SChand CLASS 9 Chapter 19 Trigonometric Ratios TEST

  TEST

Question 1

Ans: 

$\begin{aligned} A &=30^{\circ} \text { and } B=\cos A+\cos B \\ \text { L.HS } &=\sin A+\sin B=\sin 30^{\circ}+\sin 60^{\circ} \\ &=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2} \end{aligned}$

R.HS $=\cos 30^{\circ}+\cos 60^{\circ}$

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

So L.H.S. $=$ R.H.S

L.HS $\tan A+\tan B=\tan 30^{\circ}+\tan 60^{\circ}$

$=\frac{1}{\sqrt{3}}+\sqrt{3}=\frac{1+3}{\sqrt{3}}=\frac{4}{\sqrt{3}}$

L.H.S $\tan A+\tan B=\tan 30^{\circ}+\tan 60^{\circ}$

$=\frac{1}{\sqrt{3}}+\sqrt{3}=\frac{1+3}{\sqrt{3}}=\frac{4}{\sqrt{3}}$

So L.H.S = R.H.S 

So both I and II are correct(c) 

Question 2

Ans: $\cot 15^{\circ} \cot 20^{\circ} \cot 70^{\circ} \cot 75^{\circ}$

$\begin{aligned}&=\cot \left(90^{\circ}-75^{\circ}\right) \cot \left(90^{\circ}-70^{\circ}\right) \cot 70^{\circ}\operatorname{coc} 75^{\circ} \\&=\tan 75^{\circ} \tan 70^{\circ} \cot 70^{\circ} \cot 75^{\circ} \\&=\tan 75^{\circ} \times \cot 75^{\circ} \tan 70^{\circ} \times \cot 70^{\circ} \\&=1 \times 1=1\end{aligned}$

Option (c) is correct 

Question 3

Ans: (i) Sin 3A = Cos (A- 2) , 3A and (A- 2) are acute angles 

Sin 3A = Cos (A- 2) =Sin $\left\{90^{\circ}-A+2^{\circ}\right\}$

Comparing both sides, 

$\begin{gathered}3 A=90^{\circ}-A+2^{\circ} \Rightarrow 3 A+A=92^{\circ} \\\Rightarrow 4 A=92^{\circ} \Rightarrow A=\frac{92^{\circ}}{4}=23^{\circ} \\\text { So } A=23^{\circ}\end{gathered}$

Option b is correct 

(ii) $\sin (x-2 y)=\cos (4 y-x)$

$\sin (x-2 y)=\sin \left(90^{\circ}-4 y+x\right)$

comparing

$\begin{aligned}&x-2 y=90^{\circ}-4 y+x \\&4 y-2 y=90^{\circ} \Rightarrow 2 y=90^{\circ} \Rightarrow y=45^{\circ}\end{aligned}$

Now $\cot 2 y=\cot \left(2 \times 45^{\circ}\right)=\cot 90^{\circ}$

Not defined (d)

Question 5

Ans : In $\triangle A B C, \angle A B C=90^{\circ}, \angle A C B=30^{\circ}, A D=5 \mathrm{~cm}$

(IMAGE TO BE ADDED)

(i)$\sin 30^{\circ}=\frac{A D}{A C}$

$=\frac{1}{2}=\frac{5}{A C}$

$\Rightarrow A C=2 \times 5=10 \mathrm{~cm}$

(IMAGE TO BE ADDED)

(ii) $\tan A+\tan B$

$=\frac{u}{v}+\frac{v}{4}=\frac{y^{2}+v^{2}}{w}$

But $w^{2}=u^{2}+v^{2}$

So $\frac{w^{2}}{u v}$ (d)

Question 6

(IMAGE TO BE ADDED)

(i) $\cot A=\frac{8}{15}$

So In right $\triangle A B C$

$\cot A=\frac{A C}{B C}=\frac{8}{15}$

So $A C=8, B C=15$

Then $A B^{2}=A C^{2}+B C^{2}=8^{2}+15^{2}$

Then $A B^{2}=A C^{2}+B C^{2}=8^{2}+15^{2}$

$=64+225=289=(17)^{2}$

So $A B=17$

$\cos A=\frac{A C}{A D}=\frac{8}{17}$

Now $\sqrt{\frac{1-\cos A}{1+\cos A}}$

$=\sqrt{\frac{1-\frac{8}{17}}{1+\frac{8}{17}}}=\sqrt{\frac{\frac{17-8}{17}}{\frac{17+8}{17}}}$=$\sqrt{\frac{\frac{9}{17}}{\frac{25}{17}}}$

$=\sqrt{\frac{9}{17} \times \frac{17}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$ (c)

(ii)(IMAGE TO BE ADDED)

$\sin A=\frac{\sqrt{x-1}}{2 x}=\frac{B C}{A B}=\frac{\sqrt{x-1}}{\sqrt{2 x}}$

In the $\triangle A B C$

$B C=\sqrt{x-1} \quad A B=\sqrt{2 x}$

So $\begin{aligned} & A C^{2}=A B^{2}-B C^{2}=(\sqrt{2} x)^{2}-(\sqrt{x}-1)^{2} \\=& 2 x-(x-1)=2 x-x+1 \\=& x+1 \end{aligned}$

