(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)

(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) (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) : 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) (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) (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) (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) (a):  \(\text { In } \triangle A C D\)
\(\tan 45^{\circ}=\frac{C D}{A C}=1 \)
\(\therefore \quad A C=C D=20 \mathrm{~m}\)

(ii) (b): Let, BD = h m be the height of the statue.
\(\text { In } \Delta A B C, \tan 60^{\circ}=\frac{B C}{A C} \Rightarrow \frac{B D+C D}{A C}=\sqrt{3} \)
\(\Rightarrow \frac{20+h}{20}=\sqrt{3}[\text { using }(\mathrm{i})] \Rightarrow h=20(\sqrt{3}-1) \mathrm{m}\)
(iii) (d): Since, in \(\triangle A C D, \angle D A C=45^{\circ}\)
\(\therefore \quad A C=C D(\operatorname{say} x) \)
\(\text { In } \Delta B A C, \tan 60^{\circ}=\frac{B C}{A C} \)
\(\Rightarrow \frac{1.6+x}{x}=\sqrt{3} \)
\(\Rightarrow 1.6=x(\sqrt{3}-1)\)

\(\Rightarrow \quad x=\frac{1.6}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=0.8(\sqrt{3}+1) \mathrm{m}\)
(iv) (c): \(\text { In } \triangle A B C\)
\(\tan 60^{\circ}=\frac{B C}{A C} \Rightarrow \frac{39}{A C}=\sqrt{3} \)
\(\Rightarrow A C=\frac{39}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=13 \sqrt{3} \mathrm{~m}\)

(v) (a): \(\text { In } \Delta A C D, \sin 45^{\circ}=\frac{C D}{A D}\)

\(\Rightarrow \quad \frac{35}{A D}=\frac{1}{\sqrt{2}} \)
\(\Rightarrow A D=35 \sqrt{2} \mathrm{~m}\)

(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) (c): \(\text { In } \Delta O P Q\), we have
\(\tan 60^{\circ}=\frac{P Q}{P O} \)
\(\Rightarrow \sqrt{3}=\frac{20}{P O} \)
\(\Rightarrow P O=\frac{20}{\sqrt{3}} \mathrm{~m}\)
(ii) (b): In \(\Delta\)ORS, we have
\(\tan 30^{\circ}=\frac{R S}{O R} \Rightarrow \frac{1}{\sqrt{3}}=\frac{20}{O R} \Rightarrow O R=20 \sqrt{3} \mathrm{~m}\)
(iii) (d): Clearly, width of the road = PR
\(\begin{array}{l} =P O+O R=\left(\frac{20}{\sqrt{3}}+20 \sqrt{3}\right) \mathrm{m} \\ =20\left(\frac{4}{\sqrt{3}}\right) \mathrm{m}=\frac{80}{\sqrt{3}} \mathrm{~m}=46.24 \mathrm{~m} \end{array}\)
(iv) (a): \(\text { In } \Delta O P Q \text { , if } \angle P O Q=45^{\circ} \text { , then }\)
\(\tan 45^{\circ}=\frac{P Q}{P O} \Rightarrow 1=\frac{20}{P O} \Rightarrow P O=20 \mathrm{~m}\)
(v) (b)

(i) (c): Let AB be the tree of height h m and let it broken at height of x m, as shown in figure. Clearly CD = AC = (h - x) m
Now, in \(\Delta\)CBD, we have
\( \tan 30^{\circ}=\frac{x}{30} \)
\(\Rightarrow \quad x=\frac{30}{\sqrt{3}} \mathrm{~m}\)

Thus, the height of remaining part of the tree is \(\frac{30}{\sqrt{3}} \mathrm{~m}\)
(ii) (c): In this case, BD = BC = x m
\(\therefore\) If \(\theta\) be the angle of inclination, then
\( \tan \theta=\frac{B C}{B D}=1 \)
\(\Rightarrow \tan \theta=\tan 45^{\circ} \)
\(\Rightarrow \theta=45^{\circ}\)
(iii) (b): The angle of elevation and depression are always acute angles.
(iv) (d): Clearly, BD = AB - AD
\(=(10 \sqrt{3}-2 \sqrt{3}) \mathrm{m}=8 \sqrt{3} \mathrm{~m}\)
Now, in \(\Delta\)BCD,we have
\(\sin 60^{\circ}=\frac{B D}{D C} \)
\(\Rightarrow \frac{\sqrt{3}}{2}=\frac{8 \sqrt{3}}{D C} \Rightarrow D C=16 \mathrm{~m}\)
(v) (a): Here, h = 6 m, \(\theta\) = 30°
\(\therefore\) DC = AC = (6 - x) m
Now, in \(\Delta\)BCD,we have
\(\sin 30^{\circ}=\frac{B C}{C D} \)
\(\Rightarrow \frac{1}{2}=\frac{x}{6-x} \)
\(\Rightarrow 6-x=2 x\)
\(\Rightarrow 3 x=6 \Rightarrow x=2\)

