Exercise 19 C
Question 1
Ans: (i)
$\begin{aligned} \text { L.H.S } &=\sin 60^{\circ}=\frac{\sqrt{3}}{2} \\ \text { R.H.5 } &=2 \sin 30^{\circ} \cos 30^{\circ} \end{aligned}$
$\left\{\begin{array}{l}\sin 30^{\circ}=\frac{1}{2} \\ \cos 60^{\circ}=\frac{1}{2} \\ \cos 20^{\circ}=\frac{2}{2}\end{array}\right\}$
$=2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$
Hence $\sin 60^{\circ}=2 \sin 30^{\circ} \cos 30^{\circ}$
(ii)
$\begin{aligned} 3 \sin ^{2} & 45^{\circ}+2 \cos ^{2} 60^{\circ} \\=& 3\left(\frac{1}{\sqrt{2}}\right)^{2}+2 \times\left(\frac{1}{2}\right)^{2} \\=& \frac{3}{2}+2 \times \frac{1}{4} \\=& \frac{3}{2}+\frac{1}{2}=\frac{3+1}{2}=\frac{4}{2}=2 \end{aligned}$
Question 2
Ans: $\sin x^{\circ}=\cos x^{\circ} \Rightarrow \frac{\sin x}{\cos x}=1$
$\Rightarrow \tan x=1=\tan 45^{\circ}$
$\operatorname{comparing}$ we get
$x=45^{\circ}$
Question 3
Ans: (i) $\sin ^{2} 60^{\circ}+\cos ^{2} 45^{\circ}=\left(\sqrt{\frac{3}{2}}\right)^{2}+\left(\left.\frac{1}{\sqrt{2}}\right)^{2}\right.$
$\left\{\begin{array}{l}\text { if } \sin 60^{\circ}=\frac{\sqrt{3}}{2} \\ \cos 45^{\circ}=\frac{1}{\sqrt{2}}\end{array}\right\}$
$=\frac{3}{4}+\frac{1}{2}$
$=\frac{3+2}{4}=\frac{5}{4}=1 \frac{1}{4}$
(ii)$3 \cos ^{2} 30^{\circ}+\tan ^{2} 60^{\circ}=3$ $\left(\frac{\sqrt{3}}{2}\right)^{2}$ +$(\sqrt{3})^{2}$
$=3 \times \frac{3}{4}+3$
$=\frac{9}{4}+3=\frac{9+12}{4}=\frac{21}{4}=5 \frac{1}{4}$
(iii) $4 \sin ^{2} 60^{\circ}+3 \tan ^{2} 30^{\circ}-8 \sin 45^{\circ} \cos 45^{\circ}$
$=4\left(\frac{\sqrt{3}}{2}\right)^{2}+3\left(\frac{1}{\sqrt{3}}\right)^{2}-8 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$=4 \times \frac{3}{4}+3 \times \frac{1}{3}-8 \times \frac{1}{2}$
$=3+1-4-4-4=0$
(iv)
$\begin{aligned}2 \sin ^{2} 30^{\circ} &-3 \cos ^{2} 45^{\circ}+\tan ^{2} 60^{\circ} \\=& 2\left(\frac{1}{2}\right)^{2}-3\left(\frac{1}{\sqrt{2}}\right)^{2}+(\sqrt{3})^{2} \\=& 2 \times \frac{1}{4}-3 \times \frac{1}{2}+3 \\=& \frac{1}{2}-\frac{3}{2}+3 \\=& \frac{1-3+6}{2}=\frac{4}{2}=2\end{aligned}$
(v) $\frac{\sin 60^{\circ}}{\cos ^{2} 45^{\circ}}-3 \tan 30^{\circ}+5 \cos 90^{\circ}$
$=\frac{\sqrt{3}}{\left(\frac{1}{\sqrt{2}}\right)^{2}}-3 \times \frac{1}{\sqrt{3}}+5 \times 0$
$=\frac{\sqrt{3}}{2 \times \frac{1}{2}}-\frac{3}{\sqrt{3}}+0$
$=\sqrt{3}-\sqrt{3}+0=0$
(vi) $\cos 90^{\circ}+\cos ^{2} 45^{\circ} \sin 30^{\circ} \tan 45^{\circ}$
$=0+\left(\frac{1}{\sqrt{2}}\right)^{2} \times \frac{1}{2} \times 1$
$=0+\frac{1 \times 1}{2 \times 2} \times 1=0+\frac{1}{4}=\frac{1}{4}$
