Let AB be the tower and AC be the flag staff on the tower. Let D be a point on the ground such that the angles of elevation of foot A and top C of the flag staff are 45\(°\) and 60\(°\) respectively.
Then, we have AB = 20 m, \(\angle ADB=45° and \angle CDB=60°\)

In right angled \(\Delta ABD,\)
\(tan\quad 45°=\frac { AB }{ BD } \left[ \because tan\theta =\frac { P }{ B } \right] \)
\(\Rightarrow 1=\frac { 20 }{ BD } \)
\(\Rightarrow BD=20\quad m \left[ \because tan45°=1 \right] \)
and in right angled \(\Delta CBD,\)
\(tan 60°=\frac { BC }{ BD } \)
\(\sqrt { 3 } =\frac { BC }{ 20 } \left[ \because tan60°=\sqrt { 3 } \right] \)
\(\Rightarrow BC=20\sqrt { 3 } =20\times 1.732 \left[ \because \sqrt { 3 } =1,732 \right] \)
= 34.64 m (approx)
Now, AC = BC - AB = 34.64 - 20
=14.64 m (approx)
Hence, the height of the fleg staff is 14.64 m.

Let AB be the tree whose part AC breaks and touches the ground at D
Then,  BD = \(2\sqrt { 3 } m\)
and AC = CD
In right angled \(\Delta \)CBD,
\(cos 60°=\frac { B }{ H } =\frac { BD }{ CD } \)
\(\Rightarrow \frac { 1 }{ 2 } =\frac { 2\sqrt { 3 } }{ CD } \)
\(\left[ \because cos 60°=\frac { 1 }{ 2 } and BD=2\sqrt { 3 } m \right] \)

\(\Rightarrow\) \(CD=2\times 2\sqrt { 3 } =4\sqrt { 3 } \)
\(=4\times 1.732=6.928\ m [\because \sqrt { 3 } =1.732]\)
\(\therefore\)  AC = CD = 6.928 m
Again, in right angled \(\Delta \)CBD,
\(tan\quad 60°=\frac { P }{ B } =\frac { BC }{ BD } \)
\(\Rightarrow \sqrt { 3 } =\frac { BC }{ 2\sqrt { 3 } } [\because tan60°=\sqrt { 3 } and BD=2\sqrt { 3 } m]\)
\(\Rightarrow BC=\sqrt { 3 } \times 2\sqrt { 3 } =6 m\)
Now, AB = AC + BC
= 6.928 + 6 = 12.928 m (approx)
Hence, the height of the tree is 12.928 m.

In the given figure, AB is the height of the pole and AC = 20 m is the length of rope which is tied from the top of the pole to the ground at point C.
In right angled \(\Delta ABC\) ,
\(sin30°=\frac { AB }{ AC } =\frac { AB }{ 20 } \)
⇒ \(\frac { 1 }{ 2 } =\frac { AB }{ 20 } \Rightarrow AB=\frac { 20 }{ 2 } =10\quad m\)
Hence, the height of the pole is 10 m.

17.32m

Let AB be the tower of height h m and CD be the cliff of height 60\(\sqrt m\) m, such that 
\(\angle \)XDA = \(\angle \)DAE = 45^o and \(\angle \)XDB = \(\angle \)DBC = 60^o

In rt.  \(\triangle\)DCB,
\(DC\over BC\)=tan 60^o
\(\frac { 60\sqrt { 3 } }{ BC } =\sqrt { 3 } \)
BC = 60m
Now BC = AE = 60m
and DE = DC - CE (or AB)
= \(60\sqrt { 3 } -h\)
Again in rt. \(\triangle\)DEA
\(\angle \)E = 90^o
ஃ \(DE\over AE\)=tan 45^o
\(\frac { 60\sqrt { 3 } -h }{ 60 } =1\)
⇒ 60\(\sqrt3\) -  h = 60
⇒ h = 60\(\sqrt3\) - 60 
⇒ h = 60(\(\sqrt3\)-1)
Hence the required height of the tower is 60(\(\sqrt3\)-1)m

Let AB=h m be height of the tree and AC is the part of the broken tree. As ㄥCDB=450 and BD=8 m
Consider rt . angled ΔCBD, we have
tan45^o=\(\frac { CB }{ DB } \)
1=\(\frac { CB }{ 8 } \)
⇒ CB=8 m
Also, cos450=\(\frac { DB }{ DC } \)
\(\frac { 1 }{ \sqrt { 2 } } =\frac { 8 }{ DC } \)
⇒ Dc=8\(\sqrt { 2 } \)
Total height of the tree AB=AC+CB
=CD+CB
=8\(\sqrt { 2 } \)m+8m
=8(\(\sqrt { 2 } \) +1)m

Let height of lighthouse be h meters. From point O angle of elevation is ፀ and from point P is ф, OP = 240 m.
Let PB = x metres ,
tanፀ = \(\frac { 5 }{ 12 } ;tan\phi =\frac { 3 }{ 4 } \)
In right angled ΔOBA,
\(\frac { AB }{ OB } =tan\theta \Rightarrow \frac { h }{ 240+x } =\frac { 5 }{ 12 } \)  .......(i)
In right angled ΔPBA,
\(\frac { AB }{ PB } =tan\phi \Rightarrow \frac { h }{ x } =\frac { 3 }{ 4 } \).....(ii)
Dividing (i) by (ii), we get
\(\frac { h }{ 240+x } \times \frac { x }{ h } =\frac { 5 }{ 12 } \times \frac { 4 }{ 3 } \)
⇒ \(\frac { x }{ 240+x } =\frac { 5 }{ 9 } \) ⇒ 9x = 1200 + 5x
⇒ 4x = 1200 ⇒ x = 300
Putting x = 300 in (ii) we get, h = 3/4 x 300 = 225
Hence height of lighthouse is 225 metres.

