ML Aggarwal Solution Class 9 Chapter 18 Trigonometric Ratios and Standard Angles Exercise 18.1

 Exercise 18.1

Question 1

Sol :

(i)

$7 \sin 30^{\circ} \cos 60^{\circ}$

$\Rightarrow 7\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$

$\Rightarrow \frac{7}{4}$

(ii)

$3 \sin ^{2} 45^{\circ}+2 \cos ^{2} 60^{\circ}$

$\Rightarrow 3 \cdot\left(\frac{1}{\sqrt{2}}\right)^{2}+2 \cdot\left(\frac{1}{2}\right)^{2}$

$\Rightarrow  \frac{3}{2}+\frac{2}{4}$

$\Rightarrow  \frac{3}{2}+\frac{1}{2}$

$\Rightarrow  \frac{3+1}{2}$

$\Rightarrow \frac{4}{2}$

⇒2

(iii) $\cos ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}+\sin ^{2} 30^{\circ}$

$\Rightarrow\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}$

$\Rightarrow \frac{1}{2}+\frac{3}{4}+\frac{1}{4}$

$\Rightarrow \frac{2+3+1}{4}=\frac{6}{4}=\frac{3}{2}$

(iv) 

$\cos 90^{\circ}+\cos ^{2} 45^{\circ} \sin 30^{\circ} \tan 45^{\circ}$

$\Rightarrow 0+\left(\frac{1}{\sqrt{2}}\right)^{2} \cdot\left(\frac{1}{2}\right) \cdot(1)$

$\Rightarrow 0+\frac{1}{2} \cdot \frac{1}{2}$

$\Rightarrow 0+\frac{1}{4}$

$\Rightarrow  \frac{0+1}{4}=\frac{1}{4}$

Question 2

Sol :

(i)

$\frac{\sin ^{2} 45^{\circ}+\cos ^{2} 45}{\tan ^{2} 60}$

$\Rightarrow \frac{\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}}{(\sqrt{3})^{2}}$

$\Rightarrow \frac{\frac{1}{2}+\frac{1}{2}}{3}$

$\Rightarrow \frac{1+1}{2 \cdot 3}$

$\Rightarrow \frac{2}{6}$

$\Rightarrow \frac{1}{3}$

(ii) 

$\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \times \tan 60^{\circ}}$

$\Rightarrow \frac{\frac{1}{2}-1+2(1)}{\frac{1}{\sqrt{3}} \times \sqrt{3}}$

$\Rightarrow \frac{1}{2}+1$

$\Rightarrow \frac{1+2}{2}$

$\Rightarrow \frac{3}{2}$

(iii) 

⇒$\frac{4}{3} \tan ^{2} 30^{\circ}+\sin ^{2} 60^{\circ}-3 \cos ^{2} 60^{\circ}+\frac{3}{4} \tan ^{2} 60^{\circ}-2 \tan ^{2} 45^{\circ}$

⇒$\frac{4}{3}\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-3\left(\frac{1}{2}\right)^{2}+\frac{3}{4}(\sqrt{3})^{2}-2 \cdot(1)^{2}$

⇒$\frac{4}{3} \cdot \frac{1}{3}+\frac{3}{4}-\frac{3}{4}+\frac{9}{4}-2$

⇒$\frac{4}{9}+\frac{9}{4}-2$

⇒$\frac{4 \times 4+9 \times 9-2 \times 36}{36}=\frac{16+81-12}{36}$

⇒$\frac{25}{36}$

Question 3

(i) $\frac{\sin 60^{\circ}}{\cos ^{2} 45^{\circ}}-3 \tan 30^{\circ}+5 \cos 90^{\circ}$

$\Rightarrow \frac{\sqrt{3} / 2}{(1 / \sqrt{2})^{2}}-3 \frac{1}{\sqrt{3}}+5(0)$

$\Rightarrow \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}-\frac{3}{\sqrt{3}}+0$

