^0$.

Second rule: Value of $\sin \theta$ increases as value of $\theta$ increases.

Third rule: Two more important values of $\sin \theta$ are $\displaystyle\frac{1}{2}$ and $\displaystyle\frac{\sqrt{3}}{2}$ that are values of $\sin \theta$ for two angles of $30^0$ and $60^0$.

Fourth rule: As $\displaystyle\frac{1}{2} \lt \displaystyle\frac{\sqrt{3}}{2}$ and as value of $\sin \theta$ increases with increase of value of $\angle \theta$,

$\sin 30^0=\displaystyle\frac{1}{2}$ and $\sin 60^0=\displaystyle\frac{\sqrt{3}}{2}$.

Once you remember this much, recalling all other values of the trigonometric ratios would become easy.

This is the most critical rule.

Fifth rule: Value of $\sin 45^0=\displaystyle\frac{1}{\sqrt{2}}$ which is midway between its values at $30^0$ and $60^0$.

At this angle of $45^0$, $\sin \theta=\cos \theta=\displaystyle\frac{1}{\sqrt{2}}$, and so, $\tan \theta=1$.

Sixth rule: $\sin \theta$ increases to its maximum value of 1 at $\theta=90^0$.

Observe the variation of $\cos \theta$ with $\angle \theta$.

Seventh rule: Just as $\sin \theta$ starts with minimum value of 0 at $\theta=0^0$, $\cos \theta$ starts with maximum value of 1 at $\theta=0^0$ and goes on decreasing with increase of $\theta$.

Eighth rule: $\cos \theta$ also takes, the pair of values that $\sin \theta$ takes at $\theta=30^0$ and $\theta=60^0$, but at $\theta=60^0$ and $\theta=30^0$.

This is complementary nature of these two.

This complementary nature of the two trigonometric ratios hold good for $\theta=0^0$ and $\theta=90^0$ also.

Ninth rule: Value of $\tan \theta$ is just he ratio of $\sin \theta$ and $\cos \theta$.

Tenth rule: The values of the rest three trigonometric ratios are just the inverted forms of the first three.


Now we'll solve a few example problems to know how this new knowledge can be used for solving problems.

Example problems on values of trigonometric ratios for selected angles

The first is a simple example where value of one side and value of one angle of a right-angled triangle are given.

Problem example 1.

In right-angled triangle $\triangle ABC$ with right angle at B, if $\angle A=60^0$ and $AB=6$ cm, find the lengths of BC and CA.

ncert-solutions-10-maths-trigo-ch-8-pt-2-8.png

Solution example 1:

With $\angle A=60^0$, and base $AB=6$ cm, the trigonometric ratio value of $\tan A=\tan 60^0=\sqrt{3}$ would have to be used as, this ratio connects the given base with the desired height.

$\tan 60^0=\sqrt{3}=\displaystyle\frac{BC}{AB}=\frac{BC}{6}$.

So,

$BC=6\sqrt{3}$ cm.

On the other hand, to find the length of the hypotenuse, you have to use value of $\cos 60^0$ as it is the ratio of given base and desired hypotenuse.

$\cos 60^0=\displaystyle\frac{1}{2}=\frac{AB}{CA}=\frac{6}{CA}$.

So,

$CA=2\times{6}=12$ cm.

Now we'll take up a problem where lengths of two sides are given, and value of angles are to be found out.

Problem example 2.

In right-angled $\triangle ABC$, if $BC=5$ cm and $CA=10$ cm, find $\angle A$, and $\angle C$.

ncert-solutions-10-maths-trigo-ch-8-pt-2-9.png

Solution example 2.

The trigonometric ratio on $\angle A$ that connects BC and CA is $\sin A$ with height BC and hypotenuse CA,

$\sin A=\displaystyle\frac{BC}{CA}=\frac{5}{10}=\frac{1}{2}=\sin 30^0$.

$\angle A=30^0$, and $\angle C=60^0$, as their sum is $90^0$.

In the last example we'll show you how a special problem involving compound angles can be solved simply by using trigonometric ratio values of selected angles.

Problem example 3.

In a right-angled triangle $\triangle ABC$ with right angle at B, if $\sin (A+C)=1$ and $\sin (A-C)=0$, find $\sin A$, $\tan C$.

Solution example 3.

Given,

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

So,

$(A+C)=90^0$.

Also,

$\sin (A-C)=0$,

So,

$(A-C)=0^0$.

Adding the two $\angle C$ cancels out,

$2A=90^0$,

Or, $\angle A=45^0$, and

$\angle C=45^0$.