(i) (d): Clearly, \(\angle\)DAC = 60°
So,in \(\Delta\)ADC, we have
\(\tan 60^{\circ}=\frac{C D}{A D} \Rightarrow \sqrt{3}=\frac{6}{A D} \)
\(\Rightarrow A D=\frac{6}{\sqrt{3}} \mathrm{~m}\)
(ii) (b): Clearly, \(\angle\)DBC = 30°
So, in \(\Delta\)BDC,we have
\(\tan 30^{\circ}=\frac{C D}{B D} \)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{6}{B D} \)
\(\Rightarrow B D=6 \sqrt{3} \mathrm{~m}\)
(iii) (b): Width of the river = AB = AD + BD
\(\begin{array}{l} =\frac{6}{\sqrt{3}}+6 \sqrt{3} \\ =6\left(\frac{1}{\sqrt{3}}+\sqrt{3}\right)=6\left(\frac{4}{\sqrt{3}}\right)=\frac{24}{\sqrt{3}} \mathrm{~m}=13.87 \mathrm{~m} \end{array}\)
(iv) (a): The angle of elevation and angle of depression are always acute angles.
(v) (c): In \(\Delta\)BCD,if BD = 21m, then
\( \tan 30^{\circ}=\frac{C D}{B D} \)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{C D}{21} \Rightarrow C D=\frac{21 \sqrt{3}}{3}=7 \sqrt{3} \mathrm{~m}\)

(i) (b): In the right  \(\Delta\)ADQ, we have
\(\sin 30^{\circ}=\frac{D Q}{A D} \Rightarrow \frac{1}{2}=\frac{D Q}{24}\)
\(\Rightarrow \quad D Q=12 \mathrm{~m}\)
Thus, distance of paraglider from the ground is 12 m.
(ii) (a): We have PQ = BC = 6 m
Now, as DQ = 12 m
\(\therefore\) DP = DQ - PQ = 12 - 6 = 6 m
(iii) (c) : In right  \(\Delta\)BDP,we have
\(\sin 45^{\circ}=\frac{D P}{B D} \Rightarrow \frac{1}{\sqrt{2}}=\frac{6}{B D}\)
\(\Rightarrow \quad B D=6 \sqrt{2} \mathrm{~m}\)
Thus, the distance of paraglider from the girl is 6\(\sqrt{2}\) m.
(iv) (d): \(\angle\)AOP given in figure, is the angle of depression.
(v) (c): If A and B are two objects and the eye of an observer is at point 0, then line of sight will be both OA and OB.

(i) (c): Distance of first position of pigeon from the eyes of boy = AC

\(\text { In } \triangle A B C \text { , } \)
\(\sin 60^{\circ}=\frac{B C}{A C} \Rightarrow A C=\frac{C H-B H}{\sin 60^{\circ}}=\frac{54-4}{\sqrt{3} / 2}=\frac{100}{\sqrt{3}} \mathrm{~m}\)
(ii) (b): If the distance increases, then the angle of elevation decreases.
(iii) (b): Distance between boy and pole = AB
\(\text { Now, in } \triangle A B C \text { , }\)
\(\tan 60^{\circ}=\frac{B C}{A B} \Rightarrow \sqrt{3} A B=50 \Rightarrow A B=\frac{50}{\sqrt{3}} \mathrm{~m}\)
(iv) (c): \(\text { In } \Delta A E D, \tan 45^{\circ}=\frac{E D}{A D}\)
\(\Rightarrow A D=B C=50 \mathrm{~m} \quad A D \quad(\because E D=B C)\)
Now, distance between two positions of pigeon = EC = BD = AD-AB
\(=\left(50-\frac{50}{\sqrt{3}}\right) \mathrm{m}=\frac{50(1.73-1)}{1.73}=21.09 \mathrm{~m}\)
(v) (a): \(\text { Speed of pigeon }=\frac{\text { Distance covered }}{\text { Time taken }} \)
\(=\left(\frac{21.09}{8}\right) \mathrm{m} / \mathrm{sec}=2.63 \mathrm{~m} / \mathrm{sec}\)

Given, side of square top = 2 m
\(\therefore\) AB = HT = QR = CD = 2 m
Also, A C and BD are perpendicular to the ground.
So, AH = HQ = QC.  (By B.P.T. Theorem)
(i) (b):\(\text { In } \triangle A E C\)
\(\sin 60^{\circ}=\frac{A C}{A E} \Rightarrow \frac{\sqrt{3}}{2}=\frac{6}{A E} \Rightarrow A E=6.93 \mathrm{~m}\)
\(\therefore\) Length of each leg i.e., AE = BF = 6.93 m.
(ii) (c): \(\text { In } \Delta A G H, \tan 60^{\circ}=\frac{A H}{G H} \Rightarrow \sqrt{3}=\frac{2}{G H}\)
\(\Rightarrow G H=1.15 \mathrm{~m}\)
(iii) (a) :Length of second step = GH + HT + TU
= 1.15 + 2 + 1.15 = 4.3 m
(iv) (b):\(\text { In } \Delta A P Q\)
\(\tan 60^{\circ}=\frac{A Q}{P Q} \Rightarrow \sqrt{3}=\frac{4}{P Q} \Rightarrow P Q=\frac{4}{\sqrt{3}} \mathrm{~m}=2.31 \mathrm{~m}\)
(v) (c): Length of first step = PQ + QR + RS
=2.31 + 2 + 2.31 = 6.62 m