(i) (c): \(\text { In } \Delta A B C, \frac{A B}{B C}=\tan 60^{\circ}\)
\(\Rightarrow \quad A B=25 \times \sqrt{3}\)
\(\therefore\) Height of building is 25\(\sqrt{3}\) m .
(ii) (b): \(\text { In } \Delta A B D, \frac{A B}{B D}=\tan 30^{\circ}\)
\(\Rightarrow \frac{25 \sqrt{3}}{B D}=\frac{1}{\sqrt{3}} \Rightarrow B D=75 \mathrm{~m}\)
\(\therefore\) Distance between two positions of car = (75 - 25) m = 50m.
(iii) (d): Time taken to cover 50 m distance = 6 sec.
\(\therefore\) Time taken to cover 25 m distance = 3 sec.
\(\therefore\) Total time taken by car = 6 sec + 3 sec = 9 sec
(iv) (c): \(\text { In } \Delta A B C, \frac{B C}{A C}=\cos 60^{\circ}\)
\(\Rightarrow \quad \frac{25}{A C}=\frac{1}{2} \)
\(\Rightarrow A C=50 \mathrm{~m}\)
(v) (a)

(i) (c) : Let AC be the length of the ladder.

\(\text { In } \Delta A B C, \frac{B C}{A C}=\sin 30^{\circ} \)
\(\Rightarrow \frac{6}{A C}=\frac{1}{2} \Rightarrow A C=12 \mathrm{~m}\)
(ii) (b): \(\text { In } \Delta A B C, \frac{A B}{A C}=\cos 30^{\circ}\)
\(\Rightarrow \frac{5}{A C}=\frac{\sqrt{3}}{2} \Rightarrow A C=\frac{10}{\sqrt{3}} \mathrm{~m}\)

(iii) (a) : Let BC be the height of window from ground.

\(\text { In } \Delta A B C, \frac{B C}{A B}=\tan 60^{\circ} \)
\(\Rightarrow \frac{B C}{2.5}=\sqrt{3} \)
\(\Rightarrow B C=2.5 \times 1.73=4.325 \mathrm{~m}\)
(iv) (d): Let AB be the horizontal distance between the foot of ladder and wall.

\(\text { In } \Delta A B C, \frac{B C}{A B}=\tan 45^{\circ} \)
\(\Rightarrow \quad \frac{8}{A B}=1 \Rightarrow A B=8 \mathrm{~m}\)
(v) (b): Let the required distance be x.
\(\text { In } \Delta A B C,(9)^{2}=x^{2}+(6)^{2}\)
[By Pythagoras theorem]

\(\Rightarrow 81-36=x^{2} \Rightarrow 45=x^{2} \)
\(\Rightarrow \quad x=3 \sqrt{5} \mathrm{~m}\)

(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) (b): The person who makes small angle of elevation is more closer to the balloon.
\(\therefore\) Radlra is more closer to the balloon.
(ii) (b): \(\text { In } \Delta E F D, \tan 30^{\circ}=\frac{E D}{D F}\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{D F} \)
\(\Rightarrow \quad D F=h \sqrt{3} \mathrm{~m}\)
(iii) (a): In \(\Delta\)GCE,
\(\begin{array}{l} \tan 60^{\circ}=\frac{E C}{G C}=\frac{h+4}{D F} \\ \Rightarrow \quad \sqrt{3}=\frac{h+4}{\sqrt{3} h} \Rightarrow 3 h=h+4 \Rightarrow h=2 \end{array}\)
(iv) (c): Height of the balloon from the ground = BE = BC + CD + DE = 2 + 4 + 2 = 8 m
(v) (b)

(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)