Examples - Chapter 8 Introduction To Trigonometry class 10 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 8 - Introduction To Trigonometry - Ncert Solutions class 10 - Maths

Example $1:$

Given $\tan A=\frac{4}{3}$, find the other trigonometric ratios of the angle $\mathrm{A}$.
Solution :

Let us first draw a right $\triangle \mathrm{ABC}$ (see Fig 8.8).
Now, we know that $\tan \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{4}{3}$.
Therefore, if $\mathrm{BC}=4 k$, then $\mathrm{AB}=3 k$, where $k$ is a positive number.
Now, by using the Pythagoras Theorem, we have

So,
$
\begin{aligned}
\mathrm{AC}^2 & =\mathrm{AB}^2+\mathrm{BC}^2=(4 k)^2+(3 k)^2=25 k^2 \\
\mathrm{AC} & =5 k
\end{aligned}
$

Now, we can write all the trigonometric ratios using their definitions.
$
\begin{aligned}
\sin \mathrm{A} & =\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{4 k}{5 k}=\frac{4}{5} \\
\cos \mathrm{A} & =\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3 k}{5 k}=\frac{3}{5}
\end{aligned}
$

Therefore, $\cot A=\frac{1}{\tan A}=\frac{3}{4}, \operatorname{cosec} A=\frac{1}{\sin A}=\frac{5}{4}$ and $\sec A=\frac{1}{\cos A}=\frac{5}{3}$.

Example 2 :

If $\angle \mathrm{B}$ and $\angle \mathrm{Q}$ are acute angles such that $\sin B=\sin Q$, then prove that $\angle \mathrm{B}=\angle \mathrm{Q}$.
Solution :

Let us consider two right triangles $A B C$ and $P Q R$ where $\sin B=\sin Q$ (see Fig. 8.9).

We have
$
\sin B=\frac{A C}{A B}
$
and
$
\sin Q=\frac{P R}{P Q}
$

Then
$
\begin{aligned}
& \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\mathrm{PR}}{\mathrm{PQ}} \\
& \frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AB}}{\mathrm{PQ}}=k, \text { say }
\end{aligned}
$

Now, using Pythagoras theorem,
and
$
\begin{aligned}
& \mathrm{BC}=\sqrt{\mathrm{AB}^2-\mathrm{AC}^2} \\
& \mathrm{QR}=\sqrt{\mathrm{PQ}^2-\mathrm{PR}^2}
\end{aligned}
$

So, $\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\sqrt{\mathrm{AB}^2-\mathrm{AC}^2}}{\sqrt{\mathrm{PQ}^2-\mathrm{PR}^2}}=\frac{\sqrt{k^2 \mathrm{PQ}^2-k^2 \mathrm{PR}^2}}{\sqrt{\mathrm{PQ}^2-\mathrm{PR}^2}}=\frac{k \sqrt{\mathrm{PQ}^2-\mathrm{PR}^2}}{\sqrt{\mathrm{PQ}^2-\mathrm{PR}^2}}=k$
From (1) and (2), we have
$
\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}
$

Then, by using Theorem 6.4, $\triangle \mathrm{ACB} \sim \Delta \mathrm{PRQ}$ and therefore, $\angle \mathrm{B}=\angle \mathrm{Q}$.

Example 3 :

Consider $\triangle \mathrm{ACB}$, right-angled at $\mathrm{C}$, in which $A B=29$ units, $B C=21$ units and $\angle A B C=\theta$ (see Fig. 8.10). Determine the values of
(i) $\cos ^2 \theta+\sin ^2 \theta$,
(ii) $\cos ^2 \theta-\sin ^2 \theta$.

Solution :

In $\triangle \mathrm{ACB}$, we have
$
\mathrm{AC}=\sqrt{\mathrm{AB}^2-\mathrm{BC}^2}=\sqrt{(29)^2-(21)^2}
$

$
=\sqrt{(29-21)(29+21)}=\sqrt{(8)(50)}=\sqrt{400}=20 \text { units }
$

So, $\quad \sin \theta=\frac{A C}{A B}=\frac{20}{29}, \cos \theta=\frac{B C}{A B}=\frac{21}{29}$.
Now, (i) $\cos ^2 \theta+\sin ^2 \theta=\left(\frac{20}{29}\right)^2+\left(\frac{21}{29}\right)^2=\frac{20^2+21^2}{29^2}=\frac{400+441}{841}=1$,
and (ii) $\cos ^2 \theta-\sin ^2 \theta=\left(\frac{21}{29}\right)^2-\left(\frac{20}{29}\right)^2=\frac{(21+20)(21-20)}{29^2}=\frac{41}{841}$.