$\Rightarrow  \sqrt{3}-\frac{3}{\sqrt{3}}$

$\Rightarrow  \frac{\sqrt{3} \cdot \sqrt{3}-3}{\sqrt{3}}$

$\Rightarrow  \frac{3-3}{\sqrt{3}}$

⇒0

(ii)

$2 \sqrt{2} \cos 45^{\circ} \cos 60^{\circ}+2 \sqrt{3} \sin 30^{\circ} \tan 60^{\circ}-\cos 0^{\circ}$

$\Rightarrow 2 \sqrt{2} \frac{1}{\sqrt{2}} \cdot \frac{1}{2}+2 \sqrt{3} \cdot \frac{1}{2} \cdot \sqrt{3}-1$

⇒1+3-1

⇒3

(iii) 

$\frac{4}{5} \tan ^{2} 60^{\circ}-\frac{2}{\sin ^{2} 30^{\circ}}-\frac{3}{4} \tan ^{2} 30^{\circ}$

$\Rightarrow \frac{4}{5}(\sqrt{3})^{2}-\frac{2}{\left(\frac{1}{2}\right)^{2}}-\frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)^{2}$

$\Rightarrow \quad \frac{4}{5} \cdot 3-\frac{2}{\left(\frac{1}{4}\right)}-\frac{3}{4} \cdot \frac{1}{3}$

$\Rightarrow \quad \frac{12}{5}-8-\frac{1}{4}$

$\Rightarrow \frac{12 \times 4-8 \times 5 \times 4-1 \times 5}{5 \times 4}$

⇒$\frac{48-160-5}{20}$

⇒$-\frac{117}{20}$

⇒$-5 \frac{17}{20}$

Question 4

Sol :
(i) 

LHS

$\Rightarrow \cos ^{2} 30+\sin 30+\tan ^{2} 45^{\circ}$

$\Rightarrow\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{2}+1^{2}$

$\Rightarrow \frac{3}{4}+\frac{1}{2}+1$

$\Rightarrow \frac{3+2+4}{4}$

$\Rightarrow \frac{9}{4}$

$\Rightarrow 2 \frac{1}{4}$

Question 5

Sol :

(i) Given : x=30°

LHS⇒tan 2x

⇒tan2(20°)

⇒tan 60°

⇒√3

RHS$\Rightarrow \frac{2 \tan x}{1-\tan ^{2} x}$

$\Rightarrow \frac{2 \cdot \tan 30^{\circ}}{1-\tan ^{2} 30}$

$\Rightarrow \frac{2 \cdot \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}$

$\Rightarrow \frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$

$\Rightarrow \frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}$

$\Rightarrow \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$

$\Rightarrow  \frac{3}{\sqrt{3}} \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow \frac{3 \sqrt{3}}{3}$

⇒√3

∴LHS=RHS

(ii)

Given : x=15°

LHS⇒4sinx.cos4x.sin6x

⇒4sin(2×15).cos(4×15).sin(6×15)

⇒4sin30°×cos 60°×sin 90°

⇒$4 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 1$

⇒1

∴LHS=RHS

Question 6

Sol :

(i)

$\sqrt{\frac{1-\cos ^{2} 30^{\circ}}{1-\sin ^{2} 30^{\circ}}}$  $\left(\because \sin ^{2} A+\cos ^{2} A=1\right)$

$\Rightarrow \sqrt{\frac{\sin ^{2} 30}{\cos ^{2} 30^{\circ}}}$

$\Rightarrow \sqrt{\frac{\left(\frac{1}{2}\right)^{2}}{\left(\frac{\sqrt{3}}{2}\right)^{2}}}$

$\Rightarrow \sqrt{\frac{\frac{1}{4}}{\frac{3}{4}}}$

$\Rightarrow  \frac{1}{\sqrt{3}}$

(ii) 

⇒$\frac{\sin 45^{\circ} \cdot \cos 45^{\circ} \cdot \cos 60^{\circ}}{\sin 60^{\circ} \cos 30^{\circ} \cdot \tan 45^{\circ}}$