(vii) $\cos ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 30^{\circ}$
$=\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}$
$=\frac{1}{2}+\frac{3}{4}+\frac{1}{4}=\frac{2+3+1}{4}=\frac{6}{4}$
$=\frac{3}{2}=1 \frac{1}{2}$
(viii)$\sin ^{2} 30^{\circ}+\cos ^{2} 60^{\circ}$
$=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}$
$=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
(ix) $\frac{\sin ^{2} 45^{\circ}+\cos ^{2} 45^{\circ}}{\tan ^{2} 60^{\circ}}=\frac{\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}}{(\sqrt{3})^{2}}$
$=\frac{\frac{1}{2}+\frac{1}{2}}{3}=\frac{1}{3}$
(x) $\frac{5 \sin ^{2} 30^{\circ}+\cos ^{2} 45^{\circ}-4 \tan ^{2} 30^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\tan 45^{\circ}}$
$5 \frac{5\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}-4\left(\frac{1}{\sqrt{3}}\right)^{2}}{2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}+1}$
$\frac{\frac{5}{4}+\frac{1}{2}-\frac{4}{3}}{\frac{\sqrt{3}}{2}+1}=\frac{\frac{15+6-16}{12}}{\frac{\sqrt{3}+2}{2}}$
$=\frac{5}{12} \times \frac{2}{\sqrt{3}+2}$
$=\frac{5}{6(\sqrt{3}+2)}=\frac{5}{6(2+\sqrt{3})}$
(xi) $2 \sqrt{2} \cos 45^{\circ} \cdot \cos 60^{\circ}+2 \sqrt{3} \sin 30^{\circ} \cdot \tan 60^{\circ}-\cos 0^{\circ}$
$\begin{aligned}=& 2 \sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2}+2 \sqrt{3} \times \frac{1}{2} \times\frac{\sqrt{3}}{1}-1 \\=& 1+3-1=4-1=3\end{aligned}$
(xii) $\frac{4}{3} \tan ^{2} 30^{\circ}+\sin ^{2} 60^{\circ}-3 \cos ^{2} 60^{\circ}+\frac{3}{4} \tan ^{2}$ $60^{\circ}-2 \tan ^{2} 45^{\circ}$
$=\frac{4}{3}\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-3\left(\frac{1}{2}\right)^{2}+\frac{3}{4}(\sqrt{3})^{2}-2(1)^{2}$
$=\frac{4}{3} \times \frac{1}{3}+\frac{3}{4}-3 \times \frac{1}{4}+\frac{3}{4} \times 3-2$
$=\frac{4}{9}+\frac{3}{4}-\frac{3}{4}+\frac{9}{4}-\frac{2}{1}$
$=\frac{16+27-27+81-72}{36}=\frac{25}{36}$
(xiii)$\left(\cos 0^{\circ}+\sin 45^{\circ}+\sin 30^{\circ}\right) \times\left(\sin 30^{\circ}-\cos 45^{\circ}+\cos 60^{\circ}\right)$
$=\left(1+\frac{1}{\sqrt{2}}+\frac{1}{2}\right)\left(1-\frac{1}{\sqrt{2}}+\frac{1}{2}\right)$
$=\left(\frac{3}{2}+\frac{1}{2}\right)\left(\frac{3}{2}-\frac{1}{\sqrt{2}}\right)$
$=\left(\frac{3}{2}\right)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{3}{4}-\frac{1}{2}$
$\left\{\because(a+b)(a-b)^{2}=a^{2}-b^{2}\right\}$
$=\frac{9-2}{4}=\frac{7}{4}=1 \frac{3}{4}$
Question 4
Ans: (IMAGE TO BE ADDED)
In $\triangle A D C, H D=A C= X
Draw $A D \perp B C$ which bisects $B C$ at $D$
