Let us assume that  the lighthouse AB be 00 m and C and D be the positions of the man when angle of elevation changes from 60\(\unicode{xb0} \) to 45\(\unicode{xb0} \)
The man has covered a distance CD in 2 minutes.
∴ Speed = Distance/Time = CD 2 minutes
Consider art. ΔABC, we have
tan60\(\unicode{xb0} \) =\(\frac { AB }{ BC } \)
⇒ \(\sqrt { 3 } =\frac { 100 }{ BC } \)   [cross-multiply]
⇒ BC = \(\frac { 100 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } \)
⇒ BC =\(\frac { 100\sqrt { 3 } }{ 3 } \) ....(i)
⇒ 1 = \(\frac { 100 }{ DB } \)
⇒ DB = 100 m
As CD = DB - BC
=\(\left( 100-\frac { 100\sqrt { 3 } }{ 3 } \right) \)m  [using (i)]
=\(100\left[ 1-\frac { \sqrt { 3 } }{ 3 } \right] \)m
∴ Speed = \(\frac { CD }{ 2 } =\frac { 100\left[ \frac { 3-\sqrt { 3 } }{ 2 } \right] }{ 2 } \)
⇒ Speed = \(50\left[ \frac { 3-\sqrt { 3 } }{ 3 } \right] \)m/minute
Hence the required speed of the boat is \(\frac { 50 }{ 3 } (3-\sqrt { 3 } )\) m/minutes.

Let AB = h m be height of tower and CB and DB are its shadows when sun's altitudes are 45^o and 60^o respectively. As ㄥACB = 45^o and ㄥADB = 60^o
CD = 20 m and BD = y m 
Consider a rt.ΔABD, we have

\(\frac{AB}{BD}\)=tan 60⁰
\(\Rightarrow \frac { h }{ y } =\sqrt { 3 } \)
\(\Rightarrow h=\sqrt { 3 } y\)....(i)
Consider a rt.ΔABC, we have
\(\frac{AB}{BC}\)=tan 45⁰
\(\frac { h }{ 20+y } =1\)
h=20+y
\(\Rightarrow h=20+\frac { h }{ \sqrt { 3 } } \)
\(h-\frac { h }{ \sqrt { 3 } } =20\)
\(h\left( \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } } \right) =20\)
\(h=\frac { 20\sqrt { 3 } }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } =\frac { 60+20\sqrt { 3 } }{ 3-1 } \)
\(=\frac { 2(30+10\sqrt { 3 } ) }{ 2 } =30+10\sqrt { 3 } \)
Hence, height of the tower=\(10(3+\sqrt { 3 } )m\)

Let height of ballon =h m In rt. ΔCDC,
\(\frac { CD }{ DC } \)=tan60^o
\(\frac { h-2 }{ DC } =\sqrt { 3 } \)
\(\frac { h-2 }{ \sqrt { 3 } } \)=DC
In rt. ΔCEF, \(\frac { CE }{ EF } \)=tan30^o
\(\frac { h-6 }{ EF } =\frac { 1 }{ \sqrt { 3 } } \)
\(\sqrt { 3 } \)(h-6)=EF
According to question
DC=EF
\(\frac { h-2 }{ \sqrt { 3 } } =\sqrt { 3 } (h-6)\)
h-2=3(h-6)
h-2=3h-18
18-2=3h-h
16=2h
8=h
Height of the ballon above the ground is 8 m.

Given: A tower PQ subtending angle α at the point A. Point B is b m vertically above A. From B angle of depression of Q is β.
To prove:
PQ=height of the tower =b tanα cotβ
Proof: Let AQ=x
ㄥEBQ=β
EB||QA
⇒ ㄥBQA=β      [Alternate angles]
In right angled ∆BAQ,
\(\frac { AB }{ AQ } =\frac { b }{ x } \)=tanβ
⇒ \(\frac { b }{ x } \)=tanβ
⇒ x=b cotβ
\(\frac { PQ }{ QA } =\frac { h }{ x } \)=tanα
⇒ h=x tanα
=b cotβ tanα=b tanα cotβ

Let aeroplane is at A and B, C are two consecutive kilometre stones such that
ㄥXAB=α and ㄥYAC=β
∴ ㄥABD=α and ㄥACD=β
Let BD=x km
AD is height of aeroplane
In ΔADB, \(\frac { AD }{ BD } \)=tanα
⇒ \(\frac { AD }{ x } =tan\alpha \quad \Rightarrow \quad x=\frac { AD }{ tan\alpha } \)       ........(i)
In ∆ADC, \(\frac { AD }{ DC } \)=tanβ
⇒ \(\frac { AD }{ 1-x } =tan\beta \)
⇒ \(\frac { AD }{ 1-\frac { AD }{ tan\alpha } } =tan\beta \)      [From (i)]
⇒ \(\frac { AD\quad tan\alpha }{ tan\alpha -AD } =tan\beta \)
⇒ AD tanα=tanα.tanβ-AD tanβ
⇒ AD tanα+AD tanβ=tanα.tanβ
⇒ AD=\(\frac { tan\alpha .tan\beta }{ tan\alpha +tan\beta } \) Hence proved.

