Exercise 8.1 (Revised) - Chapter 8 - Introduction To Trigonometry - Ncert Solutions class 10 - Maths

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NCERT Class 10 Maths: Chapter 8 - Introduction to Trigonometry Solutions

Ex 8.1 Question 1.

In $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}, \mathrm{BC}=7 \mathrm{~cm}$. Determine:
(i) $\sin A \cos A$
(ii) $\sin C \cos C$

Answer.

Let us draw a right angled triangle $\mathrm{ABC}$, right angled at $\mathrm{B}$.
Using Pythagoras theorem,

Let $\mathrm{AC}=24 \mathrm{k}$ and $\mathrm{BC}=7 \mathrm{k}$

Using Pythagoras theorem,
$
\begin{aligned}
& A C^2=A B^2+B C^2 \\
& =(24)^2+(7)^2=576+49=625 \\
& \Rightarrow \mathrm{AC}=25 \mathrm{~cm}
\end{aligned}
$
(i) $\sin \mathrm{A}=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{25}, \cos \mathrm{A}=\frac{\mathrm{B}}{\mathrm{H}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24}{25}$
(ii) $\sin \mathrm{C}=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24}{25}, \cos \mathrm{C}=\frac{\mathrm{B}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{25}$

Ex 8.1 Question 2.

$\text {In adjoining figure, find } \tan P-\cot R \text { : }$

Answer.

In triangle PQR, Using Pythagoras theorem,
$
\begin{aligned}
& P R^2=P Q^2+Q R^2 \\
& \Rightarrow(13)^2=(12)^2+Q R^2 \\
& \Rightarrow Q R^2=169-144=25 \\
& \Rightarrow Q R=5 \mathrm{~cm} \\
& \therefore \tan P-\cot R=\frac{P}{B}-\frac{B}{P}=\frac{Q R}{P Q}-\frac{Q R}{P Q}=\frac{5}{12}-\frac{5}{12}=0
\end{aligned}
$
Ex 8.1 Question 3.

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

Answer.

Given: A triangle $\mathrm{ABC}$ in which $\angle \mathrm{B}=90^{\circ}$

$\text { Let } \mathrm{BC}=3 k \text { and } \mathrm{AC}=4 k$

Then, Using Pythagoras theorem,
$
\begin{aligned}
& \mathrm{AB}=\sqrt{(\mathrm{AC})^2-(\mathrm{BC})^2}=\sqrt{(4 k)^2-(3 k)^2} \\
& =\sqrt{16 k^2-9 k^2}=k \sqrt{7} \\
& \therefore \cos \mathrm{A}=\frac{\mathrm{B}}{\mathrm{H}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{k \sqrt{7}}{4 k}=\frac{\sqrt{7}}{4} \\
& \tan \mathrm{A}=\frac{\mathrm{P}}{\mathrm{B}}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 k}{k \sqrt{7}}=\frac{3}{\sqrt{7}}
\end{aligned}
$
Ex 8.1 Question 4.

Given $15 \cot A=8$, find $\sin A$ and $\sec A$

Answer.

Given: A triangle $\mathrm{ABC}$ in which $\angle \mathrm{B}=90^{\circ}$
$15 \cot A=8$
$\Rightarrow \cot A=\frac{8}{15}$

Let $\mathrm{AB}=8 k$ and $\mathrm{BC}=15 k$
Then using Pythagoras theorem,
$
\begin{aligned}
& \mathrm{AC}=\sqrt{(\mathrm{AB})^2+(\mathrm{BC})^2} \\
& =\sqrt{(8 k)^2+(15 k)^2} \\
& =\sqrt{64 k^2+225 k^2}
\end{aligned}
$

$
\begin{aligned}
& =\sqrt{289 k^2}=17 k \\
& \therefore \sin \mathrm{A}=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{15 k}{17 k}=\frac{15}{17} \\
& \sec \mathrm{A}=\frac{\mathrm{H}}{\mathrm{B}}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{17 k}{8 k}=\frac{17}{8}
\end{aligned}
$
Ex 8.1 Question 5.

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

Answer.

