SChand CLASS 9 Chapter 19 Trigonometric Ratios Exercise 19(B)

 Exercise 19 B

Question 1

Ans: (IMAGE TO BE ADDED)

$\tan \theta=\frac{5}{12}=\frac{\text { perpendicular }}{\text { Base }}$

In the $\triangle A B C, \angle B=90^{\circ}$

$\begin{aligned}&A B=12: B C=5, \text { then } \\&A C^{2}=A B^{2}+B C^{2}\end{aligned}$

$\begin{aligned}&=(12)^{2}+(s)^{2}=144+25=169 \\&=(13)^{2}\end{aligned}$

So $A C=13$

$\begin{aligned} \text { Now } \sin \theta &=\frac{\text { perpendicular }}{\text { Hypotenuse }}=\frac{B C}{1 A C}=\frac{5}{13} \\ \text { and } \cos \theta &=\frac{\text { Bare }}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{12}{13} \end{aligned}$

Question 2

Ans: (IMAGE TO BE ADDED)

$\sin \theta=\frac{3}{5}=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{A C}{A B}$

So In $\triangle A B C$

$A C=3 . A B=5$

Now $\begin{aligned} A B^{2} &=B C^{2}+A C^{3} \Rightarrow(5)^{2}=B C^{2}+(3)^{2} \\ \Rightarrow B C^{2} &=(5)^{2} -(3)^{2}=25-9=16 \\ &=(4)^{2} \end{aligned}$

So $B C=4$

(i) $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{B C}{A B}=\frac{4}{5}$

(ii) and $\tan \theta=\frac{\text { perpendicular }}{\text { Base }}=\frac{A C}{B C}=\frac{3}{4}$.

Question 3

Ans:  (IMAGE TO BE ADDED)

(i) In right is $A B C, \angle A=90^{\circ}$

$A B=7 \mathrm{~cm}, A C=24 \mathrm{~cm}$

So 

$\begin{aligned} & B C^{2}=A B^{2}+A C^{2} \\=&(7)^{2}+(24)^{2} \\=& 49+576=625 \\=&(25)^{2} \end{aligned}$

So

 $\begin{aligned} & B C=25 \mathrm{~cm} \\ \text { Now } \tan B &=\frac{A C}{A B}=\frac{24}{7} \\ \tan C &=\frac{A B}{A C}=\frac{7}{24} \\ \sin B &=\frac{A C}{B C}=\frac{24}{25} \\ \sin C &=\frac{A B}{B C}=\frac{7}{25} \\ \cos B &=\frac{A B}{B C}=\frac{7}{25} \\ \cos C &=\frac{A C}{B C}=\frac{24}{25} \end{aligned}$

(ii)  (IMAGE TO BE ADDED)

In $\angle A B C, \angle A=90$

So $\begin{aligned} A B &=12 \mathrm{~cm}, A C=9 \mathrm{~cm} \\ B C^{2} &=A B^{2}+A C^{2} \\=&(12)^{2}+(9)^{2} \\=& 144+81=225 \\ &=(15)^{2} \end{aligned}$

$\text { So } B C=15 \mathrm{~cm}$

Now $\tan B=\frac{A C}{A B}=\frac{12}{9}$ or $\frac{4}{3}$

$\tan C=\frac{A B}{A C}=\frac{9}{12}$ or $\frac{3}{4}$

$\sin B=\frac{A C}{B C}=\frac{9}{15}$ or $\frac{3}{5}$

$\sin C=\frac{A B}{B C}=\frac{12}{15}$ or $\frac{4}{5}$

$\cos B=\frac{A B}{B C}=\frac{12}{15}$ or $\frac{4}{5}$

$\cos C=\frac{A B}{B C}=\frac{9}{15}$ or $\frac{3}{5}$

Question 4

Sol: (IMAGE TO BE ADDED)