if $\angle B=45^{\circ}$, then $\angle C=45^{\circ}$
and $\angle A=90^{\circ}$
So $\angle B A D=\angle C A D=45^{\circ}$
So $B D=D C=A D=\frac{1}{2} B C$
if $\angle A=90^{\circ}$
So $B C^{2}=A B^{2}+n C^{2}$
$=x^{2}+x^{2}=2 x^{2}$
$B C=\sqrt{2} x$
So $B D=D C=A D=\frac{\sqrt{2} x}{2}$
Now(i) $\sin 45^{\circ}=\frac{A D}{A D}=\frac{\frac{\sqrt{2} x}{2}}{x}$
$=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
$\cos 45^{\circ}=\frac{B D}{A D}=\frac{\sqrt{2 x}}{\frac{2}{x}}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
$\tan 45^{\circ}=\frac{B D}{B D}=\frac{\frac{\sqrt{2}}{2} x}{\frac{\sqrt{2}}{2} x}=1$
Question 5
Ans: $\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} x \tan 60^{\circ}}=\frac{\frac{1}{2}-1+2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}}$
$=\frac{\frac{1}{2}-1+2}{1}=\frac{1}{2}-1+2$
$=1 \frac{1}{2}=1.5$
Question 6
Ans: (ii) $\sin 60^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{\sqrt{3}}{2}$
$\sin 60^{\circ}=\sqrt{\frac{3}{2}}$
and $\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}$
Hence $\sin 60^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{\sqrt{3}}{2}$
(ii) $\cos 60^{\circ}=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{1}{2}$
$\cos 60^{\circ}=\frac{1}{2}$
and $\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}$
$=\frac{\frac{2}{3}}{\frac{4}{3}}=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}$
Hence $\cos 60^{\circ}=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{1}{2}$
(iii)
$\begin{gathered}\cos 60^{\circ}=\cos ^{2} 30^{\circ}-\sin ^{2} 30^{\circ} \\\cos 60^{\circ}=\frac{1}{2}\end{gathered}$
$\begin{aligned}&=\frac{3}{4}-\frac{1}{4} \\&=\frac{2}{4}=\frac{1}{2}\end{aligned}$
Hence $\cos 60^{\circ}=\cos ^{2} 30^{\circ}-\sin ^{2} 30^{\circ}$
(iv)$\cos 60^{\circ}=1-2 \sin ^{2} 30^{\circ}=2 \cos ^{2} 30^{\circ}-1$
$\cos 60^{\circ}=\frac{1}{2}$
$1-2 \sin ^{2} 30^{\circ}=1-2 \times\left(\frac{1}{2}\right)^{2}=1-2 \times \frac{1}{4}$
$=1-\frac{1}{2}=\frac{1}{2}$
and $2 \cos ^{2} 30^{\circ}-1=2\left(\frac{\sqrt{3}}{2}\right)^{2}-1$
$=2 \times \frac{3}{4}-1$
$=\frac{3}{2}-1=\frac{1}{2}$
Hence
$\begin{aligned} \cos 60^{\circ} &=1-2 \sin ^{2} 30^{\circ} \\ &=2 \cos ^{2} 30^{\circ}-1 \end{aligned}$
Question 7
Ans:(i)
$\begin{aligned} & \text { L.H.S }=\cos ^{2} 30^{\circ}+\sin ^{2} 30^{\circ}+\tan ^{2} 45^{\circ} \\ &=\left(-\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+(1)^{2} \\ &=\frac{3}{4}+\frac{1}{4}+1=1+1=2=R.H.S\end{aligned}$
(ii)
