Let C be the position of the kite and AC be the length of the string which makes an angle of 60° on the ground. The height of the kite from the ground is BC =60 m.
In right angled \(\Delta\)ABC,

\(\begin{aligned} \sin 60^{\circ} & =\frac{P}{H}=\frac{B C}{A C} \\ \end{aligned}\)
\(​​\begin{aligned} \Rightarrow \quad \frac{\sqrt{3}}{2} & =\frac{60}{A C} \end{aligned}\)     \(\left[\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]\)
\(\begin{aligned} \therefore \quad A C & =\frac{60 \times 2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \end{aligned}\)     [rationalising]
\(\begin{aligned} =\frac{120 \sqrt{3}}{3}=40 \sqrt{3} \mathrm{~m} \end{aligned}\)
Hence, the length of the string is 40\(\sqrt3\)m.

Let BC be the height of the tower which is standing on the ground. Let A be a point on the ground which is 30 m away from the foot of tower.

Then, AB = 30 m and \(\angle\)BAC = 30°.
In right angled \(\Delta\)ABC,
\(\begin{array}{rlrl} \tan 30^{\circ} =\frac{B C}{A B}&&{\left[\because \tan \theta=\frac{P}{B}\right]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{B C}{30} & {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad B C & =\frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=10 \sqrt{3} \mathrm{~m} \end{array}\)   [rationalising]
Hence, the height of the tower is 10\(\sqrt3\) m.

Case I slide for the children below age of 5yr.
Let BC = 1.5 m be the height of the slide and let slide AC is inclined at an angle of 30° to the ground.

In right angled \(\Delta\)ABC,
\(\sin 30^{\circ}=\frac{B C}{A C} \quad\left[\because \sin \theta=\frac{P}{H}\right]\)
\(\Rightarrow \quad \frac{1}{2}=\frac{1.5}{A C} \quad\left[\because \sin 30^{\circ}=\frac{1}{2}\right]\)
\(\Rightarrow\) AC = 3 m
Case II Slide for the elder children.
Let RQ = 3 m be the height of the slide and let slide PR is inclined at an angle of 60° to the ground.

In right angled \(\Delta\)PQR,
\(\begin{aligned} \sin 60^{\circ} & =\frac{R Q}{P R} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{\sqrt{3}}{2} & =\frac{3}{P R} \\ \end{aligned}\)      \(\left[\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]\)
\(\begin{aligned} \Rightarrow \quad P R & =\frac{3 \times 2}{\sqrt{3}}=\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \end{aligned}\)
\(\begin{aligned} =2 \sqrt{3} \mathrm{~m} \end{aligned}\) [by rationalising]
Hence, length of slides in each case are 3m and 2\(\sqrt3\)m.

Let AB is tower and BC=x

∵ BE=CD ⇒BE=10 m
Also BC=DE=x m. Take AE=y m
In right ΔAED, \(\frac { AE }{ DE } \)=tan 30^o
⇒ \(\frac { y }{ x } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \sqrt { 3 } y\)     ....(i)
In right ΔABC, \(\frac { AB }{ BC } \)=tan 60^o
⇒ \(\frac { y+10 }{ 2 } =\sqrt { 3 } \)
y+10=\(\sqrt { 3 } \)x
y+10=\(\sqrt { 3 } \)(\(\sqrt { 3 } \)y)
⇒ y+10=3y
⇒ y=5m
Height of tower=AE+BE=5+10
=15 m