⇒$\frac{\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2}}{\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \cdot 1}$

⇒$\frac{\frac{1}{2}}{\frac{3}{2}}$

⇒$\frac{1}{3}$

Question 7

Sol :

Given : θ=30°

(i)

LHS⇒$\sin 2 \theta=\sin 2.30^{\circ}$

=sin 60

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

RHS⇒2sinθcosθ$=2 \cdot \sin 30^{\circ} \cdot \cos 30^{\circ}$

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

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

∴LHS=RHS

(ii) 

LHS

$\Rightarrow \cos 2 \theta=\cos 2.30^{\circ}$

$=\cos 60^{\circ}$

$=\frac{1}{2}$

RHS

$\Rightarrow 2 \cos ^{2} \theta-1=2 \cdot \cos ^{2} \cdot 30^{\circ}-1$

$=2 \cdot\left(\frac{\sqrt{3}}{2}\right)^{2}-1$

$=2\frac{\sqrt{3}}{4}-1$

$=\frac{3-2}{2}$

$=\frac{1}{2}$

∴LHS=RHS

(iii) 

LHS⇒sin3θ

$=\sin 3\times 30^{\circ}$

$=\sin 90^{\circ}$

=1

RHS

$\Rightarrow 3 \sin \theta-4 \sin ^{3} \theta$

$\Rightarrow 3 \cdot \sin 30^{\circ}-4 \sin ^{3} 30^{\circ}$

$\Rightarrow 3 \cdot \frac{1}{2}-4\left(\frac{1}{2}\right)^{3}$

$\Rightarrow \frac{3}{2}-\frac{4}{8}$

$\Rightarrow \frac{3-1}{2}$

$\Rightarrow \frac{2}{2}=1$

∴LHS=RHS

(iv)

LHS⇒cos3θ⇒cos3×30°

⇒cos 90°

⇒0

RHS⇒$4 \cos ^{3} \theta-3 \cos \theta$

⇒$4 \cdot \cos ^{3} 30^{\circ}-3 \cos 30^{\circ}$

⇒$4 \cdot\left(\frac{\sqrt{3}}{2}\right)^{3}-3 \cdot \frac{\sqrt{3}}{2}$

⇒$4 \cdot \frac{3 \sqrt{3}}{8}-\frac{3 \sqrt{3}}{2}$

⇒$\frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}$

⇒0

∴LHS=RHS

Question 8

Sol :

Given : θ=30°

⇒2sinθ : sin2θ

⇒$\frac{2 \sin \theta}{\sin 2 \theta}=\frac{2 \cdot \sin 30^{\circ}}{\sin 2\times 30^{\circ}}$

$=\frac{2 \times \frac{1}{2}}{\sin 60}$

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

$\frac{2 \sin \theta}{\sin 2 \theta}=\frac{2}{\sqrt{3}}$

∴2sinθ : sin2θ=2 : √3

Question 9

Sol :

Given : A=30° ; B=60°

LHS⇒sin(A+B)

⇒sin(30°+60°)

⇒sin 90°

⇒1

RHS⇒sinA+sinB

⇒sin30°+sin60°

⇒$\frac{1}{2}+\frac{\sqrt{3}}{2}$

⇒$\frac{1+\sqrt{3}}{2}$

∴LHS≠RHS

ie., $\sin (A+B) \neq \sin A+\sin B$.