Distance covered in 15 seconds=AB Speed=1200 km/hr
∴ AB=1200 x 15/3600 =5 km
Let height=x km
In rt. ΔBDE,
\(\frac { BD }{ ED } =tan45^{ 0 }\Rightarrow \frac { x }{ y } \)=1 ⇒ x=y
In rt.ΔACE,
\(\frac { AC }{ EC } =tan{ 30 }^{ 0 }\Rightarrow \frac { x }{ y+5 } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ \(\frac { x }{ x+5 } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ \(\sqrt { 3 } \)x=x+5 ⇒ (\(\sqrt { 3 } \)-1)x=5
∴ x=\(\frac { 5 }{ \sqrt { 3 } -1 } =\frac { 5(\sqrt { 3 } +1) }{ 2 } \)=6.83 km.

Let AB is the height of the mountain and C is the foot of the mountain
ㄥACB=45^o
ㄥBAC=45^o
After moving 1000 along CD at an angle of 300 with the horizontal the angle of inclination at D=600
⇒ ㄥADF=60^o
⇒ Draw DE⏊BC
In right ΔDEC, \(\frac { DE }{ DC } \)=sin 30^o
⇒ \(\frac { DE }{ 1000 } =\frac { 1 }{ 2 } \Rightarrow DE=\frac { 1 }{ 2 } \times 1000\)=500 m
DE=BF=500 m
Now ㄥACD=45^o-30^o=15^o
In right ΔAFD, ㄥADF=60^o
ㄥFAD=30^o
ㄥDAC=45^o-30^o=15^o
∴ In ΔCDA,
AD=CD
⇒ AD=1000 m
In right ΔADF, \(\frac { AF }{ AD } \)=sin 60^o
⇒ \(\frac { AF }{ 1000 } =\frac { \sqrt { 3 } }{ 2 } \Rightarrow AF=500\sqrt { 3 } \)m
Now AF=AF+BF
=(500\(\sqrt { 3 } \)+500)m=500(\(\sqrt { 3 } \)+1)m
=500(1.732+1)=500 x 2.732=1366 m.

Let height of the cloud C from lake be h m . A is position of the point 60 m above the lake. D is the reflection of the cloud in lake .
Let AE=x m, CF=h m, CE=(h-60)m
DE=(60+h)m.
In right angled triangle AEC
\(\frac { AE }{ EC } \)=cot 30⁰
AE=(h-60)\(\sqrt { 3 } \)           .........(i)
In right angled triangle AED,
\(\frac { AE }{ ED } =cot60^{ 0 }\quad \Rightarrow \quad AE=\frac { h+60 }{ \sqrt { 3 } } \)            .............(ii)
From (i) and (ii), we get
\((h-60)\sqrt { 3 } =\frac { h+60 }{ \sqrt { 3 } } \)
⇒ 3h-180=h+60 ⇒ 2h=240 ⇒ h=120 m.
∴ Height of the cloud above the lake is 120 m.

Speed of jet fighter=720 km/h=200 m/s
∴ Distance covered in 15 seconds
=200 x 15=3000 m
PB=3000 m
PB=QC=3000 m
In right ΔPQA,
\(\frac { PQ }{ AQ } \)=tan60^o
⇒ \(\frac { x }{ y } =\sqrt { 3 } \Rightarrow x=\sqrt { 3 } y\)    ....(i)
In right ∆BCA,
\(\frac { BC }{ AC } =tan30^{ 0 }\Rightarrow \frac { x }{ y+3000 } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ x=\(\frac { y+3000 }{ \sqrt { 3 } } \) .......(ii)
From (i) and (ii),  \(\sqrt { 3 } y=\frac { y+3000 }{ \sqrt { 3 } } \qquad \)
⇒ 3y=y+3000 ⇒ 2y=3000 ⇒ y=1500 m
x=1500 x \(\sqrt { 3 } \)m
x=1500\(\sqrt { 3 } \)=1500 x 1.732 m=2598 m

Draw YL parallel to XP intersecting PQ at L
⇒ PXYL is a rectangle
⇒ PL=XY=40 m. LY=PX=x m opposite sides of a rectangle
Let QL, be h m and PX be x m
In ΔQLY, \(\frac { QL }{ LY } \)=tan45⁰
\(\frac { h }{ x } \)-tan45⁰ ⇒ h=x m
In ΔQPX, \(\frac { QP }{ PX } \)=tan 45⁰
⇒ \(\frac { h }{ x } \)=tan 450 ⇒ h=x m
In ΔQPX,       \(\frac { QP }{ PX } \)=tan60⁰
⇒ \(\frac { h+40 }{ x } =\sqrt { 3 } \) ⇒ h+40=\(\sqrt { 3 } \)x
Using (i), h+40=\(\sqrt { 3 } \)h
⇒ h=\(\frac { 40 }{ (\sqrt { 3 } -1) } =\frac { 40(\sqrt { 3 } +1) }{ (\sqrt { 3 } -1)(\sqrt { 3 } +1) } \)
=\(\frac { 10(1.732+1) }{ (\sqrt { 3 } )^{ 2 }-1^{ 2 } } =\frac { 40\times 2.732 }{ 3-1 } =\frac { 40\times 2.732 }{ 2 } \)=54.64
Height of the tower PQ=h+40=40+54.64=94.64 m
XQ²=PQ²+XP²      [By Pythagoras theorem]
=(94.64)²+(54.64)²
XQ=\(\sqrt { ({ 94.64 })^{ 2 }+({ 54.64) }^{ 2 } } =\sqrt { 8956.7296+2985.5296 } \)
=\(\sqrt { 11942.2592 } \)=109.2806=109.28 m=109.3 m