33.33 m

\(20(\sqrt{3}-1)m\)

Let AB be the tower of height h m, P and Q be the two points in line with the foot of the tower AB, such that
ㄥAPB=∝ and ㄥAQB=β
Consider rt. ㄥed ΔQBA,
\(\frac { AB }{ QB } \)=tan β
QB=AB.\(\frac { 1 }{ tan\beta } \)
=AB cot β....(i)
Again consider rt ㄥed ΔPBA,
\(\frac { AB }{ PB } \)=tan α
\(\frac { AB }{ PQ+QB } \)=tan α
⇒ PQ+QB=AB.\(\frac { 1 }{ tan\alpha } \)
=AB cot α....(ii)
Subtracting (i) from (ii), we obtain
PQ=AB cot α - AB cot β
PQ=AB (cot α - cot β)
PQ=h(cot α - cot β)

Let the height of the ballon at P be h meters. Let A and B be the two cars. Thus, AB=100 m
ㄥPAQ=45^o and ㄥPBQ=60^o
Consider right-angled ΔAQP
\(\frac { PQ }{ AQ } \)=tan^o
⇒ \(\frac { PQ }{ AQ } \)=1
⇒ PQ=AQ= h m
Now, consider right-angled ΔPBQ, we have
\(\frac { PQ }{ BQ } \)=tan 60^o
\(\frac { PQ }{ BQ } \)=\(\\ \\ \\ \\ \\ \\ \\ \sqrt { 3 } \\ \)
\(\frac { h }{ h-100 } =\sqrt { 3 } \)
⇒ h=\(\frac { \sqrt { 3 } (100) }{ \sqrt { 3 } -1 } \)
⇒ h=\(\frac { \sqrt { 3 } (100) }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } \)
⇒ h=\(\frac { \sqrt { 3 } (100)(\sqrt { 3 } +1) }{ 2 } \)
=50(3+\(\sqrt { 3 } \))
Hence, the height of the ballon is 50(3+\(\sqrt { 3 } \)) m.

Here, length of the ladder is 15 m and angle of elevation is 90^o - 60^o i.e., 30^o
Let h m be the height of the wall Consider right-angled ΔABC
\(\frac { BC }{ AC } \)=sin 30^o
\(\frac { h }{ 12 } =\frac { 1 }{ 2 } \)
h=\(\frac { 15 }{ 2 } \)m
Hence, the height of the wall is 15/2 m or 7.5 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

To solve the problem, we need to determine the distance traveled by the ship as observed from the top of the lighthouse. The lighthouse is 100 meters above sea level, and the angles of depression change from 30^∘ to 60^∘.
1. Understanding the Problem:
The height of the lighthouse (h) = 100 m.
Angle of depression from the lighthouse to the ship initially = 30^∘.
Angle of depression from the lighthouse to the ship finally = 60^∘.
2. Setting Up the Diagram:
Let point A be the top of the lighthouse, point B be the position of the ship when the angle of depression is 30^∘, and point C be the position of the ship when the angle of depression is 60^∘.
The distances from the base of the lighthouse to the ship at points B and C can be denoted as x and y respectively.
3. Using Trigonometric Ratios:
For the angle of depression of 30^∘:
\(\tan \left(30^{\circ}\right)=\frac{\text { height }}{\text { distance from lighthouse }}=\frac{100}{x} \)
\( \tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}} \Longrightarrow \frac{1}{\sqrt{3}}=\frac{100}{x} \Longrightarrow x=100 \sqrt{3}\)
For the angle of depression of 60^∘ :
\( \tan \left(60^{\circ}\right)=\frac{\text { height }}{\text { distance from lighthouse }}=\frac{100}{y} \)
\(\tan \left(60^{\circ}\right)=\sqrt{3} \Longrightarrow \sqrt{3}=\frac{100}{y} \Longrightarrow y=\frac{100}{\sqrt{3}}\)
4. Calculating the Distance Traveled by the Ship:
The distance traveled by the ship is the difference between the two distances:
Distance traveled \(=x-y=100 \sqrt{3}-\frac{100}{\sqrt{3}}\)
To simplify this, we need a common denominator:
\(=\frac{100 \sqrt{3} \cdot \sqrt{3}}{\sqrt{3}}-\frac{100}{\sqrt{3}}=\frac{300-100}{\sqrt{3}}=\frac{200}{\sqrt{3}}\)
5. Substituting the Value of √3 :
Now, substituting \(\sqrt{3}=1.732\) :
Distance traveled \(=\frac{200}{1.732} \approx 115.47 \mathrm{~m}\)
Final Answer: √3
The distance traveled by the ship during the period of observation is approximately 115.47 meters.

