(i) (c)
(ii) (a): Let AB be the height of pedestal.
\(\text { In } \Delta A B C, \)
\(\tan 30^{\circ}=\frac{A B}{B C} \)
\(\Rightarrow \quad A B=\frac{25}{\sqrt{3}}=\frac{25}{1.73}=14.45 \mathrm{~m}\)

(iii) (d): Let x be the distance between Arun and Pankaj.

\(\text { In } \Delta A B \dot{D}, \tan 45^{\circ}=\frac{A B}{B D}\)
\(\Rightarrow B D=30 \mathrm{~m}\)
\(\text { Now, in } \Delta \overline{A B C} \text { , }\)
\(\tan 30^{\circ}=\frac{A B}{B C}\)
\(\Rightarrow \frac{30}{30+x}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow x=30(\sqrt{3}-1)=30 \times 0.73=21.9 \mathrm{~m}\)
(iv) (c): \(\text { In } \triangle A R S\)

\(\sin 30^{\circ}=\frac{R S}{A S} \)
\(\Rightarrow \quad \frac{12}{A S} =\frac{1}{2} \Rightarrow A S=12 \times 2=24 \mathrm{~m}\)
(v) (c): \(\text { In } \Delta A B D, \frac{A B}{B D}=\tan 60^{\circ}\)

\(\begin{array}{l} \Rightarrow \frac{A B}{15}=\sqrt{3} \\ \Rightarrow A B=15 \times 1.73=25.95 \mathrm{~m} \end{array}\)

(i) (b): \(\angle X A C=45^{\circ}\) 
\(\therefore \quad \angle A C D=45^{\circ}\) [Alternate interior angles]
(ii) (b)
(iii) (c) : \(\text { In } \Delta A C D\)
\(\frac{A D}{D C}=\tan 45^{\circ} \)
\(\Rightarrow \frac{100}{D C}=1 \Rightarrow D C=100 \mathrm{~m}\)
(iv) (d): \(\text { In } \Delta A B D, \frac{A D}{B D}=\tan 30^{\circ}\)
\(\Rightarrow \quad \frac{100}{B D}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \quad B D=100 \sqrt{3} \mathrm{~m}\)
(v) (a): \(\text { In } \Delta A D C\)
\(\frac{A D}{A C}=\sin 45^{\circ} \Rightarrow \frac{100}{A C}=\frac{1}{\sqrt{2}} \Rightarrow A C=100 \sqrt{2} \mathrm{~m}\)

(i) (b): Total height of pole = 8 m
\(\therefore\) BD = AD - AB = (8 - 2)m = 6 m
(ii) (a): \(\text { In } \Delta B D C, \frac{B D}{B C}=\sin 60^{\circ}\)
\(\Rightarrow \quad \frac{6}{B C}=\frac{\sqrt{3}}{2} \)
\(\Rightarrow \quad B C=\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=4 \sqrt{3} \mathrm{~m}\)
(iii) (d): \(\text { In } \triangle B D C\)
\(\frac{B D}{C D}=\tan 60^{\circ} \Rightarrow \frac{6}{C D}=\sqrt{3} \Rightarrow C D=\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=2 \sqrt{3} \mathrm{~m}\)
(iv) (b) : \(\text { If } \Delta B C D\)
\(\frac{B D}{C D}=\tan \theta \Rightarrow 1=\tan \theta \quad[\because B D=C D] \)
\(\Rightarrow \quad \theta=45^{\circ}\)
(v) (c) : \(\operatorname{In} \Delta B D C, \angle B+\angle D+\angle C=180^{\circ}\)
\(\therefore \quad \angle B=180^{\circ}-60^{\circ}-90^{\circ}=30^{\circ}\)

(i) (d): Let h be the height of the pole.
In \(\Delta\)ABC, 

\( \frac{h}{15}=\sin 45^{\circ} \Rightarrow \frac{h}{15}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad h =\frac{15}{\sqrt{2}} \mathrm{~m}\)
(ii) (a): Let x be the required distance.
In \(\Delta\)ABC,

\(\frac{x}{15}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad x=\frac{15}{\sqrt{2}} \mathrm{~m}\)
(iii) (c) : Let h be the height of the pole.
In right triangle ABC,

\(\frac{h}{15}=\sin 30^{\circ}=\frac{1}{2}\)
\(\Rightarrow \quad h=\frac{15}{2}=7.5 \mathrm{~m}\)
(iv) (d): If 3 m rope is broken, then the length of the rope is 12 m.

\(\text { In } \Delta A B C, \frac{h}{12}=\sin 30^{\circ}=\frac{1}{2} \)
\(\Rightarrow \quad h=\frac{12}{2}=6 \mathrm{~m}\)
(v) (c)

(i) (c) : Since AE || FD
\(\therefore\) \(\angle\)EAD = \(\angle\)ADF = 30° [Alternate interior angles]

(ii) (b): Since, AE || BC
\(\therefore\) \(\angle\)EAC = \(\angle\)ACB = 60° [Alternate interior angles]
(iii) (a) : In \(\Delta\)ABC,
\(\begin{array}{l} \tan 60^{\circ}=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{50}{B C} \\ \Rightarrow \quad B C=\frac{50}{\sqrt{3}}=28.90 \mathrm{~m} \end{array}\)
(iv) (c): In \(\Delta\)ADF, \(\tan 30^{\circ}=\frac{A F}{F D}\)
\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{A B-B F}{F D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{50-C D}{\frac{50}{\sqrt{3}}}\)
\(\left[\because F D=B C=\frac{50}{\sqrt{3}}\right] \)
\(\Rightarrow \frac{50}{3}=50-C D \Rightarrow C D=50-\frac{50}{3}=\frac{100}{3}=33.33 \mathrm{~m}\)
(v) (d)