$\begin{aligned} \text { 1.H.S. } &=4\left(\sin ^{4} 30^{\circ}+\cos ^{4} 60^{\circ} 1-3\left(\cos ^{2} 45^{\circ}-\sin ^{2} 90^{\circ}\right)\right.\\ &=4\left[\left(\frac{1}{2}\right)^{4}+\left(\frac{1}{2}\right)^{4}\right]-3\left[\left(\frac{1}{\sqrt{2}}\right)^{2}-(1)^{2}\right] \\ &=4\left[\frac{1}{16}+\frac{1}{16}\right]-3\left[\frac{1}{2}-1\right] \\ &=4 \times \frac{1}{8}-3 \times\left(-\frac{1}{2}\right) \\ &=\frac{1}{2}+\frac{3}{2}=\frac{4}{2}=2=R \cdot \text { H.S. } \end{aligned}$
Question 8
Ans: $A C$ is wall and $A D$ is a pole which rests with wall at $60^{\circ}$
In right $\triangle A B C, \angle C=90^{\circ}$
$A P=15 \mathrm{~m}$
Let height of wall $=x m$ and away from foot of wall $=y m$
(IMAGE TO BE ADDED)
(i)
If $\begin{aligned} \sin 60^{\circ} &=\frac{A C}{A D} \\ \Rightarrow \frac{\sqrt{3}}{2} &=\frac{x}{15} \Rightarrow x=\frac{15 \sqrt{3}}{2} \\ \Rightarrow x &=\frac{15(1.732)}{2}=\frac{25.980}{2} \\ &=12.99 \end{aligned}$
So height of wall = 12.99m
(ii) $\cos 60^{\circ}=\frac{B C}{A B} \Rightarrow \frac{1}{2}=\frac{y}{15}$
$=y=\frac{15}{2}=7.5 \mathrm{~m}$
So Distance form the foot of wall = 7.5m
Question 9
Ans: In the $\triangle A B C, \angle B=90^{\circ}$ and
$\begin{aligned}&\angle C=\theta, A C=\operatorname{toc} \mathrm{cm}, A B=20 \mathrm{~cm} \\&\text { Herc } \sin \theta=\frac{A B}{A C}=\frac{20}{10}=\frac{1}{2} \\&=\sin 30^{\circ} \\&\text { so } \theta =30^{\circ}\end{aligned}$
Question 10
Ans: (IMAGE TO BE ADDED)
(i) In $\triangle A B C, \angle C=90^{\circ}$
$\begin{aligned}&\angle B=30 \text { and } C=-10 \\&\sin \theta=\frac{B C}{A B}=\frac{9}{C} \\&\Rightarrow \sin 30^{\circ}=\frac{a}{C} \Rightarrow \frac{1}{2}=\frac{9}{400}\end{aligned}$
$\Rightarrow a=\frac{10}{2}=20$
But $c^{2}=a^{2}+b^{2}$
$\begin{aligned}(40)^{2} &=(20)^{2}+b^{2} \Rightarrow 1600=400+b^{2} \\ \Rightarrow b^{2} &=1600-400=1200 \\ &=400 \times 3 \\ \text { so } & b=\sqrt{400 \times 3}=20 \sqrt{3}=20(1.732) \\ &=34.64=34.64 \\ a &=20 \text { and } b=34.64 \text { and } \\ & \angle B=90^{\circ}=30^{\circ}=60^{\circ} \end{aligned}$
(ii)
$\begin{aligned} \angle B=& 60^{\circ} \text { and } C=15 \\ & \sin \theta=\frac{A C}{A D} \Rightarrow \sin 60^{\circ}=\frac{9}{C} \\ \Rightarrow & \frac{3}{2}=\frac{9}{15} \Rightarrow a=\frac{15\sqrt 3}{2} \end{aligned}$
$\Rightarrow a=\frac{15(1.732)}{2}=12.99$
We know that by Pythagoras theorem
$c^{2}=a^{2}+b^{2}$
$(15)^{2}=\left(\frac{15 \sqrt{3}}{2}\right)^{2}+b^{2} \Rightarrow 225=\frac{675}{1}+b^{2}$
$\Rightarrow b^{2}=225-\frac{675}{4}$
$=\frac{900-675}{4}=\frac{225}{4}$