Consider a triangle $\mathrm{ABC}$ in which $\angle \mathrm{A}=\theta$ and $\angle \mathrm{B}=90^{\circ}$

Let $\mathrm{AB}=12 k$ and $\mathrm{BC}=5 k$
Then, using Pythagoras theorem,
$
\begin{aligned}
& \mathrm{BC}=\sqrt{(\mathrm{AC})^2-(\mathrm{AB})^2} \\
& =\sqrt{(13 k)^2-(12 k)^2} \\
& =\sqrt{169 k^2-144 k^2} \\
& =\sqrt{25 k^2}=5 k \\
& \therefore \sin \theta=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 k}{13 k}=\frac{5}{13} \\
& \cos \theta=\frac{\mathrm{B}}{\mathrm{H}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12 k}{13 k}=\frac{12}{13} \\
& \tan \theta=\frac{\mathrm{P}}{\mathrm{B}}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5 k}{12 k}=\frac{5}{12} \\
& \cot \theta=\frac{\mathrm{B}}{\mathrm{P}}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12 k}{5 k}=\frac{12}{5}
\end{aligned}
$

$
\cos e c \theta=\frac{\mathrm{H}}{\mathrm{P}}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13 k}{5 k}=\frac{13}{5}
$
Ex 8.1 Question 6.

If $\angle$ And $\angle \mathbf{B}$ are acute angles such that $\cos A=\cos B$, then show that $\angle \mathbf{A}=\angle \mathbf{B}$.

Answer.

In right triangle $\mathrm{ABC}$,

$
\begin{aligned}
& \cos A=\frac{\mathrm{AC}}{\mathrm{AB}} \text { and } \cos B=\frac{\mathrm{BC}}{\mathrm{AB}} \\
& \text { But } \cos A=\cos B \text { [Given] } \\
& \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\mathrm{BC}}{\mathrm{AB}} \\
& \Rightarrow \mathrm{AC}=\mathrm{BC} \\
& \Rightarrow \angle \mathrm{A}=\angle \mathrm{B}
\end{aligned}
$
[Angles opposite to equal sides are equal]
Ex 8.1 Question 7.

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

Answer.

Consider a triangle $\mathrm{ABC}$ in which $\angle \mathrm{A}=\theta$ and $\angle \mathrm{B}=90^{\circ}$

Let $\mathrm{AB}=7 k$ and $\mathrm{BC}=8 k$
Then, using Pythagoras theorem,
$
\begin{aligned}
& \mathrm{AC}=\sqrt{(\mathrm{BC})^2+(\mathrm{AB})^2} \\
& =\sqrt{(8 k)^2+(7 k)^2} \\
& =\sqrt{64 k^2+49 k^2} \\
& =\sqrt{113 k^2}=\sqrt{113} k \\
& \therefore \sin \theta=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{8 k}{\sqrt{113} k}=\frac{8}{\sqrt{113}} \\
& \cos \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{7 k}{\sqrt{113} k}=\frac{7}{\sqrt{113}}
\end{aligned}
$
(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{1-\sin ^2 \theta}{1-\cos ^2 \theta}$
$
=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}=\frac{113-64}{113-49}=\frac{49}{64}
$

(ii) $\cot ^2 \theta=\frac{\cos ^2 \theta}{\sin ^2 \theta}=\frac{49 / 113}{64 / 113}=\frac{49}{64}$
Ex 8.1 Question 8.

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

Answer.

Consider a triangle $\mathrm{ABC}$ in which $\angle \mathrm{B}=90^{\circ}$.

And $3 \cot A=4$
$
\Rightarrow \cot A=\frac{4}{3}
$

Let $\mathrm{AB}=4 k$ and $\mathrm{BC}=3 k$.
Then, using Pythagoras theorem,
$
\begin{aligned}
& \mathrm{AC}=\sqrt{(\mathrm{BC})^2+(\mathrm{AB})^2} \\
& =\sqrt{(3 k)^2+(4 k)^2} \\
& =\sqrt{16 k^2+9 k^2} \\
& =\sqrt{25 k^2}=5 k \\
& \therefore \sin A=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3 k}{5 k}=\frac{3}{5}
\end{aligned}
$

$
\cos A=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4 k}{5 k}=\frac{4}{5}
$

And $\tan A=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 k}{4 k}=\frac{3}{4}$
Now, L.H.S. $\frac{1-\tan ^2 A}{1+\tan ^2 A}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$
$
=\frac{16-9}{16+9}=\frac{7}{25}
$
R.H.S. $\cos ^2 A-\sin ^2 A=\left(\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2$
$
=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}
$
$\because$ L.H.S. = R.H.S.
$
\therefore \frac{1-\tan ^2 A}{1+\tan ^2 A}=\cos ^2 A-\sin ^2 A
$

Ex 8.1 Question 9.