$\sin A=\sin 45^0=\displaystyle\frac{1}{\sqrt{2}}$, and,

$\tan C=\tan 45^0=1$.

Though both the given trigonometric ratios were on compound angles $\angle (A+C)$ and $\angle (A-C)$, the problem was set up in such a way that, using simple trigonometric ratio values for selected angles, it could be solved easily.

Now it is time to take up solving the second exercise problems of chapter 8.

Problems to solve: Exercise 2 Chapter 8 NCERT solutions for Class 10 maths

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.i.

Evaluate $\sin 60^0 \cos 30^0+\sin 30^0\cos 60^0$.

Solution to Problem 1.i.

$\sin 60^0=\displaystyle\frac{\sqrt{3}}{2}$, $\cos 30^0=\displaystyle\frac{\sqrt{3}}{2}$, $\sin 30^0=\displaystyle\frac{1}{2}$, $\cos 60^0=\displaystyle\frac{1}{2}$.

The given expression values is,

$\sin 60^0 \cos 30^0+\sin 30^0\cos 60^0$

$=\displaystyle\frac{\sqrt{3}}{2}\times{\displaystyle\frac{\sqrt{3}}{2}}+\displaystyle\frac{1}{2}\times{\displaystyle\frac{1}{2}}=\displaystyle\frac{3}{4}+\displaystyle\frac{1}{4}=1$.

Answer: 1.

Problem 1.ii.

Evaluate $2\tan^2 45^0+\cos^2 30^0-\sin^2 60^0$.

Solution to Problem 1.ii.

$\tan^2 45^0=1$, $\cos^2 30^0=\displaystyle\frac{3}{4}$, and $\sin^2 60^0=\displaystyle\frac{3}{4}$.

So the value of the given expression is,

$2\tan^2 45^0+\cos^2 30^0-\sin^2 60^0$

$=2+\displaystyle\frac{3}{4}-\displaystyle\frac{3}{4}=2$.

Answer: 2.

Problem 1.iii.

Evaluate $\displaystyle\frac{\cos 45^0}{\text{sec }30^0+\text{cosec } 30^0}$.

Solution to Problem 1.iii.

$\cos 45^0=\displaystyle\frac{1}{\sqrt{2}}$, $\text{sec }30^0=\displaystyle\frac{2}{\sqrt{3}}$, and $\text{cosec }30^0=2$.

So the value of the given expression is,

$\displaystyle\frac{\cos 45^0}{\text{sec }30^0+\text{cosec } 30^0}$

$=\displaystyle\frac{1}{\sqrt{2}}\left[\displaystyle\frac{1}{\displaystyle\frac{2}{\sqrt{3}}+2}\right]$

$=\displaystyle\frac{1}{2\sqrt{2}}\left[\displaystyle\frac{\sqrt{3}}{\sqrt{3}+1}\right]$

$=\displaystyle\frac{\sqrt{3}(\sqrt{3}-1)}{4\sqrt{2}}$, rationalizing the surd fraction by multiplying and dividing by $(\sqrt{3}-1)$,

$=\displaystyle\frac{3\sqrt{2}-\sqrt{6}}{8}$, eliminating surd from denominator by multiplying and dividing by $\sqrt{2}$.

Answer: $\displaystyle\frac{3\sqrt{2}-\sqrt{6}}{8}$.

Problem 1.iv.

Evaluate $\displaystyle\frac{\sin 30^0+\tan 45^0-\text{cosec } 60^0}{\text{sec }30^0+\cos 60^0+\text{cot }45^0}$.

Solution to Problem 1.iv.

$\sin 30^0=\displaystyle\frac{1}{2}$, $\tan 45^0=1$, $\text{cosec } 60^0=\displaystyle\frac{2}{\sqrt{3}}$, and

$\text{sec }30^0=\displaystyle\frac{2}{\sqrt{3}}$, $\cos 60^0=\displaystyle\frac{1}{2}$, and $\text{cot }45^0=1$.

So the value of the given expression is,

$\displaystyle\frac{\sin 30^0+\tan 45^0-\text{cosec } 60^0}{\text{sec }30^0+\cos 60^0+\text{cot }45^0}$

$=\displaystyle\frac{\displaystyle\frac{1}{2}+1-\displaystyle\frac{2}{\sqrt{3}}}{\displaystyle\frac{2}{\sqrt{3}}+\displaystyle\frac{1}{2}+1}$

$=\displaystyle\frac{\displaystyle\frac{3}{2}-\displaystyle\frac{2}{\sqrt{3}}}{\displaystyle\frac{3}{2}+\displaystyle\frac{2}{\sqrt{3}}}$

$=\displaystyle\frac{3\sqrt{3}-4}{3\sqrt{3}+4}$

$=\displaystyle\frac{(3\sqrt{3}-4)^2}{11}$, rationalizing the surd fraction by multiplying and dividing by $(3\sqrt{3}-4)$,

$=\displaystyle\frac{43-24\sqrt{3}}{11}$.