So $b=\sqrt{\frac{225}{4}}=\frac{15}{2}=7.5$
Hence $a=12.99$ and $b=7.5$
and $\angle A=90^{\circ}-B=90^{\circ}-60^{\circ}=30^{\circ}$
(iii)
$\begin{aligned}&A=45^{\circ} \text { and } a=7 \\&\sin \theta=\frac{B C}{A C} \Rightarrow \sin 45^{\circ}=\frac{a}{C}\end{aligned}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{7}{c} \Rightarrow c=7 \sqrt{2}=7(1.414)$
$\Rightarrow c=9.898=9.9$
But $c^{2}=a^{2}+b^{2}$
$\Rightarrow(7 \sqrt{2})^{2}=(7)^{2}+b^{2} \Rightarrow 98=49+b^{2}$
$\Rightarrow b^{2}=98-49=49=(7)^{2}$
so $b=7$
Here c= 9.9 and b=7
And $\angle B=90^{\circ}-\angle A=90^{\circ}-45^{\circ}=45^{\circ}$
Question 11
Ans: $4 \sin ^{2} \theta-1=0 \Rightarrow 4 \sin ^{2} \theta=1$
$\Rightarrow \sin ^{2} \theta=\frac{1}{4}=\left(\frac{1}{2}\right)^{2}$
(i) so $\sin \theta=\frac{1}{2}=\sin 30^{\circ}$
so $\theta=30^{\circ}$
Now $\cos ^{2} \theta+\tan ^{2} \theta=\cos ^{2} 30^{\circ}+\tan ^{2} 30^{\circ}$
$=\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}$
$=\frac{3}{4}+\frac{1}{3}$
$=\frac{2+4}{12}=\frac{13}{1}=1 \frac{1}{12}$
Question 12
Ans: (i) If $\sin \theta=\cos \theta \Rightarrow \frac{310}{\cos \theta}=1$
$\begin{aligned}\Rightarrow \quad \tan \theta &=1=\tan 45^{\circ} \\\text { so } \theta &=45^{\circ} \\\text { (iil) Now } 2 \tan ^{2} \theta+\sin ^{2} \theta-1 \\=2 \tan ^{2} 45^{\circ}+\sin ^{2} 45^{\circ}-1 \\=2(1)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}-1 \\=2 \times 1+\frac{1}{2}-1=2+\frac{1}{2}-1 \\=1 \frac{1}{2}=\frac{3}{2}=1.5\end{aligned}$
Question 13
Ans: (IMAGE TO BE ADDED)
Height of kite $=75 \mathrm{~m}$
Angle of elevation $=60^{\circ}$
So JK is the sting tied to the kite
Let JK = x m
Now sin ө=\frac{K L}{J K}$
$\Rightarrow \sin 60^{\circ}=\frac{75}{x} \Rightarrow \frac{\sqrt{3}}{2}=\frac{75}{x}$
$\Rightarrow x=\frac{75 \times 2}{\sqrt{3}}=\frac{150 \times 5 \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \mathrm{~m}$
$=\frac{150 \sqrt{3}}{3}=50 \sqrt{3}=50(1.732)$
$=86.600 \mathrm{~m}=87 \mathrm{~m}$
Question 14
Ans: (i) $2 \sin \theta-1=0 \Rightarrow 2 \sin \theta=1$
$\sin \theta=\frac{1}{2}=\sin 30^{\circ}$
So $\theta=30$
$\begin{aligned} \text { (ii) Now } \cos ^{2} \theta+\tan ^{2} \theta &=\cos ^{2} 30^{\circ}+\operatorname{tan}^{2} \operatorname{30}^{2} \\ &=\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}=\frac{3}{4}+\frac{1}{3}^{\circ} \\ &=\frac{9+4}{12}=\frac{13}{12}=1 \frac{1}{12} \end{aligned}$
Question 15
(IMAGE TO BE ADDED)
Ans: In $\triangle A B C, \angle A$ is an Obtuse angle
$A D \perp B C$.