In $\triangle \mathrm{ABC}$ right angles at $\mathrm{B}$, if $\tan A=\frac{1}{\sqrt{3}}$, find value of:
(i) $\sin A \cos C+\cos A \sin C$
(ii) $\cos A \cos C-\sin A \sin C$

Answer.

Consider a triangle $\mathrm{ABC}$ in which $\angle \mathrm{B}=90^{\circ}$.
Let $\mathrm{BC}=k$ and $\mathrm{AB}=\sqrt{3} k$

Then, using Pythagoras theorem,
$
\begin{aligned}
& \mathrm{AC}=\sqrt{(\mathrm{BC})^2+(\mathrm{AB})^2} \\
& =\sqrt{(k)^2+(\sqrt{3} k)^2} \\
& =\sqrt{k^2+3 k^2}=\sqrt{4 k^2}=2 k \\
& \therefore \sin A=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2} \\
& \cos A=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}
\end{aligned}
$

For $\angle \mathrm{C}$, Base $=\mathrm{BC}$, Perpendicular $=\mathrm{AB}$ and Hypotenuse $=\mathrm{AC}$
$
\begin{aligned}
& \therefore \sin C=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2} \\
& \cos \mathrm{C}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2}
\end{aligned}
$
(i) $\sin A \cos C+\cos A \sin C=\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$
$
=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1
$

(ii) $\cos A \cos C-\sin A \sin C=\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}$
$
=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0
$
Ex 8.1 Question 10.

In $\triangle P Q R$, right angled at $Q, P R+Q R=25 \mathrm{~cm}$ and $P Q=5 \mathrm{~cm}$. Determine the values of $\sin P, \cos P$ and $\tan P$.

Answer.

In $\triangle P Q R$, right angled at $\mathrm{Q}$.

$
P R+Q R=25 \mathrm{~cm} \text { and } P Q=5 \mathrm{~cm}
$

Let $\mathrm{QR}=x \mathrm{~cm}$, then $\mathrm{PR}=(25-x) \mathrm{cm}$
Using Pythagoras theorem,
$
\begin{aligned}
& R P^2=R Q^2+Q P^2 \\
& \Rightarrow(25-x)^2=(x)^2+(5)^2 \\
& \Rightarrow 625-50 x+x^2=x^2+25 \\
& \Rightarrow-50 x=-600 \\
& \Rightarrow x=12 \\
& \therefore R Q=12 \mathrm{~cm} \text { and } \mathrm{RP}=25-12=13 \mathrm{~cm}
\end{aligned}
$

$
\begin{aligned}
& \therefore \sin P=\frac{\mathrm{RQ}}{\mathrm{RP}}=\frac{12}{13} \\
& \cos P=\frac{\mathrm{PQ}}{\mathrm{RP}}=\frac{5}{13}
\end{aligned}
$

And $\tan P=\frac{\mathrm{RQ}}{\mathrm{PQ}}=\frac{12}{5}$
Ex 8.1 Question 11.

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 $\mathrm{A}$.
(iii) $\cos A$ is the abbreviation used for the cosecant of angle $\mathrm{A}$.
(iv) $\cot A$ is the product of $\cot$ and $\mathrm{A}$.
(v) $\sin \theta=\frac{4}{3}$ for some angle $\theta$.

Answer.

(i) False because sides of a right triangle may have any length, so $\tan A$ may have any value.
(ii) True as $\sec A$ is always greater than 1 .
(iii) False as $\cos A$ is the abbreviation of cosine $\mathrm{A}$.
(iv) False as $\cot A$ is not the product of 'cot' and $\mathrm{A}$. 'cot' is separated from $\mathrm{A}$ has no meaning.
(v) False as $\sin \theta$ cannot be $>1$.