(i) In $\triangle P Q R, \angle R=90^{\circ}$

- $P Q=10 \mathrm{~cm}, Q R=8 \mathrm{~cm}$

$\begin{aligned} P Q^{2} &=Q R^{2}+P R^{2} \\(10)^{2} &=(8)^{2}+P R^{2} \Rightarrow 100=64+P R^{2} . \\ \Rightarrow P R^{2} &=100-64=36=(6)^{2} \end{aligned}$

So $P R=6 \mathrm{~cm}$

$\tan P=\frac{Q R}{P R}=\frac{8}{6}$ or $\frac{4}{3}$

$\tan \theta=\frac{P R}{Q R}=\frac{6}{8}$ or $\frac{3}{4}$

$\sin P=\frac{Q R}{P Q}=\frac{8}{10}$ or $\frac{4}{5}$

$\sin \theta=\frac{P R}{P Q}=\frac{6}{10}$ or $\frac{3}{5}$

$\cos P=\frac{P R}{P Q}=\frac{6}{10}$ or $\frac{3}{5}-$

$\cos \theta=\frac{Q R}{P Q}=\frac{8}{10}$ or $\frac{4}{5}$

(ii)(IMAGE TO BE ADDED)

In $\triangle P Q R, \angle R=90^{\circ}$

$P Q=29 \mathrm{~mm}, P R=21 \mathrm{~mm}$

So

$\begin{aligned} & P Q^{2}=P R^{2}+Q R^{2} \\ &(29)^{2}=(21)^{2}+Q R^{2} \\ \Rightarrow & 841=441+Q R^{2} \\ \Rightarrow & Q R^{2}=841-441=400=(20)^{2} \end{aligned}$

so $Q R=20 \mathrm{~mm}$

Now

$\operatorname{tap} P=\frac{Q R}{P R}=\frac{20}{21}, \tan \theta=\frac{P R}{Q R}=\frac{21}{20}$

$\sin P=\frac{Q R}{P Q}=\frac{20}{29} \sin \theta=\frac{P R}{P Q}=\frac{21}{29}$

$\cos P=\frac{P R}{P Q}=\frac{21}{29}, \cos \theta=\frac{Q R}{P Q}=\frac{20}{29}$

(iii)(IMAGE TO BE ADDED)

In $\triangle P Q, \angle R=90^{\circ}$

$P Q=37 \mathrm{~cm}, P R=3.5 \mathrm{~cm}$

$B u+P Q^{2}=P R^{2}+Q R^{2}$

$\Rightarrow(3.7)^{2}=(3.5)^{2}+Q R^{2}$

$\Rightarrow 13.69=12.25+Q R^{2}$

$Q R^{2}=13.69-12.25$

$=1.44=(1.2)^{2}$

So $Q R=1.2 \mathrm{~cm}$

Now

$\tan p=\frac{Q R}{P R}=\frac{1.2}{3.5}=\frac{12}{35}$

$\tan \theta=\frac{P R}{Q R}=\frac{3.5}{12}=\frac{35}{12}$

$\sin P=\frac{Q R}{P Q}=\frac{1.2}{3.7}=\frac{12}{37}, \sin \theta=\frac{P R}{P Q}=\frac{3.5}{3.7}=\frac{35}{37}$

$\cos P=\frac{P R}{P Q}=\frac{3.5}{3.7}=\frac{35}{37}, \cos \theta=\frac{Q R}{P Q}=\frac{1.2}{3.7}=\frac{12}{37}$

Question 5

Ans: (image to be added)

In $\triangle A B C, \angle A=\theta$ and $\angle B=90^{\circ}$

So $\sin \theta=\frac{B C}{C A}$ and $\cos \theta=\frac{A B}{C A}$

If $A B C$ is a right angled triangle

$\text { So } C A^{2}=A B^{2}+B C^{2} \ldots \text { (i) }$

Now, $\begin{aligned} & \sin ^{2} \theta+\cos ^{2} \theta=\left(\frac{B C}{C A}\right)^{2}+\left(\frac{A B}{C A}\right)^{2} \\=& \frac{B C^{2}}{C A^{2}}+\frac{A B^{2}}{C A^{2}} \\=& \frac{B C^{2}+A B^{2}}{C A^{2}}=\frac{C A^{2}}{C A^{2}} \end{aligned}$