$B D=10 \mathrm{~cm}$ and CD $=10 \sqrt{3} \mathrm{~cm}$
Let $\angle B A D=a$ and $\angle C A D=\beta$
$\angle A=a+P$
Now $\tan a=\frac{B D}{A D}=\frac{10}{10}=1=\tan 45^{\circ}$
So $a=45^{\circ}$
and $\tan β=\frac{C D}{A D}=\frac{10 \sqrt{3}}{10}=\sqrt{3}=\tan 60^{\circ}$
So $\beta=60^{\circ}$
Now $\angle A=a+\beta=45^{\circ}+60^{\circ}=105^{\circ}$
Question 16
Ans: (IMAGE TO BE ADDED)
In $\triangle A B C ; \angle C=90^{\circ}+\angle B=60^{\circ}$
$A D=15$ units
$\begin{aligned} \operatorname{so} \angle A &=180^{\circ}-\left(90^{\circ}+60^{\circ}\right) \\ &=180^{\circ}-150^{\circ}=30^{\circ} \\ & \sin 0=\frac{A C}{A D} \Rightarrow \sin 60^{\circ}=\frac{A C}{A D} \end{aligned}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{A C}{15} \Rightarrow A C=\frac{15 \sqrt{3}}{2}=\frac{15}{2} \sqrt{3}$ units
and $\cos 60^{\circ}=\frac{B C}{A D} \Rightarrow \frac{1}{2}=\frac{B C}{15}$
$\Rightarrow B C=\frac{15}{2}=7.5$ units
Hence $\angle B=30^{\circ}, D C=7.5$ units $A C=\frac{15}{2} \sqrt{3}$ units
Question 17
Ans: (IMAGE TO BE ADDED)
In rectangle $A B C D, A B=20 \mathrm{~cm}$
$\angle B A C=60^{\circ}$ and AC and BD are its two diagonals which are equal
Tan $\angle B A C=\frac{B C}{A B}$
$\Rightarrow \tan 60^{\circ}=\frac{BC}{20} \Rightarrow \sqrt{3}=\frac{B C}{20}$
$\Rightarrow B C=20 \sqrt{3} \mathrm{~cm}$
$\begin{aligned} & \text { similarly } \cos 60^{\circ}=\frac{A B}{A C}=\frac{20}{A C} \\ \Rightarrow & \frac{1}{2}=\frac{20}{A C}=A C=20 \times 2=40 \mathrm{~cm} \\ \text { if } B D=A C \\ \Rightarrow B D=40 \mathrm{~cm} \end{aligned}$
Question 18
Ans: (IMAGE TO BE ADDED)
Let AB is wall and AC is ladder which is Placed against the wall and makes an angles of 60 with the ground
Foot of the ladder as 1.5m away from the wall i.e DC =1.5m
Let h be the height of the wall then
$\tan C=\frac{A D}{D C} \Rightarrow \tan 60^{\circ}=\frac{h}{1.5}$
$\Rightarrow \sqrt{3}=\frac{h}{1.5} \Rightarrow h=1.5 \sqrt{3}$
$\Rightarrow h=(1.5)(1.731 \mathrm{~m}=2.595 \mathrm{~m}$
Question 19
Ans: BC is the electric pole and AC is the wire tied to it such that
$\angle A=45^{\circ}$ and $B C=10 \mathrm{~m}$
Let: $A C=x$
Now $\sin A=\frac{B C}{11 C}$
$\Rightarrow \sin 45^{\circ}=\frac{10}{x}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{10}{x}$
x = 10$=\sqrt{2}$ =10 $(1.419-1 \mathrm{~m}=14.14) \mathrm{~m}$
Length of wire = 14.14= 14.1m
Question 20
(IMAGE TO BE ADDED)
Ans: DB is tree which al taken from A and makes an angles of $60^{\circ}$ with the ground then $A D=A C$ and $D B=15 \mathrm{~cm}$
$A B=x$, then
$A C=A D=(15-x) m$
Now. $\sin \theta=\frac{P_{e n p}}{H y p_{0}} \Rightarrow \sin 60^{\circ}=\frac{A B}{A C}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{x}{15-x} \Rightarrow 2 x=15 \sqrt{3}-\sqrt{3} x$
$\Rightarrow 2 x+\sqrt{3} x=15-\sqrt{3}$
$\Rightarrow(2+\sqrt{3}) x=15 \sqrt{3}$
so $x=\frac{15-\sqrt{3}}{2+\sqrt{3}}=\frac{15 \sqrt{3}(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$
$=\frac{30 . \sqrt{3-45}}{4-3}$
$=\frac{30(1.73)-45}{4-3}$
$=\frac{30(1.73)-45}{1}=51.9-45=6.9 \mathrm{~m}$
So height of tree from which it was broken = 6.9m
Question 21
Ans: K is kite and KL is string tied to it which makes an angles of 60 with the ground
So $\angle L=60^{\circ}, L K=150 \mathrm{~m}$
Let $k m=h m$
$\sin \theta=\frac{p e r p}{H y p} .$
$\Rightarrow \sin 60^{\circ}=\frac{K m}{K L}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{h}{150} \Rightarrow h=\frac{150 \sqrt{3}}{2}=75 \sqrt{3} \mathrm{~m}$
Height of the kite = $75 \sqrt{3}=75 \times 1.73 \mathrm{~m}$
$=129.75=129.8 \mathrm{~m}$
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