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.

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 \(20(\sqrt{3}-1) \mathrm{m}\).

33.33 m

Let x m be the length of the side BC.
Given, \(AB=12 cm \& \angle BAC=30°\)
In right angled \(\Delta ABC,\)
   \(\tan { 30° } =\frac { BC }{ AB } \)

\(\Rightarrow \frac { 1 }{ \sqrt { 3 } } =\frac { x }{ 12 } \left[ \therefore \tan { 30° } =\frac { 1 }{ \sqrt { 3 } } \right] \)
\(\Rightarrow \sqrt { 3 } \times x=12\Rightarrow x=\frac { 12 }{ \sqrt { 3 } } \) 
\(\Rightarrow x=\frac { 12 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } \)  [by rationalising]
\(=\frac { 12\sqrt { 3 } }{ 3 }\)    
\( =4\sqrt { 3 }cm\)
Hence, the length of the side BC is \(4\sqrt { 3 }cm\) 

30⁰

60⁰

200m

(i)20m
(ii)34.64m

150 seconds

False,
Since  \(\frac { h }{ x } =\tan { { 30 }^{ 0 } } \)
\(\frac { h }{ x } =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow h=\frac { x }{ \sqrt { 3 } } \)
Let h be changed to 2h
\(\therefore \frac { 2h }{ x } =\tan { \theta } \Rightarrow \frac { 2x/\sqrt { 3 } }{ x } =\tan { \theta } \)
\(\Rightarrow \frac { 2 }{ \sqrt { 3 } } =\tan { \theta } \quad \)

Also \(\tan { { 60 }^{ 0 } } =\sqrt { 3 } \)
\(Since\frac { 2 }{ \sqrt { 3 } } \neq \sqrt { 3 } \)
\(\theta \neq { 60 }^{ 0 }\)

Let AB=h m be the height of the pole
AC=12 m is the length of rope
and ㄥACB=30^o
Consider rt. angled ΔABC, we have
sin 300=\(\frac { AB }{ AC } \)
\(\frac { 1 }{ 2 } =\frac { h }{ 12 } \)
h=\(\frac { 12 }{ 2 } \)=6 m

In art. \(\triangle\)ABC, \(\angle\)B = 90°
\(\therefore\)   AB = 2 m, BC = \(2\sqrt{3}\)m
Let  \(\angle\)ACB = \(\theta\)
\(\therefore\)  tan \(\theta={{AB}\over{BC}}={{2}\over{2\sqrt{3}}}\)
tan \(\theta ={{1}\over{\sqrt{3}}}\Rightarrow{\theta}=30°\)

52 m

20 m

\(75\sqrt { 2 } m\)

In ΔABD, \(\frac { BD }{ AB } \)=cot 60^o
⇒ \(\frac { BD }{ 50 } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow BD\frac { 50 }{ \sqrt { 3 } } \)
BD=EC⇒ EC=\(\frac { 50 }{ \sqrt { 3 } } \)
In ΔAEC, \(\frac { AE }{ EC } \)=tan 45^o
⇒ ΔAEC, \(\frac { AE }{ EC } \)=tan 45^o
⇒ AE=EC⇒ AE=\(\frac { 50 }{ \sqrt { 3 } } \)m
Now BE=AB-AE=50-\(\frac { 50 }{ \sqrt { 3 } } \)
=\(\frac { 50\sqrt { 3 } -50 }{ \sqrt { 3 } } \)
=\(\frac { 50(\sqrt { 3 } -1) }{ \sqrt { 3 } } \)
DC=BE=\(\frac { 50(\sqrt { 3 } -1) }{ \sqrt { 3 } } \)m

Let AB and CD are towers of height h₁ and h₂ respectively 
If F. is the midpoint of BD then BE = DE = x 
In right \(\Delta\)ABE
\(\frac { { h }_{ 1 } }{ x } =\tan { { 60 }^{ o } } \)
\(\Rightarrow\) \({ h }_{ 1 }=\sqrt { 3 } x\)  ...(i)
In right  \(\Delta\)CDE
\(\frac { { h }_{ 2 } }{ x } =\tan { { 30 }^{ o } } \Rightarrow { h }_{ 2 }=\frac { x }{ \sqrt { 3 } } \)
Now \(\frac { { h }_{ 1 } }{ { h }_{ 2 } } =\frac { \sqrt { 3 } x }{ \frac { x }{ \sqrt { 3 } } } =\frac { 3 }{ 1 } \)  \(\Rightarrow\) h₁ : h₂ = 3 : 1

(i) 14.66 km
(ii) Presence of mind, ability to take promote decisions.

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.