=1

Hence $\sin ^{2} \theta+\cos ^{2} \theta=1$

Question 6

Ans: In rhombus ABCD Diagonals AC and BD bisect each other at O at right angles 

So AO = OC and BO = OD 

and $\angle A O B=\angle B O C=\angle C O D=\angle D O A=90^{\circ}$

if $A C=6$ units and $B D=8$ units

So $A O=O C=\frac{6}{2}=3$ units and $B O=00=\frac{3}{2}$ $=4$ units

Now in right $\triangle D O C$

$\begin{aligned}&D C^{2}=D 0^{2}+0 C^{2} \\&=(4)^{2}+(3)^{2}=16+9=25=(5)^{2}\end{aligned}$

So $D C=5$ units

Now $\sin \angle O C D=\frac{O D}{D C}=\frac{4}{5}$

Question 7

Sol: (IMAGE TO BE ADDED)

In $\triangle A B C, \angle B=90^{\circ}$ and $\angle A=\theta$

$\begin{aligned}&\cos \theta=0.6=\frac{6}{10}=\frac{3}{5}=\frac{A B}{C A} \\&\operatorname{so} A B=3, C A=5\end{aligned}$

Now

$C A^{2}=A B^{2}+B C^{2}$

$\Rightarrow(5)^{2}=(3)^{2}+B C^{2} \Rightarrow 25=9+B C^{2}$

$\Rightarrow B C^{2}=25-9=16=(4)^{2}$

So BC=4

So $\sin \theta=\frac{B C}{C A}=\frac{4}{5}$ and $\tan \theta=\frac{B C}{A B}=\frac{4}{3}$

Now $5 \sin \theta-3 \tan \theta=5 \times \frac{4}{5}-3 \times \frac{4}{3}$

$=4-4=0$

Question 8

Ans : Let ABC is a right triangle in which $\angle A$ is an acute angle $\angle B=90^{\circ}$

(IMAGE TO BE ADDED)

$\tan A=\frac{B C}{A B}=\frac{3}{4}$

So $B C=3, A B=4$

But $A C^{2}=A D^{2}+B C^{2}$

$=(4)^{2}+(3)^{2}=16+9=25=\left(51^{2}\right.$

So $A C=5$

Now $\cos A=\frac{A B}{A C}=\frac{4}{5}$ or $0.8$

Question 9

Sol: (IMAGE TO BE ADDED)

In $A B C, \angle B=90^{\circ}$ and $\angle A=0$ 

$\sin \theta=\frac{6}{10}=\frac{B C}{A C}$

So $\begin{aligned} B C &=6 A C=10 \\ \text { But } A C^{2} &=A B^{2}+B C^{2} \\(10)^{2} &=A B^{2}+(6)^{2} \\ \Rightarrow 100 &=A B^{2}+36 \end{aligned}$

$\Rightarrow A B^{2}=100-36=64$

$\Rightarrow A B^{2}=64=(8)^{2}$

So $A B=8$

Now $\cos \theta=\frac{A B}{A C}=\frac{8}{10}=\frac{4}{5}$

and $\tan \theta=\frac{B C}{A B}=\frac{6}{8}=\frac{3}{4}$

So $\cos \theta+\tan \theta=\frac{4}{5}+\frac{3}{4}$

$=\frac{16+15}{20}=\frac{31}{20}=1 \frac{11}{20}$

Question 10

Ans: (a) In the figure,

(i) $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{A B}{B C}$

$\tan \theta=\frac{B C}{A B}$ is false

(ii) $\sec \theta=\frac{\text { Hypotenuse}}{B a s c}=\frac{A C}{B C}$

so $\sec \theta=\frac{B C}{A C}$ is false

(b) $A B: B C=\sin A: \cos A$

Now, $\sin A: \cos A=\frac{\sin A}{\cos A}=\tan A$

$=\frac{\text { Perpendicular }}{\text { Base }}=\frac{B C}{A B}$

$=B C: A B$

So $A B: B C=\sin A: \cos A$ is false

Question 11

Ans: (IMAGE TO BE ADDED)