From the point of observation (O),

plane is at A, Al = 3000 \(\sqrt{3}\)m and ㄥAOL = 60^o
After 30 seconds, plane is at B, therefore,
BM=3000\(\sqrt{3}\) m and ㄥBOM=30^o
Distance AB is covered in 30 seconds. In right-angled triangle OLA.
\(\frac { OL }{ AL } \)=cot 60^o
⇒ OL=\(3000\sqrt { 3 } \times \frac { 1 }{ \sqrt { 3 } } \)=3000 m   ....(i)
In right-angled triangle OMB,
\(\frac { OM }{ BM } \)=cot 30^o
OM=\(\\ 3000\sqrt { 3 } \times \sqrt { 3 } \)=9000 m     ......(ii)
∴ AB=lM=OM-OL
=(9000-3000)m=6000 m      [from (i) and (ii)]
Now in 30 s, distance covered=6000 m
∴ In 1 hour (3600 s), distance covered
=\(\frac { 6000 }{ 30 } \times \frac { 3600 }{ 1000 } \)km=720 km
∴ Speed of the aeroplane
=720 km/h

A is the position of the man, OA = 12m, BC is cliff. 
Let height of the cliff
BC = h m and CE = (h - 12)m
Let AE = OB = x m
In right angled triangle AEB,
\(\frac { AE }{ BE } \) = cot30^o ⇒ AE = 12 x \(\sqrt { 3 } \)
=12 x 1.732 m = 20.78 m
∴ Distance of ship from cliff = 20.78 m
In right angled triangle AEC,
\(\frac { CE }{ AE } =tan{ 60 }^{ 0 }\Rightarrow \frac { h-12 }{ 12\sqrt { 3 } } \) = \(\sqrt { 3 } \)
h - 12 = 36 ⇒ h = 48 m
∴. Height of the cliff = 48 m

Let AB is pole and BC is its shadow
∴ AB = 6m, BC = 2\(\\ \\ \\ \\ \sqrt { 3 } \) m
In right ΔABC, \(\frac { AB }{ BC } \)=tanፀ
⇒ tanፀ = \(\frac { 6 }{ 2\sqrt { 3 } } \)
⇒ tanፀ = \(\sqrt { 3 } \)
⇒ ፀ = 60^o

In right ΔABC
\(\frac { BC }{ AC } \) = cos 60\(\unicode{xb0} \)
⇒ \(\frac { 2.5 }{ AC } =\frac { 1 }{ 2 } \)
⇒ AC = 2.5 x 2 = 5 m

Let BC = 50 m be the height of the tower and AD = h m be the height of the building. Angle of elevation of the top of the building from the foot of the tower is \(\angle\)DBA = 30° and angle of elevation of the top of the tower from the foot of the building \(\angle\)CAB = 60°. Also, let AB = x m be the distance between foots of the tower and the building.
In right angled \(\Delta\)BAD, tan 30°=\(\frac{P}{B}=\frac{A D}{A B}\)
\(\begin{array}{lll} \Rightarrow & \frac{1}{\sqrt{3}}=\frac{h}{x} & {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]} \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & b=\frac{x}{\sqrt{3}} \end{array}\)   ...(i)
and in right angled \(\Delta\)CBA, tan 60° = \(\frac{B C}{A B}\)
\(\begin{array}{lll} \Rightarrow & \sqrt{3}=\frac{50}{x} & {\left[\because \tan 60^{\circ}=\sqrt{3}\right]} \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & x=\frac{50}{\sqrt{3}} \mathrm{~m} \end{array}\)
On putting \(x=\frac{50}{\sqrt{3}}\) in Eq. (i), we get
\(h=\frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}}=\frac{50}{3}=16 \frac{2}{3} \mathrm{~m}\)
Hence, the height of the building is \(16 \frac{2}{3} \mathrm{~m}\).

Hint Let a man is standing on the deck of a ship at point A such that \(A B=10 \mathrm{~m}\) and let CD be the hill.
Then, \(\angle E A D=60^{\circ}\)
and \(\angle C A E=\angle B C A=30^{\circ}\) [alternate angles]
Let \(B C=x \mathrm{~m}=A E\) and \(D E=h \mathrm{~m}\)
In right angled \(\triangle A E D\),
\(\tan 60^{\circ}=\frac{P}{B}=\frac{D E}{E A}=\frac{h}{x}\)
In right angled \(\triangle A B C\),
\(\tan 30^{\circ}  =\frac{A B}{B C}\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}  =\frac{10}{x}  {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]}\)
Ans. Distance of the hill from the ship is \(10 \sqrt{3} \mathrm{~m}\) and height of the hill is 40 m .

