Let C is cloud and R is its reflection.

ㄥDAC=30^o, ㄥDAR=60^o, let CD=x m
∴ Height of the cloud above the lake
=(x+20)m
∴ ER=(20+x)m
Now In right ΔADC,
\(\frac { CD }{ AD } =tan30^{ 0 } \Rightarrow \frac { x }{ AD } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ AD=\(\sqrt { 3 } \)x
In right, ΔADR,
\(\frac { DR }{ AD } \)=tan 60^o
⇒ \(\frac { DE+ER }{ AD } =\sqrt { 3 } \)
⇒ \(\frac { 20+20+x }{ \sqrt { 3 } x } =\sqrt { 3 } \)  (using (i))
40+x=3x
⇒ x=20 m
Now In right ΔADC, \(\frac { AC }{ DC } \)=cosec 30^o
\(\frac { AC }{ 20 } \)=2 ⇒ AC=40 m
∴ Distance of the cloud from A=40 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.

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.

Given: A boy is standing at a distance of 100 m from the bird flying at an elevation of 30^o. A girl is standing on the roof of 20 m high building finds the angle of elevation of the bird to be 45^o Boy and girl are on opposite side of the bird.
To find: Distance between the bird and the girl i.e., BE.
Solution: In ΔACB,
⇒ \(\frac { h }{ 100 } \) = sin 300 ⇒ h = \(\frac { 1 }{ 2 } \) x 100 = 50 m
⇒ BF = h - 20 = (50 - 20) m = 30 m
In ΔBFE,\(\frac { 30 }{ BE } =sin{ 45 }^{ 0 }\Rightarrow \frac { 30 }{ BE } =\frac { 1 }{ \sqrt { 2 } } \Rightarrow 30\sqrt { 2 } \) = BE
BE = 30 x 1.414 = 42.420 = 42.42 m

Let the height of the tower AB = h m,
We have PB = 4 m, QB = 9 m
Let ㄥAQB = θ   [Both are complementary angles]
Then ㄥAPB = 90\(\unicode{xb0} \) ^- θ
In ΔABP, \(\frac { AB }{ PB } \) = tan (90\(\unicode{xb0} \)- θ)
⇒ h/4 = cotθ  ....(i)
In ΔABQ, \(\frac { AB }{ QB } \) = tanፀ
h/9 = tanθ
h = 9tanθ  ....(ii)
From equation (i) and (ii) we get
h x h = 4 cotθ x 9 tanθ
⇒ h² = 36 cotθ x tanθ = 36 1/tanθ x tanθ
⇒ h² = 36 ⇒ h = 6m
Hence, the height of the tower is 6 m.

Let CD = 75 m be the height of the light house from the sea level. Let A and B be the positions of two ships on the sea level. From point D of the lighthouse, the angles of depression of two ships A and B are

\(\angle\)ODA = 30°  and \(\angle\)ODB = 45°.
\(\therefore\) \(\angle\)CAD = \(\angle\)ODA = 30°    [alternate angles]
and \(\angle\)CBD = \(\angle\)ODB = 45° [alternate angles]
Let distance between two ships be AB = y m and BC = x m.
In right angled \(\Delta\)ACD,
\(\begin{array}{ll} & \tan 30^{\circ}=\frac{P}{B}=\frac{C D}{A C} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow \quad & \frac{1}{\sqrt{3}}=\frac{75}{A B+B C} \\ \end{array}\)
\(\begin{array}{ll} & {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}} \text { and } A C=A B+B C\right]} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow \quad & \frac{1}{\sqrt{3}}=\frac{75}{y+x} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow \quad & x+y=75 \sqrt{3} \end{array}\)   ....(i)
and in right angled \(\Delta\)DCB, tan 45° \(=\frac{C D}{B C}\)
\(\Rightarrow \quad 1=\frac{75}{x} \quad\left[\because \tan 45^{\circ}=1\right]\)
\(\Rightarrow\) x = 75 m
On putting x = 75 m in Eq. (i), we get
\(75+y=75 \sqrt{3} \Rightarrow y=75(\sqrt{3}-1) \mathrm{m}\)
Hence, the distance between two ships is 75(\(\sqrt{3}-1\))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 \(0.8(\sqrt{3}+1) \mathrm{m}\).