(a) (i) $\sin \phi=\frac{C D}{B C}=\frac{5}{13}$

Draw DE $110 C$ So that

(ii) 

$\begin{aligned} B O &=B C=12 \text { and } \\ E D &=D C=5 \\ \text { So } A E &=A B-E B=14-5=9 \\ \text { Now } \tan \theta &=\frac{D E}{A E}=\frac{12}{9}=\frac{4}{3} \end{aligned}$

(b) In right $\triangle A E D$

$A D^{2}=A E^{2}+E D^{2}$

$=(9)^{2}+(12)^{2}=81+144$

$=225=(15)^{2}$

So $A D=15$

$\sin \theta=\frac{D E}{A D} \Rightarrow A D=\frac{D E}{\sin \theta}=\frac{12}{\sin \theta}$

and $\cos \theta=\frac{A E}{A D} \Rightarrow A D=\frac{A E}{\cos \theta}=\frac{9}{\cos \theta}$

So $A D=\frac{12}{\sin \theta}$ or $\frac{9}{\cos \theta}$

Question 12

Ans: In $\triangle A B C, \angle B=90^{\circ} \angle C=0$

$A B=2, B C=1$ units

But $A C^{2}=B C^{2}+A B^{2}$

$=(1)^{2}+(2)^{2}=1+4=5$

So $A C=\sqrt{5}$

Now $\sin \theta=\frac{A B}{A C}=\frac{2}{\sqrt{5}}$ and $\tan \theta=\frac{A B}{B C}=\frac{2}{1}$

So $\sin ^{2} \theta+\tan ^{2} \theta=\left(\frac{2}{\sqrt{5}}\right)^{2}+\left(\frac{2}{1}\right)^{2}=\frac{4}{5}+\frac{4}{1}$

$=\frac{4+20}{5}=\frac{24}{5}=4 \frac{4}{5}$

Question 13

Ans: (i) In $\triangle A B C, \angle A=0$ and $\angle B=90$

(IMAGE TO BE ADDED)

$\sin \theta=\frac{12}{13}=\frac{B C}{A C}$

So $B C=12$ and $A C=13$

$B u t A C^{2}=A B^{2}+B C^{2}$

$\Rightarrow(13)^{2}=A B^{2}+(12)^{2}$

$\Rightarrow 169=A B^{2}+144$

$\Rightarrow A B^{2}=169=144=25=(5)^{2}$

So $A B=5$

Now $\cos \theta=\frac{A B}{A C}=\frac{5}{13}$

and $\tan \theta=\frac{B C}{A D}=\frac{12}{5}$

So 

$\begin{aligned} \cos \theta+\tan \theta &=\frac{5}{13}+\frac{12}{5} \\ &=\frac{25+156}{65}=\frac{181}{65}=2 \frac{51}{65} \end{aligned}$

(ii) (IMAGE TO BE ADDED)

In $\triangle A B C, \angle A=0, \angle B=90^{\circ}$

$\cos \theta=\frac{12}{13}=\frac{A B}{A C}$

So $\begin{aligned} & A B=12, A C=13 \\ & B \text { ut } A C^{2}=A B^{2}+B C^{2} \\ \Rightarrow &(13)^{2}=(12)^{2}+B C^{2} \\ \Rightarrow & 169=144+B C^{2} \\ \Rightarrow & B C^{2}=169-144=25=(5)^{2} \\ & \text { so } B C=5 \end{aligned}$

 Now $\sin \theta=\frac{B C}{A C}=\frac{5}{13}$

and $\tan \theta=\frac{13 C}{A C}=\frac{5}{12}$

and $2 \sin \theta-4 \tan \theta=2 \times \frac{5}{13}-4 \times \frac{5}{13}$

$=\frac{10}{13}-\frac{5}{3}$

$=\frac{30-65}{39}=\frac{-35}{39}$

Question 14

Ans: $\tan \theta-\frac{5}{12}$

Now $\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}$ =$\frac{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}$