Let P and Q be the positions of two aeroplanes when Q is vertically below P and OP=4000m
Given, the angles of elevation of P and Q from a point An on the ground be  \({ 60 }^{ \circ }\) and \({ 45 }^{ \circ }\), respectively.
In right angled \(\Delta AOP,\)

\(\tan { { 60 }^{ \circ } } =\frac { OP }{ AO }\)
\( \\ \Rightarrow \sqrt { 3 } =\frac { 4000 }{ \sqrt { 3 } }\)
\(\Rightarrow AO=\frac { 4000 }{ \sqrt { 3 } } ...(i)\)
In right angled \(\Delta AOQ, \tan { { 45 }^{ \circ } } =\frac { 4000 }{ \sqrt { 3 } } m\)
\(\Rightarrow 1=\frac { OQ }{ OA } \)
\(\\ \Rightarrow OA=OQ...(ii)\)
From Eqs. (i) and (ii), we get \(OQ=\frac { 4000 }{ \sqrt { 3 } } m\)
∴ Vertical distance between the aeroplanes,
\(PQ=OP-OQ\)
\(=4000-\frac { 4000 }{ \sqrt { 3 } }\)
\(=4000\left( 1-\frac { 1 }{ \sqrt { 3 } } \right) =4000\left( \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } } \right)\)
\( =4000\left( \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } \right)\)
\(=4000\left( \frac { 3-\sqrt { 3 } }{ 3 } \right)\)
\(=4000\left( \frac { 3-1.732 }{ 3 } \right) [\because \sqrt { 3 } =1.732]\)

Let AB is the building

∴ AB=60 m and CD is the light house.
ㄥEAC=30^o
and ㄥEAD=60^o
∴ ㄥADB=60^o
∴ AE||BD
In right ΔABD,
\(\frac { BD }{ AB } \)=cot 60⁰⇒ \(\frac { BD }{ 60 } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ BD=\(\frac { 60 }{ \sqrt { 3 } } \)   m=20\(\sqrt { 3 } \) m.
∵ BD=AE
∴ AE=20\(\sqrt { 3 } \) m
Now in right ΔCEA, tan 300=\(\frac { CE }{ AE } \)
⇒ \(\frac { 1 }{ \sqrt { 3 } } =\frac { CE }{ 20\sqrt { 3 } } \)⇒ CE=20 m
(i) Difference between the heights of the light house and the building=CE=20 m
(ii) The distance between the light house and the building =BD=20\(\sqrt { 3 } \) m

Let BE = 22 m be the height of the tower and AD = 1.5 m be the height of the observer. The point D be the observer's eye. Draw DC ∥ AB.

Then, AB = 20.5 m = DC and EC = BE - BC
= BE - AD = 22 - 1.5 = 20.5 m [∵ BC = AD]

Let 0 be the angle of elevation make by observer's eye to θ the top of the tower i.e. \(\angle E D C=\theta .\)

In right angled △DCE,

\(\tan \theta=\frac{P}{B}=\frac{C E}{D C}=\frac{20.5}{20.5}\)

⇒ tan θ = 1 ⇒ tan = tan 45° ⇒ 0 = 45°

Let AB be the tower and BC be its shadow.

Given, AB = BC

In right angled △ABC,

\(\tan C=\frac{\text { Perpendicular }}{\text { Base }}=\frac{A B}{B C}=1\)

⇒ tan C = tan45° [∵ tan45° = 1]

∴ C = 45°

In Figure, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river. P is a point on the bridge at a height of 3 m, i.e., DP = 3 m. We are interested to determine the width of the river, which is the length of the side AB of the D APB. 

Now, AB = AD + DB
In right Δ APD, ∠ A = 30°.
So, tan 30°\(=\frac{P D}{A D}\)
i.e., \(\frac{1}{\sqrt{3}}=\frac{3}{\mathrm{AD}} \text { or } \mathrm{AD}=3 \sqrt{3} \mathrm{~m}\) 
Also, in right Δ PBD, ∠ B = 45°. So, BD = PD = 3 m.
Now, AB = BD + AD = 3 + 3\(\sqrt3\) = 3 (1 + \(\sqrt3\) ) m.
Therefore, the width of the river is 3 ( \(\sqrt3\) + 1) m.

Let the length of the ladder be l m

i.e. OP = OR = l m and OA = x m, OC = y m

(i) In right angled △OAP,

\(\begin{aligned} \sin 60^{\circ} & =\frac{A P}{O P} \\ \Rightarrow \quad \frac{\sqrt{3}}{2} & =\frac{12 \sqrt{3}}{O P} \Rightarrow l=O P=24 \mathrm{~m} \end{aligned}\)

Hence, the length of the ladder = 24 m

(ii) Again, in right angled △OAP,

\(\begin{array}{rlrl} & & \tan 60^{\circ} & =\frac{A P}{O A} \\ \Rightarrow & \sqrt{3} & =\frac{12 \sqrt{3}}{x} \Rightarrow x=12 \mathrm{~m} \end{array}\)

Hence, the distance of the building Y from point O is 12 m.

(iii) (a) In right angled △OCR,

\(\begin{aligned} & & \cos 45^{\circ} & =\frac{O C}{O R} \\ \Rightarrow & & \frac{1}{\sqrt{2}} & =\frac{O C}{24} \\ \Rightarrow & & y & =O C=12 \sqrt{2} \mathrm{~m} \end{aligned}\)

Horizontal distance between the buildings \(=(x+y)=12(\sqrt{2}+1) \mathrm{m}\)

Or

(b) Again, in right angled △OCR,

\(\begin{aligned} & & \tan 45^{\circ} & =\frac{R C}{O C} \\ \Rightarrow & & 1 & =\frac{R C}{12 \sqrt{2}} \\ \Rightarrow & & x & =R C=12 \sqrt{2} \mathrm{~m} \end{aligned}\)

Height of building x is \(12 \sqrt{2} \mathrm{~m} .\)