Example 4 :

In a right triangle $A B C$, right-angled at $B$, if $\tan A=1$, then verify that
$
2 \sin A \cos A=1 \text {. }
$

Solution : In $\triangle A B C, \tan A=\frac{B C}{A B}=1 \quad$ (see Fig 8.11)
i.e.,
$
\mathrm{BC}=\mathrm{AB}
$

Let $\mathrm{AB}=\mathrm{BC}=k$, where $k$ is a positive number.

Now,
$
\begin{aligned}
\mathrm{AC} & =\sqrt{\mathrm{AB}^2+\mathrm{BC}^2} \\
& =\sqrt{(k)^2+(k)^2}=k \sqrt{2}
\end{aligned}
$

Therefore,
$
\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{1}{\sqrt{2}} \text { and } \cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1}{\sqrt{2}}
$

So, $\quad 2 \sin A \cos A=2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=1$, which is the required value.

Example 5 :

In $\Delta \mathrm{OPQ}$, right-angled at $\mathrm{P}$, $\mathrm{OP}=7 \mathrm{~cm}$ and $\mathrm{OQ}-\mathrm{PQ}=1 \mathrm{~cm}$ (see Fig. 8.12). Determine the values of $\sin Q$ and $\cos Q$.

Solution :

In $\Delta \mathrm{OPQ}$, we have
$
\begin{array}{rlrl}
\mathrm{OQ}^2 & =\mathrm{OP}^2+\mathrm{PQ}^2 & \\
& & & \\
\text { i.e., } & (1+\mathrm{PQ})^2 & =\mathrm{OP}^2+\mathrm{PQ}^2 \quad \text { (Why?) } \\
\text { i.e., } & 1+\mathrm{PQ}^2+2 \mathrm{PQ} & =\mathrm{OP}^2+\mathrm{PQ}^2 & \\
\text { i.e., } & 1+2 \mathrm{PQ} & =7^2 \quad(\text { Why?) } \\
\text { i.e., } & \mathrm{PQ} & =24 \mathrm{~cm} \text { and } \mathrm{OQ}=1+\mathrm{PQ}=25 \mathrm{~cm} \\
\text { So, } & \sin \mathrm{Q} & =\frac{7}{25} \text { and } \cos \mathrm{Q}=\frac{24}{25} .
\end{array}
$

Example 6 :

In $\triangle \mathrm{ABC}$, right-angled at $\mathrm{B}$, $\mathrm{AB}=5 \mathrm{~cm}$ and $\angle \mathrm{ACB}=30^{\circ}$ (see Fig. 8.19). Determine the lengths of the sides $\mathrm{BC}$ and $\mathrm{AC}$.

Solution :

To find the length of the side $B C$, we will choose the trigonometric ratio involving $\mathrm{BC}$ and the given side $A B$. Since $B C$ is the side adjacent to angle $C$ and $A B$ is the side opposite to angle $C$, therefore

$\begin{array}{ll}
\qquad \frac{\mathrm{AB}}{\mathrm{BC}}=\tan \mathrm{C} \\
\text { i.e., } & \frac{5}{\mathrm{BC}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\
\text { which gives } \quad & \mathrm{BC}=5 \sqrt{3} \mathrm{~cm}
\end{array}$

To find the length of the side $A C$, we consider
$
\sin 30^{\circ}=\frac{A B}{A C} \quad \text { (Why?) }
$
i.e.,
$
\frac{1}{2}=\frac{5}{\mathrm{AC}}
$
i.e.,
$
\mathrm{AC}=10 \mathrm{~cm}
$

Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above,
i.e.,
$
\mathrm{AC}=\sqrt{\mathrm{AB}^2+\mathrm{BC}^2}=\sqrt{5^2+(5 \sqrt{3})^2} \mathrm{~cm}=10 \mathrm{~cm} .
$

Example 7 :

In $\triangle \mathrm{PQR}$, right-angled at $\mathrm{Q}$ (see Fig. 8.20), $\mathrm{PQ}=3 \mathrm{~cm}$ and $\mathrm{PR}=6 \mathrm{~cm}$. Determine $\angle \mathrm{QPR}$ and $\angle \mathrm{PRQ}$.
Solution :

Given $\mathrm{PQ}=3 \mathrm{~cm}$ and $\mathrm{PR}=6 \mathrm{~cm}$.
Therefore, $\quad \frac{P Q}{P R}=\sin R$

or
$
\sin R=\frac{3}{6}=\frac{1}{2}
$

So,
$
\begin{array}{ll}
\text { So, } & \angle \mathrm{PRQ}=30^{\circ} \\
\text { and therefore, } & \angle \mathrm{QPR}=60^{\circ} . \text { (Why?) }
\end{array}
$

You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.

Example 8 :

If $\sin (A-B)=\frac{1}{2}, \cos (A+B)=\frac{1}{2}, 0^{\circ}B$, find $A$ and $\mathrm{B}$.

Solution :

Since, $\sin (A-B)=\frac{1}{2}$, therefore, $A-B=30^{\circ}$ (Why?)
Also, since $\cos (A+B)=\frac{1}{2}$, therefore, $A+B=60^{\circ}$ (Why?)
Solving (1) and (2), we get : $A=45^{\circ}$ and $B=15^{\circ}$.

Example 9 :

Express the ratios $\cos \mathrm{A}, \tan \mathrm{A}$ and $\sec \mathrm{A}$ in terms of $\sin \mathrm{A}$.
Solution :

Since $\cos ^2 A+\sin ^2 A=1$, therefore,
$
\cos ^2 \mathrm{~A}=1-\sin ^2 \mathrm{~A} \text {, i.e., } \cos \mathrm{A}= \pm \sqrt{1-\sin ^2 \mathrm{~A}}
$

This gives $\quad \cos \mathrm{A}=\sqrt{1-\sin ^2 \mathrm{~A}} \quad$ (Why?)
Hence, $\quad \tan A=\frac{\sin A}{\cos A}=\frac{\sin A}{\sqrt{1-\sin ^2 A}}$ and $\sec A=\frac{1}{\cos A}=\frac{1}{\sqrt{1-\sin ^2 A}}$

Example 10 :

Prove that $\sec \mathrm{A}(1-\sin \mathrm{A})(\sec \mathrm{A}+\tan \mathrm{A})=1$.
Solution :
$
\text { LHS }=\sec A(1-\sin A)(\sec A+\tan A)=\left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)
$

$\begin{aligned}
& =\frac{(1-\sin A)(1+\sin A)}{\cos ^2 A}=\frac{1-\sin ^2 A}{\cos ^2 A} \\
& =\frac{\cos ^2 A}{\cos ^2 A}=1=\text { RHS }
\end{aligned}$

Example 11 :

Prove that $\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}$

Solution :
$
\begin{aligned}
& \text { Solution : LHS }=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A} \\
& =\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\text { RHS } \\
&
\end{aligned}
$

Example 12 :

Prove that $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$, using the identity $\sec ^2 \theta=1+\tan ^2 \theta$.

Solution :

Since we will apply the identity involving $\sec \theta$ and $\tan \theta$, let us first convert the LHS (of the identity we need to prove) in terms of $\sec \theta$ and $\tan \theta$ by dividing numerator and denominator by $\cos \theta$.
$
\begin{aligned}
\text { LHS } & =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\
& =\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)} \\
& =\frac{\left(\tan ^2 \theta-\sec ^2 \theta\right)-(\tan \theta-\sec \theta)}{\{\tan \theta-\sec \theta+1\}(\tan \theta-\sec \theta)} \\
& =\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)}
\end{aligned}
$

$
=\frac{-1}{\tan \theta-\sec \theta}=\frac{1}{\sec \theta-\tan \theta},
$
which is the RHS of the identity, we are required to prove.