Given: AB the lighthouse 100 m above sea level and C is a ship sailing towards AB. 
⇒ ㄥEAC = 30\(\unicode{xb0} \)
After travelling from C to C' angle of depression changes from ㄥEAC = 30\(\unicode{xb0} \) to ㄥEAC' = 60\(\unicode{xb0} \)
To Find: CC'
Solution: AE||BC
[Line of sight and line of horizontal]
⇒ ㄥACC' = ㄥEAC = 30\(\unicode{xb0} \) [Alternate angles]
ㄥAC'B = ㄥEAC' = 60\(\unicode{xb0} \) [Alternate angles]
In right ΔABC', \(\frac { AB }{ BC } \) = tan30^o
\(\frac { 100 }{ BC } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow BC=100\sqrt { 3 } \)
CC' = BC -  BC'
=\(\left( 100\sqrt { 3 } -\frac { 100 }{ \sqrt { 3 } } \right) \)m
=\(\frac { 100\times 3-100 }{ \sqrt { 3 } } \)
=\(\frac { 200 }{ \sqrt { 3 } } m=\frac { 200\sqrt { 3 } }{ 3 } m\) = 115.466 m

Let CD = h m be the height of the tower. At point D of the tower, a man is standing and observes the car at an angle of depression of 30°. After six seconds, the angle of depression of the car is 60°.
i.e. \(\angle\)ODA = 30° and \(\angle\)ODB = 60°
\(\Rightarrow\) \(\angle\)DAC = \(\angle\)ODA = 30°  [alternate angles]
and \(\angle\)DBC = \(\angle\)ODB = 60°  [alternate angles]
Let AB = y m and BC = x m
In right angled \(\Delta\)BCD,

\(\begin{array}{rlrl} \tan 60^{\circ} & =\frac{P}{B}=\frac{C D}{B C} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \sqrt{3} & =\frac{h}{x} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad h & =\sqrt{3} x & {\left[\because \tan 60^{\circ}=\sqrt{3}\right]} \end{array}\)
\(\Rightarrow \quad h = \sqrt3 x\)....(i)
In right angled \(\Delta\)ACD,
\(\begin{array}{rlrl} \tan 30^{\circ} & =\frac{C D}{A C}=\frac{C D}{A B+B C} & & {[\because A C=A B+B C]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{h}{x+y} & & {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad x+y & =h \sqrt{3} & \end{array}\)
\(\Rightarrow \quad x + y=\sqrt3 x(\sqrt3)\) [from Eq. (i)]
\(\Rightarrow\) x + y = 3x  .....(ii)
It is given that a car moves from point A to B in six seconds. Let its speed be k km/s.
\(\therefore \quad \text { Time }=\frac{\text { Distance }}{\text { Speed }}\)
\(\Rightarrow \quad 6=\frac{y}{k} \Rightarrow y=6 k\)
On putting y = 6k in Eq. (ii), we get
x + 6k = 3x \(\Rightarrow\) 6k - 2x \(\Rightarrow\) x = 3k
\(\therefore \quad \text { Time }=\frac{\text { Distance }}{\text { Speed }}=\frac{x}{k}=\frac{3 k}{k}=3 \mathrm{~s}\)
Hence, the car moves from point B to point C in 3s.

Let AD = 7 m be the height of the building and BC = h m be the height of the cable tower. From the top of the building D, the angles of elevation and depression are \(\angle\)CDE = 60^o and \(\angle\)EDB = 45^o    
From the point D, draw a line DE || AB.
Then, \(\angle\)EDB = \(\angle\)ABD = 45^o  [alternate angles]

