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

 

 Exercise 19 A 

Question 1

Ans: $\begin{aligned}&\theta=\frac{\text { perpendicular }}{\text { Bare }} \\&\cos \theta=\frac{\text { base }}{\text { Hypotenuse }} \\&\sin \theta=\frac{\text { perpendicular }}{\text { Hypotenuse }} \\&\cot \theta=\frac{1}{\tan \theta}=\frac{\text { Base }}{\text { [Perpendicular }} \\&\sec \theta=\frac{1}{\cos \theta}=\frac{\text { Hypotenuse }}{\text { Bare }} \\&\text { ard } \operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{\text { Hypotenuse }}{\text { Perpendicular }}\end{aligned}$

Now, we shall complete the give

(IMAGE TO BE ADDED)

Question 2

Ans: $A B=5, B C=12$ and hyp. . $A C=13$ and $\angle A C B=\theta$

$\begin{aligned}&\text { So } \sin \theta=\frac{A B}{A C}=\frac{5}{13} \\&\cos \theta=\frac{B C}{A C}=\frac{12}{13} \\&\tan \theta=\frac{A B}{B C}=\frac{5}{12} \\&\sin ^{2} \theta=\left|\frac{5}{13}\right|^{2}=\frac{25}{159} \\&\cos ^{2} \theta=\left|\frac{12}{15}\right|^{2}=\frac{144}{169} \\&\tan ^{2} \theta=\left(\frac{5}{12}\right)^{2}=\frac{25}{144}\end{aligned}$

Question 3

Ans: $\angle A B C=\theta$ and $\angle C=90^{\circ}$

$A B=5 \cdot B C=3$ and $A C=4$

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

$\tan ^{2} \theta=\left(\frac{4}{3}\right)^{2}=\frac{16}{9}$

$\tan ^{3} \theta=\left(\frac{4}{3}\right)^{2}=\frac{64}{27}$

if $\sin \theta=\frac{A C}{A B}=\frac{4}{5}$

So $\sin ^{3} \theta=\left(\frac{4}{5}\right)^{2}=\frac{64}{125}$

and $\cos \theta=\frac{D C}{A B}=\frac{3}{5}$

So $\cos ^{3} \theta=\left(\frac{3}{5}\right)^{3}=\frac{27}{125}$

Question 4

Ans: In the Figure

$\angle$ OMP, $\angle O R O$ and $\angle O O M$

(a) $\sin R QM=\frac{R M}{Q M}$ (b) $\sin Q M P=\frac{P O}{P M}$

(c) $\sin O Q K=\frac{O R}{O Q}$ (d) $\Cos Q m P=\frac{Q M}{P m}$

(e) $\tan R Q m=\frac{R m}{Q R} \quad$ (f) $\Cot$ MOP $=\frac{o m}{m p}$

(g) $\mathrm{Sec} \quad R O Q=\frac{O Q}{O R}$.