Here, AB is the chimney, CD the observer and ∠ ADE the angle of elevation (see Figure). In this case, ADE is a triangle, right-angled at E and we are required to find the height of the chimney.
We have AB = AE + BE = AE + 1.5 and DE = CB = 28.5 m
To determine AE, we choose a trigonometric ratio, which involves both AE and DE. Let us choose the tangent of the angle of elevation.

Now, \(\tan 45^{\circ}=\frac{\mathrm{AE}}{\mathrm{DE}}\) 
i.e., \(1=\frac{\mathrm{AE}}{28.5}\) 
Therefore, AE = 28.5
So the height of the chimney (AB) = (28.5 + 1.5) m = 30 m.

Let us assume that the height of the tower AC be h m and ∠ABC=60^o , ∠ADC=30^o
As BC=4m
DC=9m   (given)

Consider a rt△ACB, we have
tan 60^o = \(\frac{AC}{BC}\)
⇒ \(\sqrt{3} = \frac{h}{4}\) --- (i)
Again, consider a rt. △ACD, we have
tan 30⁰ = \(\frac{AC}{DC}\)
⇒ \(\frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 9 } \)  --- (ii)
Now, multiply (i) and (ii), we have
\(1=\frac{h}{4}\times\frac{h}{9}\)
⇒ 1= \(\frac {h^{2}} {36}\)
⇒ h²=36
⇒ h=\(\sqrt{36}\)=6 m
⇒ h=6m
Hence, height of the tower is 6m.

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 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.

28

51.96m

\(75\sqrt { 2 } m\)

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

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.

Trigonometric ratio involving AB, BC, OD, OA and angles is tanθ. [Refer AB, BC, OA and OD from the figure.]
Distance travelled by the balloon OB = AB - OA
From the figure, OD = BC, and it can be calculated as
88.2 m - 1.2 m = 87 m --- (1)
In ΔAOD,
tan 60° = OD/OA
√3 = 87/OA
OA = 87 / √3
= 87 × √3 / √3 × √3
= (87 × √3) / 3
= 29√3 m

25.36 m

Let AB be the pole and C be the point 1.3 m below the top A of the pole. Also, let D be the point on the ground, where ladder CD be placed.
So, \(\angle CDB=60°and BC=(5-1.3)m=3.7m\quad \)
Now, \(\frac { BC }{ CD } =sin\quad 60°\)
\(\Rightarrow \frac { 3.7 }{ CD } =\frac { \sqrt { 3 } }{ 2 } \)
\(\Rightarrow CD=\frac { 7.4 }{ \sqrt { 3 } } m\)
So, length of the ladder = \(\frac { 7.4 }{ \sqrt { 3 } } m\)
\(=\frac { 7.4\sqrt { 3 } }{ 3 } \) m      

Also,  \(\frac { BC }{ BD } =tan\quad 60°\)
 \(\Rightarrow \frac { 3.7 }{ BD } =\sqrt { 3 } \Rightarrow BD=\frac { 3.7 }{ \sqrt { 3 } } m\)
Thus, foot of the ladder must be placed \(\frac { 3.7 }{ \sqrt { 3 } } m, i.e.\frac { 3.7\sqrt { 3 } }{ 3 } \)
m away from the foot of the pole.
Value Women can work in any field without gender bias.
Also, it indicates 'dignity for labour'.

(i) In right angled \(\Delta\)OPB,
\(\begin{aligned} & \cos 30^{\circ}=\frac{O P}{O B}=\frac{36}{O B} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad O B & =\frac{36}{\cos 30^{\circ}}=\frac{36}{\frac{\sqrt{3}}{2}}=\frac{72}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad & O B=24 \sqrt{3} \mathrm{~cm} \end{aligned}\)
In right angled \(\Delta\)OPB,
\(\begin{aligned} \tan 30^{\circ} & =\frac{B P}{O P} \\ \end{aligned} \)
\(\begin{aligned} \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{B P}{36} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad B P & =\frac{36}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \end{aligned}\)
\(\begin{aligned} =12 \sqrt{3} \mathrm{~cm} \end{aligned}\)
In right angled \(\Delta\)APO,
\(\begin{aligned} & \tan 45^{\circ}=\frac{A P}{O P} \\ \end{aligned}\)
\(\Rightarrow\) OP = AP   [\(\because\) tan 45° = 1]
\(\Rightarrow\) AP = 36 cm   ....(i)
\(\therefore\) Distance AB = AP - BP
= 36 - 12\(\sqrt3\)
=12(3-\(\sqrt3\)) cm
(iii) Height of the section A from the base of the tower
= AP = 36 cm                  [using Eq (i)]
Or
In right angled \(\Delta\)OPB,
\(\begin{aligned} \tan 30^{\circ} & =\frac{B P}{O P} \\ \end{aligned}\)
\(\Rightarrow\)  BP = OP tan 30^\(\circ\)
\(=36 \times \frac{1}{\sqrt{3}}=12 \sqrt{3} \mathrm{~cm}\)
\(\therefore\) Area of \(\Delta\)OPB = \(\begin{aligned} & =\frac{1}{2} \times O P \times B P \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 36 \times 12 \sqrt{3} \\ \end{aligned}\)
\(\begin{aligned} & =216 \sqrt{3} \mathrm{~cm}^2 \end{aligned}\)

(i) (b): \(\angle X A C=45^{\circ}\) 
\(\therefore \quad \angle A C D=45^{\circ}\) [Alternate interior angles]
(ii) (b)
(iii) (c) : \(\text { In } \Delta A C D\)
\(\frac{A D}{D C}=\tan 45^{\circ} \)
\(\Rightarrow \frac{100}{D C}=1 \Rightarrow D C=100 \mathrm{~m}\)
(iv) (d): \(\text { In } \Delta A B D, \frac{A D}{B D}=\tan 30^{\circ}\)
\(\Rightarrow \quad \frac{100}{B D}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \quad B D=100 \sqrt{3} \mathrm{~m}\)
(v) (a): \(\text { In } \Delta A D C\)
\(\frac{A D}{A C}=\sin 45^{\circ} \Rightarrow \frac{100}{A C}=\frac{1}{\sqrt{2}} \Rightarrow A C=100 \sqrt{2} \mathrm{~m}\)