Also, let AB = DE = x m be the distance between building and tower.
In right angled \(\Delta\)BAD,
\(\begin{array}{rlrl} \tan 45^{\circ} & =\frac{P}{B}=\frac{A D}{A B} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad 1 & =\frac{7}{x} & {\left[\because \tan 45^{\circ}=1\right]} \end{array}\)
\(\Rightarrow\) x = 7 m   ....(i)
and in right angled \(\Delta\)CED,
\(\begin{aligned} & \tan 60^{\circ}=\frac{C E}{D E}=\frac{C B-B E}{A B}[\because C E=C B-B E] \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \sqrt{3}=\frac{b-7}{x} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right\} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad b-7=x \sqrt{3} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad h=x \sqrt{3}+7 \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad h=7 \sqrt{3}+7 \quad \text { [from Eq. (i)] } \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad h=7(\sqrt{3}+1) \mathrm{m} \\ \end{aligned}\)
Hence, the height of the tower is \(7(\sqrt{3}+1) \mathrm{m} .\)

Let BC be the building, AB be the transmission tower and D be the point on the ground from where the angles of elevations are to be measured.

\(\begin{array}{rlrl} \text { In } \triangle B C D, & \tan 45^{\circ} =\frac{B C}{C D} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow 1 =\frac{20}{C D} \Rightarrow C D=20 \mathrm{~m} \end{array}\)
\(\begin{array}{llrl} \text { In } \triangle A C D, & \tan 60^{\circ} =\frac{A C}{C D} \end{array}\)
\(\begin{array}{llrl} \Rightarrow \sqrt{3} =\frac{A B+B C}{C D} \\ \end{array}\)
\(\begin{array}{llrl} \Rightarrow \sqrt{3} =\frac{A B+20}{20} \\ \end{array}\)
\(\Rightarrow A B =20 \sqrt{3}-20=20(\sqrt{3}-1) \mathrm{m}\)
Thus, the height of the tower is \(20(\sqrt{3}-1) \mathrm{m}\).

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.

Let C be the top of the pole BC
Height of pole=12 m
The distance covered by the fireman to reach the top of the pole =AC
In right-angled triangle ABC, we have
\(\frac { BC }{ AC } \)=sin 30^o
⇒ \(\frac { 12 }{ AC } =\frac { 1 }{ 2 } \) ⇒ AC=24
Hence, distance covered by Mahesh is 24 metres.
(i) Humanity and act of bravery.

Here, the height of the pole be h  m and length of its shadow is  \(\sqrt { 3 } \)  h .
Let ፀ be the elevation of the sun. Consider right-angle ΔPQR
tanፀ=\(\frac { h }{ \sqrt { 3 } h } =\frac { 1 }{ \sqrt { 3 } } \)
= tan 30\(\unicode{xb0} \)
ፀ = 30\(\unicode{xb0} \)
Hence, the angle of elevation of the sun is 30\(\unicode{xb0} \).

87 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

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, \(\frac { AB }{ BC } \) = tan30^o
⇒ \(\frac { AB }{ 20\sqrt { 3 } } =\frac { 1 }{ \sqrt { 3 } } \) ⇒ AB = 20 m
Height of tower AD = AB + BD
= 20 + 1.7 = 21.7 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 \(16 \frac{2}{3} \mathrm{~m}\).

Let A be the centre of the balloon (spherical) whose radius is r.
Let AB be the height of the ballon i.e., h units, such that
ㄥDPC=ፀ, ㄥAPB-ф
Since ΔPDA and ΔPCA are congruent, therefore
ㄥAPC=ㄥAPD=ፀ/2
In rt. ㄥed ΔPCA,
\(\frac { AC }{ AP } =sin\frac { \theta }{ 2 } \)
⇒ AP=\(AC\frac { 1 }{ sin\frac { \theta }{ 2 } } =r.cosec\frac { \theta }{ 2 } \) or
Consider rt . ㄥed ΔPBA, we have
\(\frac { AB }{ AP } \)=sin ф
AB=AP.sinф
h=r.cosec\( \frac { \theta }{ 2 } \).sinф
h=r sinф.cosec\( \frac { \theta }{ 2 } \).