Question 10

Given : A=60° ; B=30°

Sol :

(i)

LHS⇒sin(A+B)=sin(60+30)

=sin 90=1

RHS⇒sinAcosB+cosAsinB

⇒sin60°cos30°+cos60°sin30°

⇒$\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \cdot \frac{1}{2}$

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

=1

(ii)

A=60° ; B=30°

LHS⇒cos(A+B)=cos(60°+30°)

=cos90°=0

RHS⇒cosA.cosB-sinAsinB

⇒cos60°.cos30°-sin60°.sin30°

⇒$\frac{1}{2} \cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \cdot \frac{1}{2}$

⇒$\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

⇒0

∴LHS=RHS

i.e. cos(A+B)=cosA.cosB-sinAsinB

(iii) 

LHS

⇒sin(A-B)=sin(60°-30°)

=sin30°

$=\frac{1}{2}$

RHS

⇒sinA.cosB-cosAsinB

⇒sin60°.cos30°-cos60°.sin30°

⇒$\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2}. \frac{1}{2}$

⇒$\frac{3}{2}-\frac{1}{2}$

⇒$\frac{3-1}{2}=\frac{1}{2}$=RHS

∴sin(A-B)=sinAcosB-cosAsinB

(iv)

A=60° ; B=30°

LHS⇒tan(A-B)=tan(60°-30°)

=tan 30°

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

RHS⇒$\frac{\tan A-\tan B}{1+\tan A \cdot \tan B}$

$\Rightarrow \frac{\operatorname{lan} 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}$

$\Rightarrow \frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \frac{1}{\sqrt{3}}}$

$\Rightarrow \frac{\frac{\sqrt{3} \cdot \sqrt{3}-1}{\sqrt{3}}}{2}$

$\Rightarrow \frac{\frac{3-1}{\sqrt{3}}}{2}$

$\Rightarrow \frac{\frac{2}{\sqrt{3}}}{2}$

$\Rightarrow \frac{1}{\sqrt{3}}$

∴LHS=RHS

i.e. $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$

Question 11

Sol :

(i) Given : 2sin2θ=√3

$\sin 2 \theta=\frac{\sqrt{3}}{2}$

sin2θ=sin60°

2θ=60°

$\theta=\frac{60}{2}$

θ=30°

(ii)

Given : 

$\cos \left(20^{\circ}+x\right)=\sin 60^{\circ}$

$\cos (20+x)=\frac{\sqrt{3}}{2}$

$\cos (20+x)=\cos 30^{\circ}$

$20+x=30^{\circ}$

$x=30-20$

$x=10^{\circ}$

(iii)

Given 

$3 \sin ^{2} \theta =2 \frac{1}{4}$

$3 \sin ^{2} \theta =\frac{9}{4}$

$\sin ^{2} \theta =\frac{9}{4\times 3}$

$\sin ^{2} \theta =\frac{3}{4}$

$\sin \theta =\sqrt{\frac{3}{4}}$

$\sin \theta =\frac{\sqrt{3}}{2}$

$\sin \theta =\frac{\sqrt{3}}{2}$

$\sin \theta=\sin 60^{\circ}$

$\theta =60^{\circ}$

Question 12

Sol :

Given : 

$\operatorname{sen} \theta=\cos \theta$

$\frac{\sin \theta}{\cos \theta}=1$

$\operatorname{tan} \theta=1$

$\theta=45^{\circ}$

$\Rightarrow 2 \tan ^{2} \theta+\sin ^{2} \theta-1$

$\Rightarrow  2 \cdot \tan ^{2} 45+\sin ^{2} 45-1$

$\Rightarrow   2 \cdot(1)+\left(\frac{1}{\sqrt{2}}\right)^{2}-1$

$\Rightarrow  2+\frac{1}{2}-1$

$\Rightarrow  \frac{4+1-2}{2}$

$\Rightarrow   \frac{3}{2}$

Question 13

Sol :

(i) From figure 

$\operatorname{tan} x^{\circ}=\frac{\sqrt{3}}{1}$

$\operatorname{tan} x^{\circ}=\sqrt{3}$

(ii) x=60

$\sin \theta =\frac{\sqrt{3}}{2}$

$\sin \theta =\sin 60^{\circ}$

$\theta=60^{\circ}$

Question 12

Sol :

Given : 