Let A be the window and CE be the opposite house
CD=AB=60 m [Opposite side of a rectangle] ........(i)
In rt. ΔABC, tan45^o=\(\frac { 60 }{ BC } \)
⇒ 1=\(\frac { 60 }{ BC } \)
⇒ BC=60 m        .........(ii)
AD=BC [Opposite sides of a rectangle]
∴ AD=60 m [From (ii)]        ......(iii)
In rt. ΔADE, tan600=\(\frac { DE }{ AD } \)
⇒ \(\sqrt { 3 } =\frac { DE }{ 60 } \)    [From (iii)]
⇒ DE=60\(\sqrt { 3 } \) m
∴ Height of the opposite house
CE=CD+DE=60+60\(\sqrt { 3 } \)
=60(1+\(\sqrt { 3 } \))m.

Let AB is statue, BC is pedestal and BC=x m, CD=y m.
In right ΔBCD, ΔBCD, \(\frac { BC }{ CD } \)=tan 45^o
⇒, \(\frac { x }{ y } \)=1 ⇒ x=y     .....(i)
In right ΔACD, \(\frac { AC }{ CD } \)=tan 600
⇒ \(\frac { x+1.46 }{ y } =\sqrt { 3 } \)
⇒ \(\frac { x+1.46 }{ x } \)=1.73    [Using (i)]
⇒ x+1.46=1.73x ⇒ 0.73x=1.46 ⇒ x=2

Let CD be a tower of height x m and BC is a pole and ㄥBAD=600 and ㄥPCA=450 ㄥPCA=ㄥCAD=45⁰
In right ΔCDA, \(\frac { CD }{ DA } \)=tan 45^o
⇒ \(\frac { x }{ DA } \)=1 ⇒ DA= x m
In right ΔBDA, \(\frac { BD }{ DA } \)=tan 60^o
⇒ \(\frac { 5+x }{ x } =\sqrt { 3 } \)
⇒ 5+x=\(\sqrt { 3 } \)x ⇒ 5=(\(\sqrt { 3 } \)-1)x
⇒ x=\(\frac { 5 }{ \sqrt { 3 } -1 } m=\frac { 5(\sqrt { 3 } +1) }{ 2 } m=\frac { 5(1.73+1) }{ 2 } m=\frac { 13.65 }{ 2 } \)m=6.82 m

Let 'h' m be height of the tower PQ
AB=20 m. Let BQ=x m
In rt. ΔPQB
\(\frac { PQ }{ BQ } \)=tan 60^o  ⇒ \(\frac { h }{ x } =\sqrt { 3 } \Rightarrow \sqrt { 3 } x\) .........(i)
In rt. ΔPQA,
\(\frac { PQ }{ AQ } \)=tan 30^o   ⇒ \(\frac { h }{ 20+x } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \sqrt { 3 } \)h=20+x     ....(ii)
From (i) and (ii)
\(\sqrt { 3 } .\sqrt { 3 } \)x=20+x ⇒ x=10
∴ Distance of tower fro A is 20 m+10 m=30 m
Put x=10 in (i), we get
h=\(\sqrt { 3 } \) x 10=1.732 x 10=1.732 m

Let a man is standing on the deck of a ship at point A such that AB = 10 m and let CD be the hill.
Then, ∠EAD = 60° and ∠CAE = ∠BCA = 30° [alternate angles]
Let BC = x m = AE and DE = h m

In right angled ΔAED, 
\(\tan 60^{\circ}=\frac{P}{B}=\frac{D E}{E A}=\frac{h}{x}\)
In right angled ΔABC,
\(\tan 30^{\circ}=\frac{A B}{B C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{10}{x}\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]\)
distance of the hill from the ship is 10\(\sqrt3\) m and height of the hill is 40 m.