Answer: $\displaystyle\frac{43-24\sqrt{3}}{11}$.

Problem 1.v.

Evaluate $\displaystyle\frac{5\cos^2 60^0+4\text{sec}^2 30^0-\tan^2 45^0}{\sin^2 30^0+\cos^2 30^0}$.

Solution to Problem 1.v.

$\cos 60^0=\displaystyle\frac{1}{2}$, $\text{sec }30^0=\displaystyle\frac{2}{\sqrt{3}}$, $\tan 45^0=1$,

$\sin 30^0=\displaystyle\frac{1}{2}$, and $\cos 30^0=\displaystyle\frac{\sqrt{3}}{2}$.

So the value of the given expression is,

$\displaystyle\frac{5\cos^2 60^0+4\text{sec}^2 30^0-\tan^2 45^0}{\sin^2 30^0+\cos^2 30^0}$

$\displaystyle\frac{\displaystyle\frac{5}{4}+\displaystyle\frac{16}{3}-1}{\displaystyle\frac{1}{4}+\displaystyle\frac{3}{4}}$

$=\displaystyle\frac{67}{12}$.

Answer: $\displaystyle\frac{67}{12}$.

Problem 2.i.

Choose the correct option and justify your choice.

$\displaystyle\frac{2\tan 30^0}{1 + \tan^2 30^0}$ is equal to,

$\text{a. }\sin 60^0 \qquad \text{b. }\cos 60^0 \qquad \text{c. }\tan 60^0 \qquad \text{d. }\sin 30^0$.

Solution to Problem 2.i.

$\tan 30^0=\displaystyle\frac{1}{\sqrt{3}}$.

So the given expression value is,

$\displaystyle\frac{2\tan 30^0}{1 + \tan^2 30^0}$

$=\displaystyle\frac{\displaystyle\frac{2}{\sqrt{3}}}{1 + \displaystyle\frac{1}{3}}$

$=\displaystyle\frac{\displaystyle\frac{2}{\sqrt{3}}}{\displaystyle\frac{4}{3}}$

$=\displaystyle\frac{\sqrt{3}}{2}=\sin 60^0$.

Answer: Option a. $\sin 60^0$.

Problem 2.ii.

Choose the correct option and justify your choice.

$\displaystyle\frac{1-\tan^2 45^0}{1 + \tan^2 45^0}$ is equal to,

$\text{a. }\tan 90^0 \qquad \text{b. }1 \qquad  \text{c. }\sin 45^0 \qquad \text{d. }0$.

Solution to Problem 2.ii.

$\tan 45^0=1$.

So the numerator of the given expression,

$1-\tan^2 45^0=1-1=0$.

And the value of the expression is 0.

Answer: Option d. 0

Problem 2.iii.

Choose the correct option and justify your choice.

$\sin 2A=2\sin A$ is true when $\angle A$ is equal to,

$\text{a. }0^0 \qquad \text{b. }30^0 \qquad \text{c. }45^0 \qquad \text{d. }60^0$.

Solution to Problem 2.iii.

Only when $\angle A=0^0$,

$\sin 2A=\sin 0^0=0=2\sin A=0$,

and the given equation is satisfied.

Answer: Option a. $0^0$.

Solution by mathematical reasoning:

If the given equation is to be satisfied then—if $\angle A$ is doubled, the value of $\sin A$ also has to be doubled.

But except at the starting point of $\angle A=0^0$, as the angle increases, the rate of increase of $\sin A$ does not increase proportionally—in fact the rate of increase of $\sin A$ gradually reduces as value of $\angle A$ increases linearly.

For example at $30^0$, $\sin A=\sin 30^0=\displaystyle\frac{1}{2}$.

When $\angle A$ is increased $\displaystyle\frac{3}{2}$ times to $\angle A=45^0$, the value of,

$\sin 45^0=\displaystyle\frac{1}{\sqrt{2}} \lt 1.5\times{\displaystyle\frac{1}{2}}=\displaystyle\frac{3}{4}$.

Similarly when $\angle A=30^0$ is doubled to $60^0$, the value of $\sin A$ doesn't become 1 from $\displaystyle\frac{1}{2}$ as it were at $\angle A=30^0$, it becomes $\displaystyle\frac{\sqrt{3}}{2} \lt 1$.

Only at $\angle A=0^0$, the given equation is satisfied.