$\operatorname{sin} \theta=\cos \theta$

$\frac{\sin \theta}{\cos \theta}=1$

$\tan \theta=1$

$\therefore \quad \theta=45^{\circ}$

$\Rightarrow 2 \tan ^{2} \theta+\sin ^{2} \theta-1$

$\Rightarrow 2 \tan ^{2} 45+\sin ^{2} 45-1$

$\Rightarrow 2 \cdot(1)+\left(\frac{1}{\sqrt{2}}\right)^{2}-1$

$\Rightarrow 2+\frac{1}{2}-1$

$\Rightarrow \frac{4+1-2}{2}$

$\Rightarrow \frac{3}{2}$

Question 13

Sol :

(i) From figure

$\tan x^{\circ}=\frac{\sqrt{3}}{1}$

$\operatorname{tan} x^{\circ}=\sqrt{3}$

(ii) x=60°

Question 15

Sol :

Given : $\tan 3 x=\sin 45^{\circ} \cdot \cos 45^{\circ}+\sin 30^{\circ}$

$\tan 3 x=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2}$

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

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

$=\frac{2}{2}$

tan 3x=1

tan 3x=tan 45°

3x=45°

x=15°

Question 16

Sol :

(i) Given : $4 \cos ^{2} x-1 =0$

$4 \cos ^{2} x =1$

$\cos ^{2} x =\frac{1}{4}$

$\cos ^{3} x =\frac{1}{\sqrt{4}}$

$\cos x =\frac{1}{2}$

cos x=cos 60°

x=60°

(ii) $\sin ^{2} x+\cos ^{2} x \Rightarrow \sin ^{2} 60^{\circ}+\cos ^{2} 60^{\circ}$

$\Rightarrow\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}$

$\Rightarrow \frac{3}{4}+\frac{1}{4}$

$\Rightarrow \frac{3+1}{4}$

$\Rightarrow \frac{4}{4}$

=1

(iii) $\cos ^{2} x^{\circ}-\sin ^{2} x^{\circ}$

$\Rightarrow \cos ^{2} 60^{\circ}-\sin ^{2} 60^{\circ}$

$\Rightarrow \left(\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}$

$\Rightarrow  \frac{1}{4}-\frac{3}{4}$

$\Rightarrow \frac{1-3}{4}$

$\Rightarrow  \frac{-2}{4}$

$\Rightarrow  \frac{-1}{2}$

Question 17

Sol :

(i)

Given : secθ=cosecθ

$\frac{1}{\cos \theta}=\frac{1}{\sin \theta}$

$\sin \theta=\cos \theta$

$\frac{\sin \theta}{\cos \theta}=1$

$\tan \theta=1$

$\theta=45^{\circ}$

(ii) 

Given $\tan \theta=\cot \theta$

$\frac{\tan \theta}{\cot \theta}=1$ $\left(\because \tan \theta=\frac{1}{\cot \theta}\right)$

$\tan ^{2} \theta=1$

$\therefore \theta=45^{\circ}$

Question 18

Sol :

Given : sin3x=1

sin3x=sin90°

3x=90°

$x=\frac{90^{\circ}}{3}$

$x=30^{\circ}$

(i) sinx⇒$\sin 30=\frac{1}{2}$

(ii) cos2x⇒cos 2×30°

⇒cos 60°

⇒$\frac{1}{2}$

(iii) $\tan ^{2} x-\sec ^{2} x=\tan ^{2} 30^{\circ}-\sec ^{2} 30^{\circ}$

$=\left(\frac{1}{\sqrt{3}}\right)^{2}-\left(\frac{2}{\sqrt{3}}\right)^{2}$

$=\frac{1}{3}-\frac{4}{3}$

$=\frac{1-4}{3}=\frac{-3}{3}$

=-1

Question 19

Sol :

Given : $3 \tan ^{2} \theta-1=0$

$3 \tan ^{2} \theta=1$

$\tan ^{2} \theta=\frac{1}{3}$

$\tan \theta=\frac{1}{\sqrt{3}}$

$\theta=30^{\circ}$

∴cos2θ=cos2×30°

=cos60°

$=\frac{1}{2}$

Question 20

Sol :