So $A C=\sqrt{x+1}$

Now $\tan A=\frac{B C}{A B}=\frac{\sqrt{x-1}}{\sqrt{x+1}}$

$=\sqrt{\frac{x-1}{x+1}}$

Question 7

Ans: 

$\begin{aligned} \sin ^{3} 60^{\circ} \cot 30^{\circ}-2 \sec ^{2} 45^{\circ}+3 \cos 60^{\circ} \tan 45^{\circ} &-\tan ^{2} 60^{\circ} \\=&\left(\frac{\sqrt{3}}{2}\right)^{2} \times \sqrt{3}-2(\sqrt{2})^{2}+3 \times \frac{1}{2} \times 1-(\sqrt{3})^{2} \\=& \frac{3 \sqrt{3}}{8} \times \sqrt{3}-2 \times 2+\frac{3}{2}-3 \\=& \frac{9}{8}-4+\frac{3}{2}-3=\frac{9}{8}+\frac{3}{2}-7 \\=& \frac{9+12}{8}-7=\frac{21}{8}-7=\frac{21-56}{8} \\=& \frac{-35}{8} \quad \text { (b) } \end{aligned}$

Question 8

Ans: $x+y=90^{\circ}, \frac{\sin x}{\sin y}=\frac{\sqrt{3}}{1}$

$y=91-x^{\circ}$

$\frac{\sin x}{\sin \left(90^{\circ}-x\right.}=\frac{\sqrt{3}}{1}$

$\Rightarrow \frac{\sin x}{\cos x}=\frac{\sqrt{3}}{1}+\operatorname{an} x=\sqrt{3}$

$\Rightarrow \tan x=\tan 60^{\circ}$

so $x=60^{\circ}$

and $y=90^{\circ}-60^{\circ}=30^{\circ}$

so $x: y=60^{\circ}: 30^{\circ}$

$\Rightarrow x=y=2: 1$

(ii)

 $\begin{aligned} \sin x=\cos y &=x \text { and } y \text { are acute angle } \\ \sin x &=\cos y=\sin \left(90^{\circ}-y\right) \\ \text { so } x &=90^{\circ}-y \quad x+y=90^{\circ}=\frac{\pi}{2} \end{aligned}$

Question 9

Ans: $\frac{5 \sin 75^{\circ} \sin 77^{\circ}+2 \cos 13^{\circ} \cos 15^{\circ}}{\cos 15^{\circ} \sin 77^{\circ}}-\frac{7 \sin 81^{\circ}}{\sin 9^{\circ}}$

$=\frac{5 \sin 75^{\circ} \sin 77^{\circ}+2 \cos \left(90^{\circ}-77^{\circ}\right) \cdot \cos \left(90^{\circ}-75^{\circ}\right)}{\cos 15^{\circ} \sin 77^{\circ}}$ -$\frac{7 \sin 81^{\circ}}{\cos \left(90^{\circ}-81^{\circ}\right)}$

$=\frac{5 \sin 75^{\circ} \sin 77^{\circ}+2 \sin 77^{\circ} \sin 75^{\circ}}{\sin 75^{\circ} \sin 77^{\circ}}-\frac{7 \sin 81^{\circ}}{\cos 81^{\circ}}$

$=\frac{7 \sin 75^{\circ} \sin 77}{\sin 75^{\circ} \sin 77^{\circ}}-7=7-7=0$ (b)

(ii)$\cos ^{2} \frac{\pi}{8}+4 \cos ^{2} \frac{\pi}{7}-\sec \frac{\pi}{3}+5 \tan ^{2}-\frac{\pi}{3}+\sin ^{2} \frac{\pi}{8}$

$\Rightarrow \cos ^{2} \frac{45^{\circ}}{2}+4 \cos ^{2} 45^{\circ}-\sec 60^{\circ}+5 \tan 260^{\circ}+\sin ^{2} \frac{45}{8}$

$=\left(\sin ^{2} \frac{45^{\circ}}{2}+\cos ^{2} \frac{45^{\circ}}{2}\right)+4\left(\frac{1}{\sqrt{2}}\right)^{2-2}+5(\sqrt{3})^{2}$

$=1+4 \times \frac{1}{2}-2+15$

$=16+2-2=18-2=16 \quad$ (C)

Question 10

(IMAGE TO BE ADDED)

Sol: In the given figure

A rockets is launcher from p 20 km to then 80 km to B making an angle of 30 with PA. 

We have to find pc from A draw AD||PC, Then 

AD = PC

Then $\angle B A D=90^{\circ}-30^{\circ}=60^{\circ}$

$\cos \theta=\frac{A D}{A B} \Rightarrow \cos 60^{\circ}=\frac{A D}{A D}$

$\Rightarrow \frac{1}{2}=\frac{P C}{\operatorname{lo}}=P C=\frac{80}{2}=40 \mathrm{~km}$

and $\begin{aligned} \sin 60^{\circ} &=\frac{B D}{A B} \Rightarrow \frac{\sqrt{3}}{2}=\frac{B D}{80} \\ \Rightarrow B D &=\frac{80 \sqrt{3}}{2}=40 \sqrt{3} \mathrm{~km} \\ \text { So } B C &=B D+D C=B D+A P \\ &=(40 \sqrt{3}+20) \mathrm{km} \end{aligned}$