Problem 2.iv.

Choose the correct option and justify your choice.

$\displaystyle\frac{2\tan 30^0}{1 - \tan^2 30^0}$ is equal to,

$\text{a. }\cos 60^0 \qquad \text{b.  }\sin 60^0 \qquad \text{c.  }\tan 60^0 \qquad \text{d.  }\sin 30^0$.

Solution to Problem 2.iv.

$\tan 30^0=\displaystyle\frac{1}{\sqrt{3}}$.

So the value of the given equation becomes,

$\displaystyle\frac{2\tan 30^0}{1 - \tan^2 30^0}$

$=\displaystyle\frac{\displaystyle\frac{2}{\sqrt{3}}}{1 - \displaystyle\frac{1}{3}}$

$=\displaystyle\frac{\displaystyle\frac{2}{\sqrt{3}}}{\displaystyle\frac{2}{3}}$

$=\sqrt{3}=\tan 60^0$.

Answer: Option c. $\tan 60^0$.

Problem 3.

If $\tan (A+B)=\sqrt{3}$, and $\tan(A-B)=\displaystyle\frac{1}{\sqrt{3}}$; $0^0 \lt A+B \leq 90^0$; $A \gt B$, find $A$ and $B$.

Solution to Problem 3.

With the constraints given on values of angles A and B, both A and B are acute angles so that the trigonometric ratio values in a right-angled triangle would be applicable.

Given,

$\tan (A+B)=\sqrt{3}=\tan 60^0$,

So, $A+B=60^0$.

Also given,

$\tan(A-B)=\displaystyle\frac{1}{\sqrt{3}}=\tan 30^0$.

So $A-B=30^0$.

Adding the two equations,

$2A=90^0$,

So, $A=45^0$, and $B=60^0-45^0=15^0$.

Answer: $\angle A=45^0$, $\angle B=15^0$.

Problem 4.i.

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

$\sin (A+B)=\sin A + \sin B$.

False, as sum of two values of $\sin \theta$ for two values of $\theta$ may exceed 1 on the RHS. But as the LHS is another Sine, it cannot exceed 1.

For example, $\angle A=30^0$ and $\angle B=60^0$ would make the RHS,

$\sin 30^0+\sin 60^0=\displaystyle\frac{1}{2}+\displaystyle\frac{\sqrt{3}}{2}=\displaystyle\frac{\sqrt{3}+1}{2} \gt 1$, 

as numerator $\sqrt{3}+1=1.7+1=2.7 \gt 2$, the denominator.

Answer: False.

Problem 4.ii

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

The value of $\sin \theta$ increases as $\theta$ increases.

True.

$\sin 0^0=0$,

$\sin 30^0=\displaystyle\frac{1}{2}$,

$\sin 45^0=\displaystyle\frac{1}{\sqrt{2}}$,

$\sin 60^0=\displaystyle\frac{\sqrt{3}}{2}$, and

$\sin 90^0=1$.

All five angles from the first to the fifth are gradually increasing as are the values of Sine.

Answer: True.

Problem 4.iii.

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

The value of $\cos \theta$ increases as $\theta$ increases.

False.

When $\theta=30^0$, $\cos \theta=\cos 30^0=\displaystyle\frac{\sqrt{3}}{2}$.

As $\angle \theta$ increases to $60^0$, the value of $\cos \theta$ becomes,

$\cos \theta=\cos 60^0=\displaystyle\frac{1}{2} \lt \displaystyle\frac{\sqrt{3}}{2}$, the value of $\cos 30^0$.

The fact is just the opposite—as $\theta$ increases, value of $\cos \theta$ gradually decreases.

Answer: False.

Problem 4.iv.

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

$\sin \theta=\cos \theta$ for all values of $\theta$.

False.

$\sin 0^0=0$, while $\cos 0^0=1$,

$\sin 30^0=\displaystyle\frac{1}{2}$ while $\cos 30^0=\displaystyle\frac{\sqrt{3}}{2}$.

Only when $\text{base}=\text{height}$, that is, the right-angled triangle is an isosceles triangle, both the angles other than the right angle are $45^0$, and

$\sin 45^0=\displaystyle\frac{1}{\sqrt{2}}=\cos 45^0$.

Answer: False.

Problem 4.v.

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

$\text{cot }A$ is not defined for $\angle A=0^0$.

Solution to Problem 4.v

For $\angle A=0^0$,

$\text{cot }A=\displaystyle\frac{\cos 0^0}{\sin 0^0}$.

As the denominator $\sin 0^0$ is 0, $\text{cot }A$ at $\angle A=0^0$ is undefined.

Answer: True.


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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