Given : sinx+cosy=1

∵x=30°

∴sin30°+cosy=1

$\frac{1}{2}+\cos y=1$

$\cos y=1-\frac{1}{2}$

$\cos y=\frac{1}{2}$

y=60°

Question 21

Sol :

Given : sin(A+B)$=\frac{\sqrt{3}}{2}$=cos(A-B)

⇒sin(A+B)$=\frac{\sqrt{3}}{2}$

⇒sin(A+B)=sin+60°

⇒A+B=60°...(i)

⇒cos(A-B)$=\frac{\sqrt{3}}{2}$

⇒cos(A-B)=cos30°

⇒A-B=30°...(ii)

From (i) and (ii)

$\begin{aligned} A+B &=60^{\circ} \\ A-B &=30^{\circ} \\\hline 2 A &=90^{\circ} \end{aligned}$

A=45°

From (i)

⇒A+B=60°

⇒45°+B=60°

⇒B=60°-45°=15°

∴A=45° ; B=15°

Question 22

Sol :

From figure ΔBOC

$\sin 60 =\frac{B O}{B C}$

$\frac{\sqrt{3}}{2} =\frac{B O}{8}$

$BO=\frac{8 \cdot \sqrt{3}}{2}$

BO=4√3

Similarly from ΔOCD

$\sin 60=\frac{O D}{C D}$

$\frac{\sqrt{3}}{2}=\frac{O D}{8}$

OD=4√3

∴BD=DO+OD=4√3+4√3=8√3

and from ΔBOC : cos60$=\frac{O C}{B C}$

$\frac{0 C}{8}=\frac{1}{2}$

OC=4

From ΔAOB ; cos60$=\frac{O A}{A B}$

$\frac{1}{2}=\frac{OA}{8}$

OA=4

∴AC=OA+OC

=4+4=8

Question 23

Sol :

Given : AB=6

∠C=90°

∠B=60°

From Figure 

$\sin 60=\frac{A C}{A B}$

$\frac{\sqrt{3}}{2}=\frac{A C}{6}$

$A C=\frac{6 \cdot \sqrt{3}}{2}$

AC=3√3

From Figure 

$\cos 60=\frac{B C}{A B}$

$\frac{1}{2}=\frac{B C}{6}$

$B C=\frac{6}{2}$

BC=3

Question 24

QC=QB-CB

∵CB=AP=1.8 m

QC=13.8-1.8=12 m

From figure 

$\operatorname{tan} 30^{\circ}=\frac{Q C}{P C}$

$\frac{1}{\sqrt{3}}=\frac{12}{P C}$

PC=12√3

PC=distance of man from building is 12√3

Question 25

ΔABC ; ∠B=45°

∠C=60°

BC=8 m

From figure ΔABD

$\operatorname{tan} 45^{\circ}=\frac{A D}{B D}$

$1=\frac{A D}{B D}$

AD=BD...(i)

In ΔADC ; $\operatorname{Tan} 60^{\circ}=\frac{A D}{D C}$

$\sqrt{3}=\frac{A D}{D C}$

$\Rightarrow D C=\frac{A D}{\sqrt{3}}$

∵BC=8

⇒BD+DC=8

⇒$A D+\frac{A D}{\sqrt{3}}=8$

⇒$(\sqrt{3}+1) A D=8 \sqrt{3}$

⇒$A D=\frac{8 \sqrt{3}}{\sqrt{3+1}}$

⇒$A D=\frac{8 \sqrt{3}}{\sqrt{3}+1} \cdot \frac{\sqrt{3}-1}{\sqrt{3}-1}$

$=\frac{8 \sqrt{3}(\sqrt{3}-1)}{3-1}$

$=\frac{8 \times 3-8 \sqrt{3}}{2}$

$=\frac{24-8 \sqrt{3}}{2}$

$=12-4 \sqrt{3}$

$AD=4(3-\sqrt{3})$