Let AD = h km be the height of aeroplane above the road and B, C are positions of two stones on the road. so ∠ABD = 60^o, ∠ACB = 30^o, BC = 1 km

Let BD = x km, then CD = (1 —x) km 
Consider a rt. ΔABD, we have
\(\frac { AD }{ BD } ={ tan\quad 60 }^{ o }\)
\(\Rightarrow \frac { h }{ x } =\sqrt { 3 } \)
\(\Rightarrow h=\sqrt { 3 } x\)    ...(i)
Now, consider a rt. ΔACD, we have
\(\frac { AD }{ CD } ={ tan\quad 30 }^{ o }\)
\(\Rightarrow \frac { h }{ 1-x } =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \sqrt { 3 } h=1-x\)
\(\Rightarrow \sqrt { 3 } \times \sqrt { 3 } x=1-x\)     [using (i)]
⇒ 3x=1-x
⇒ 4x=1
\(\Rightarrow x=\frac { 1 }{ 4 } km\)
Putting the value of x in (i), we have
\(h=\sqrt { 3 } \times \frac { 1 }{ 4 } =\frac { \sqrt { 3 } }{ 4 } km\)
∴ Height of aeroplane =\(\frac { \sqrt { 3 } }{ 4 } km\)

Let AB = h m be height of tower and CB and DB are its shadow when sun's altitudes are 30^o and 45^o respectively. As ㄥACB= 30^o and ㄥABC = 45^o

CD=2x m and BD=y m
consider rt.ΔABD, we have
\(\frac{AB}{BD}\)=tan 45^o
\(\Rightarrow\)\(\frac {h}{y}\)=1
\(\Rightarrow\)h=y ....(i)
consider rt.ΔABC, we have
\(\frac{AB}{BC}\)=tan 30^o
\(\frac { h }{ 2x+y } =\frac { 1 }{ \sqrt { 3 } } \)
\(\sqrt { 3 } h=2x+h\because From\ (i)\quad h=y]\)
\(h(\sqrt { 3 } -1)=2x\)
\(h=\frac { 2x }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } \)
\(h=\frac { 2x(\sqrt { 3 } +1) }{ 2 } =x(\sqrt { 3 } +1)\)
\(\therefore\) Height of the tower=\((\sqrt { 3 } +1)xm\)

Let AB=h m be height of tower and BC=30 m be length of its shadow when sun's elevation is 300.Let BD=x m be length of shadow when sun's elevation is 60^o
As ㄥACB=30^o and ㄥADB=60^o
Consider rt . ΔABC, we have
\(\frac { AB }{ BC } \)=tan300
⇒ \(\frac { h }{ 30 } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow h=\frac { 30 }{ \sqrt { 3 } } \)
⇒ AB=\(\frac { 30 }{ \sqrt { 3 } } \)m
Consider rt . ΔABD, we have
\(\frac { AB }{ BD } \)=tan600 ⇒ \(\frac { h }{ BD } =\sqrt { 3 } \)
⇒ \(\frac { \frac { 30 }{ \sqrt { 3 } } }{ BD } =\frac { \sqrt { 3 } }{ 1 } \)    [∵ h=\(\frac { 30 }{ \sqrt { 3 } } \)]
∴ Length of shadow=10 m

Let us assume that PQ is the building of height 7 m and RS is the cable tower of height h m , such that ㄥRPT=60^o
∴ ㄥTPS=45^o=ㄥQSP
∴ PQ=7m=TS
∴ RT=RS-TS=(h-7)m
Let QS=x m
Consider a rt . ΔPQS. we have
tan45^o=\(\frac { PQ }{ QS } \)
⇒ 1=\(\frac { 7 }{ x } \)
⇒ x=7        ...........(i)
Now, PT=QS=x m=7 m
Again, consider art. ΔPTR, we have
tan60⁰=\(\frac { RT }{ PT } \)
⇒ \(\sqrt { 3 } =\frac { h-7 }{ 7 } \)   [cross-multiply]
⇒ h-7=7\(\sqrt { 3 } \)
⇒ h=7+7\(\sqrt { 3 } \)
⇒ h=7(\(\sqrt { 3 } \)+1)m
Hence, the height of the cable tower is 7(\(\sqrt { 3 } \) +1)m.

