Let AB=60 m is height of building and CD is lamp post
(i) In rt ΔABD,  \(\frac { AB }{ BD } \)=tan60⁰
⇒ \(\frac { 60 }{ BD } =\sqrt { 3 } \Rightarrow \frac { 60 }{ \sqrt { 3 } } \)=BD
⇒ BD=\(\frac { 60\times \sqrt { 3 } }{ 3 } =20\sqrt { 3 } \)m
⇒ BD=20 x 1.732=34.64 m
(ii) In rt. ΔAEC, \(\frac { AE }{ EC } \)=tan30⁰
⇒ \(\frac { AE }{ 20\sqrt { 3 } } =\frac { 1 }{ \sqrt { 3 } } \)            [∵ EC=BD]
⇒ AE=20 m
and EB=AB-AE=60-20=40 m
Also EB=CD
⇒ CD=40 m
∴ Height of lamp post =40 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

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 = h m be the height of the pedestal and CD = 1.6m be the length of the statue, which is standing on the pedestal.
Again, let point A be a fixed point on the ground such that the angles of elevation of the top of the statue and bottom of the statue (i.e. top of the pedestal) are
\(\angle D A B=60^{\circ} \text { and } \angle C A B=45^{\circ}\)
Also, let AB = x m.
In right angled \(\Delta\)ABD, \(\tan 60^{\circ}=\frac{P}{B}=\frac{B D}{A B}\)
\(\begin{array}{lll} \Rightarrow & \sqrt{3}=\frac{B C+C D}{x} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right] \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & \sqrt{3}=\frac{h+1.6}{x} & \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & h=\sqrt{3} x-1.6 \end{array}\) ...(i)
In right angled \(\Delta\)CBA, tan 45° \(=\frac{B C}{A B}\)
\(\begin{array}{ll} \Rightarrow & 1=\frac{h}{x} \quad\left[\because \tan 45^{\circ}=1\right] \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & x=h \end{array}\)
On putting x = h in Eq. (i), we get
\(h=\sqrt{3} h-1.6 \Rightarrow h(\sqrt{3}-1)=1.6\)
\(\Rightarrow \quad h=\frac{1.6}{(\sqrt{3}-1)} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)  [rationalising] 
\(\begin{aligned} & =\frac{1.6(\sqrt{3}+1)}{(\sqrt{3})^2-(1)^2}\left[\because(a+b)(a-b)=a^2-b^2\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1.6}{2}(\sqrt{3}+1)=0.8(\sqrt{3}+1) \mathrm{m} \end{aligned}\)
Hence, the height of the pedestal is \(0.8(\sqrt{3}+1) \mathrm{m}\).

Let AB = 30 m be the height of the building and DC = 1.5 m be the height of the boy. The point D be the boy's eye.
Draw the line DF || CA.
Then, CD = AF = 1.5 m
The angle of elevation is \(\angle\)BDF = 30°.
Let he walked DE = xm towards the building. Then, the angle of elevation is \(\angle\)BEF = 60°.
Let EF = y m
Now, BF = AB - AF = 30 - 1.5 = 28.5 m.

In right angled \(\Delta\)BFD,
\(\begin{aligned} & \tan 30^{\circ}=\frac{P}{B}=\frac{B F}{D F}=\frac{B F}{D E+E F} \quad[\because D F=D E+E F] \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{28.5}{x+y} \quad\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right] \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad x+y=28.5 \sqrt{3} \mathrm{~m} \\ \end{aligned}\)    ...(i)
and in right angled \(\Delta\)BFE,
\(\begin{aligned} \tan 60^{\circ} & =\frac{B F}{E F} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \sqrt{3} & =\frac{28.5}{y} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right] \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y & =\frac{28.5}{\sqrt{3}} \mathrm{~m} \end{aligned}\)
On putting \(y=\frac{28.5}{\sqrt{3}} \text { in Eq. (i), we get } x+\frac{28.5}{\sqrt{3}}=28.5 \sqrt{3}\)
\(\begin{aligned} \Rightarrow \quad x & =28.5\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=28.5\left(\frac{3-1}{\sqrt{3}}\right) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{28.5 \times 2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \end{aligned}\)     [rationalising]
\(\begin{aligned} =\frac{57 \sqrt{3}}{3}=19 \sqrt{3} \mathrm{~m} \end{aligned}\)
Hence, the distance he walked towards the building, is 19\(\sqrt3\)m.

In rt. ΔABC, tan60^o = \(\frac { AB }{ BC } =\frac { h }{ x } \)
⇒ \(\sqrt { 3 } =\frac { h }{ x } \Rightarrow h=\sqrt { 3 } x\)
In rt. ∆ABD, tanፀ = \(\frac { AB }{ BD } \)
⇒  tanፀ = \(\frac { h }{ 3x } \)
⇒  tanፀ = \(\frac { \sqrt { 3 } x }{ 3x } =\frac { 1 }{ \sqrt { 3 } } \)⇒ ፀ = 30^o​​​

Given: AB be the tower and AC the flag pole on top of the tower.
ㄥCEB=60^o, ㄥAEB=30^o
To find Height of the tower and the height of the flag pole. Let height of the tower and flag pole be h m and h'm respectively.
Solution: In right ∆ABE
=\(\frac { AB }{ BE } =tan{ 30 }^{ 0 }\Rightarrow \frac { h }{ 9 } =\frac { 1 }{ \sqrt { 3 } } \)
h=\(\frac { 9 }{ \sqrt { 3 } } m=\frac { 9 }{ \sqrt { 3 } } =\frac { 9\sqrt { 3 } }{ 3 } m=3\sqrt { 3 } =5.196\)m  .....(i)
In right ΔCBE,
\(\frac { CB }{ BE } =tan600\Rightarrow \frac { h+{ h }^{ ' } }{ BE } =tan60^{ 0 }\)
\(\frac { h+h' }{ BE } =\sqrt { 3 } \)
⇒ h+h'=\(9\sqrt { 3 } \)               .......(ii)
\(h'=\frac { 27-9 }{ \sqrt { 3 } } =\frac { 18\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } =\frac { 18\sqrt { 3 } }{ 3 } =6\sqrt { 3 } \)m
=6 x 1.732 m=10.392 m
Height of flag pole mounted on tower=10.392 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

100m

36.6m

\({40\over\sqrt{3}}m\)

30⁰

40m

60^o

\(75\sqrt { 3 } m\)

Let PQ is tower and QR is its shadow. 

In right \(\Delta\) PQR, 
\(\frac { PQ }{ QR } =\tan { { 60 }^{ o } } \)
\(\Rightarrow\)  \(\frac { 30 }{ QR } =\sqrt { 3 } \)
\(\Rightarrow\)  \(QR=\frac { 30 }{ \sqrt { 3 } } \)
\(=10\sqrt { 3 } \) m

\(35\left( 1+\sqrt { 2 } \right) m\)

25m