$=\frac{1-\tan \theta}{1+\tan \theta}=\frac{1-\frac{5}{12}}{1+\frac{5}{12}}$

$\frac{12-5}{\frac{12}{\frac{12+5}{12}}}=\frac{\frac{7}{12}}{\frac{17}{12}}$

$=\frac{7}{12} \times \frac{12}{17}=\frac{7}{17}$

Question 15

Ans: $5 \sin \theta=4 \Rightarrow \sin \theta=\frac{4}{5}$

Now $\frac{1+\sin 0}{1-\sin 0}=\frac{1+\frac{4}{5}}{1-\frac{4}{5}}=\frac{\frac{5+4}{5}}{\frac{5-4}{5}}$

$=\frac{\frac{9}{5}}{\frac{1}{5}}=\frac{9}{5} \times \frac{5}{1}=9$

Question 16

Ans: $b \tan \theta=a \Rightarrow \tan \theta=\frac{a}{b}$

Now $\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}$

$=\frac{1+\tan \theta}{1-\tan \theta}=\frac{1+\frac{a}{b}}{1-\frac{a}{b}}$

$=\frac{\frac{b+a}{b}}{\frac{b-a}{b}}=\frac{b+a}{b} \times \frac{b}{b-a}$

$=\frac{b+a}{b-a}$

Question 17

Ans: $5 \tan \theta=4 \Rightarrow \tan \theta=\frac{4}{5}$

Now $\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta}$=$\frac{5 \frac{\sin \theta}{\cos \theta}-3 \frac{\cos \theta}{\cos \theta}}{5 \frac{\sin \theta}{\cos \theta}+2 \frac{\cos \theta}{\cos \theta}}$

$=\frac{5 \tan \theta-3}{5 \tan \theta+2}$

$=\frac{5 \times \frac{4}{3}-3}{5 \times \frac{4}{5}+2}=\frac{4-3}{4+2}=\frac{1}{6}$

Question 18

Ans: (image to be added)

$13 \sin A=5 \Rightarrow \sin A=\frac{5}{13}$

In right $\triangle A B C, \angle B=90^{\circ}$

$\sin A=\frac{5}{13}=\frac{B C}{A C}$

So $B C=5, A C =13$

But $A C^{2}=A B^{2}+B C^{2} \Rightarrow(13)^{2}=A D^{2}+(5)^{2}$

$\Rightarrow 169=A B^{2}+25$

$\Rightarrow A B^{2}=169-25=144=(12)^{2}$

So $A B=12$

Now $\cos A=\frac{A B}{A C}=\frac{12}{13}$ and $\tan A=\frac{B C}{A B}=\frac{5}{12}$

Now $\frac{5 \sin A-2 \cos A}{\tan A}=\frac{5 \times \frac{5}{13}-2 \times \frac{12}{13}}{\frac{5}{12}}$

$\frac{\frac{25}{13}-\frac{24}{13}}{\frac{5}{12}}=\frac{\frac{1}{13}}{\frac{5}{12}}=\frac{1}{13}\times \frac{12}{5}$

$=\frac{12}{65}$

Question 19

Ans: $5 \cos A-12 \sin A=0$

$\Rightarrow 5 \cos A=12 \sin A$

$\Rightarrow \frac{\sin A}{\cos A}=\frac{5}{12}$

Now, $\frac{\sin A+\cos A}{2 \cos A-\sin A}=\frac{\frac{\sin A}{\cos A}+\frac{\cos A}{\cos A}}{\frac{2 \cos A}{\cos A}-\frac{\sin A}{\cos A}}$

$=\frac{\tan A+1}{2 \tan A}=\frac{\frac{5}{12}+1}{2-\frac{5}{12}}$

$=\frac{\frac{5+12}{12}}{\frac{24-5}{12}}=\frac{\frac{17}{12}}{\frac{19}{12}}$

=$\frac{17}{12} \times \frac{12}{19}=\frac{17}{19}$