Let A and B are the objects p m apart, DC is the tower of height h. Let BC=x m.
In ∆ACD, \(\frac { CD }{ AC } \)=tanβ
⇒ \(\frac { h }{ p+x } =tan\beta \)       ...........(i)
In ΔBCD, \(\frac { CD }{ BC } \)=tan α
⇒ \(\frac { h }{ x } =tan\alpha \quad \Rightarrow \quad x=\frac { h }{ tan\alpha } \)
Substituting in (i), we get
\(\frac { h }{ p+\frac { h }{ tan\alpha } } =tan\beta \)
⇒ h=\(p\quad tan\beta +\frac { h\quad tan\beta }{ tan\alpha } \)
⇒ h(tanα=p tanα tanβ+h tanβ
⇒ h(tanα-tanβ)=p tanα tanβ
h=\(\frac { p\quad tan\alpha \quad tan\beta }{ tan\alpha -tan\beta } \)
When p=50 m, ∝=60⁰, β=30⁰
h=\(\frac { 50\times tan60^{ 0 }.tan{ 30 }^{ 0 } }{ tan{ 60 }^{ 0 }-tan30^{ 0 } } \)
=\(\frac { 50\times \sqrt { 3 } \times \frac { 1 }{ \sqrt { 3 } } }{ \sqrt { 3 } -\frac { 1 }{ \sqrt { 3 } } } =\frac { 50\sqrt { 3 } }{ 2 } =25\sqrt { 3 } \)m

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.

Let AB is the leaning tower, BC=a and BD=bLet AE is the perpendicular distance of A from the horizontal ground.
In right ∆AED.
⇒ \(\frac { DE }{ AE } \)=cotβ
⇒ \(\frac { h+x }{ h } \)=cotβ
⇒ h cotβ=b+x     .......(i)
⇒ h=\(\frac { h+x }{ cos\beta } \)           .......(ii)
In right ∆AEC,
\(\frac { CE }{ AE } \)=cotα
⇒ a+x=h cotα ........(iii)
and \(\frac { a+x }{ h } \)=h          ..........(iv)
From (ii) and (iv)
\(\frac { b+x }{ cos\beta } =\frac { a+x }{ cot\alpha } \)
⇒ b cotα+x cotα=a cotβ-x cotβ
⇒ x(cotα-cotβ)=a cotβ-b cotα
⇒ x=\(\frac { a\quad cot\beta \quad -\quad b\quad cot\alpha }{ cot\alpha \quad -\quad cot\beta } \)
In right ∆AEB,
cotፀ=\(\frac { BE }{ AE } \)
⇒ cotፀ=x/h
⇒ \(cos\theta =\frac { \frac { a\quad cot\beta -b\quad cot\alpha }{ cot\alpha -cot\beta } }{ h } \)
⇒ \(\frac { a\quad cot\beta -b\quad cot\alpha }{ h\quad cot\alpha -h\quad cot\beta } \)
⇒ cotθ=\(\frac { a\quad cot\beta -b\quad cot\alpha \qquad }{ a-b } \)
=\(\frac { b\quad cot\alpha -a\quad cot\beta }{ b-a } \).

Let AB is path
In rt. ΔDAC, \(\frac { DC }{ AD } \)=cosec 60⁰
⇒ \(\frac { DC }{ 90 } =\frac { 2 }{ \sqrt { 3 } } \)
DC=\(\frac { 2 }{ \sqrt { 3 } } \times 90\quad m\quad =\frac { 180 }{ \sqrt { 3 } } \)m
Now DC=CE     [length of ladder]
∴ CE=\(\frac { 180 }{ \sqrt { 3 } } \)m
In rt. ΔEBC, \(\frac { BE }{ CE } \)=sin45⁰
⇒ BE=\(\frac { 1 }{ \sqrt { 2 } } \times \frac { 180 }{ \sqrt { 3 } } \)m
⇒ BE=73.47 m.

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 bird is at A and after 2 seconds it reaches at E.
∴ Distance covered=AE
In right ΔABC, \(\frac { BC }{ AB } \)=cot 45^o
\(\frac { BC }{ 80 } \)=1
⇒ BC=80 m
In right ΔEDC, \(\frac { DC }{ DE } \)=cot 30^o
⇒ DC=80 x \(\sqrt { 3 } \)        [∵ DE=AB]
Now, BD=CD-BC
=\(80\sqrt { 3 } -80=80\sqrt { 3 } -1)\)=80 x 0.732=58.56 m
Now, BD=AE=58.56 m
∴ Speed of bird=\(\frac { 58.56 }{ 2 } \)=29.28 m/sec.