20 m

\({15\over2}{\sqrt {3}} \ m \)

Let AB is tower and C is a point on the ground such that BC = 100 m
In right \(\Delta\)ABC,
\(\frac { AB }{ BC } =\tan { { 45 }^{ o } } \)
\(\Rightarrow\)  \(\frac { AB }{ 100 } =1\)
\(\Rightarrow\)   AB = 100 m

A is the position of the man, OA = 12m, BC is cliff. 
Let height of the cliff
BC = h m and CE = (h - 12)m
Let AE = OB = x m
In right angled triangle AEB,
\(\frac { AE }{ BE } \) = cot30^o ⇒ AE = 12 x \(\sqrt { 3 } \)
=12 x 1.732 m = 20.78 m
∴ Distance of ship from cliff = 20.78 m
In right angled triangle AEC,
\(\frac { CE }{ AE } =tan{ 60 }^{ 0 }\Rightarrow \frac { h-12 }{ 12\sqrt { 3 } } \) = \(\sqrt { 3 } \)
h - 12 = 36 ⇒ h = 48 m
∴. Height of the cliff = 48 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

Let AB is the tower and C is the point 20 m away from the base of the tower 
∴ BC = 20m, ∠ACB = 45^o

In right ΔABC, tan 45^o = \(\frac{AB}{BC}\)
⇒ I = \(\frac{AB}{20}\) ∴ AB = 20 m

\(\frac { 20 }{ \sqrt { 3 } } \)

Let aeroplane be at B and let the two ships at C and D, such that their angles of depression from B are \(30°\)and\(60°\), respectively.Then, angles of elevation of D and C from B are \(30°\)and \(60°\)respectively.

We have, AB=1200 m
Here, one side AB is common in both triangles.
Let AC=x m and CD= y m
In \(\Delta BAC\) , we have
\(tan\quad 60°=\frac { A }{ B } \Rightarrow \sqrt { 3 } =\frac { 1200 }{ x } \)
⇒ \(x=\frac { 1200 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 1200\sqrt { 3 } }{ 3 } =400\sqrt { 3 } \)  ....(i)
In \(\Delta BAD\) , We have
\(tan\quad 30°=\frac { AB }{ AD }=\frac { AB }{ DC+CA }\)    [∵ AD=DC+CA]
On putting the value of x from Eq.(i) in Eq. (ii) we get
\(y=1200\sqrt { 3 } -400\sqrt { 3 } \)
⇒ y = 1800\(\sqrt { 3 } \)
⇒ y = 800 x 1.732 \(\left[ \because \sqrt { 3 } =1.732 \right] \)
\(\Rightarrow y=1385.6m\)
Hence, the distance between both ships is 1385. 6m.

Let A be the window and CE be the opposite house
CD=AB=60 m [Opposite side of a rectangle] ........(i)
In rt. ΔABC, tan45^o=\(\frac { 60 }{ BC } \)
⇒ 1=\(\frac { 60 }{ BC } \)
⇒ BC=60 m        .........(ii)
AD=BC [Opposite sides of a rectangle]
∴ AD=60 m [From (ii)]        ......(iii)
In rt. ΔADE, tan600=\(\frac { DE }{ AD } \)
⇒ \(\sqrt { 3 } =\frac { DE }{ 60 } \)    [From (iii)]
⇒ DE=60\(\sqrt { 3 } \) m
∴ Height of the opposite house
CE=CD+DE=60+60\(\sqrt { 3 } \)
=60(1+\(\sqrt { 3 } \))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