Heights and Distances Questions in English

$(15\sqrt{3})$ First,let us represent the problem with a right-angled triangle $ABC$ (as shown in the figure),where $AB$ is the height of the tower,$BC = 15\, m$ is the distance of the point from the foot of the tower,and $\angle ACB = 60^{\circ}$ is the angle of elevation.
In the right-angled triangle $ABC$,we have:
$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}$
$\tan 60^{\circ} = \frac{AB}{BC}$
Since $\tan 60^{\circ} = \sqrt{3}$,we substitute the values:
$\sqrt{3} = \frac{AB}{15}$
$AB = 15\sqrt{3}\, m$
Thus,the height of the tower is $15\sqrt{3}\, m$.

(N/A) In the figure,the electrician is required to reach the point $B$ on the pole $AD$.
So,
$BD = AD - AB = (5 - 1.3)\, m = 3.7\, m$
Here,$BC$ represents the ladder. We need to find its length,i.e.,the hypotenuse of the right triangle $BDC$.
Now,we consider the trigonometric ratio $\sin 60^{\circ}$.
So,$\frac{BD}{BC} = \sin 60^{\circ}$ or $\frac{3.7}{BC} = \frac{\sqrt{3}}{2}$
Therefore,$BC = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{1.73} \approx 4.28\, m$.
i.e.,the length of the ladder should be $4.28\, m$.
Now,$\frac{DC}{BD} = \cot 60^{\circ} = \frac{1}{\sqrt{3}}$.
i.e.,$DC = \frac{3.7}{\sqrt{3}} = \frac{3.7}{1.73} \approx 2.14\, m$.
Therefore,she should place the foot of the ladder at a distance of $2.14\, m$ from the pole.

(C) Let $AB$ be the chimney,$CD$ be the observer,and $\angle ADE$ be the angle of elevation. In this case,$\triangle ADE$ is a right-angled triangle at $E$,and we need to find the height of the chimney.
We have $AB = AE + BE = AE + 1.5\, m$.
Also,$DE = CB = 28.5\, m$.
To determine $AE$,we use the tangent trigonometric ratio,which involves both $AE$ and $DE$:
$\tan 45^{\circ} = \frac{AE}{DE}$
Since $\tan 45^{\circ} = 1$,we have:
$1 = \frac{AE}{28.5}$
$AE = 28.5\, m$.
Therefore,the height of the chimney $AB = AE + BE = 28.5\, m + 1.5\, m = 30\, m$.

(N/A) In the figure,$AB$ denotes the height of the building,$BD$ the flagstaff,and $P$ the given point. Note that there are two right-angled triangles,$\triangle PAB$ and $\triangle PAD$. We are required to find the length of the flagstaff,i.e.,$DB$,and the distance of the building from the point $P$,i.e.,$PA$.
Since we know the height of the building $AB = 10 \, m$,we will first consider the right $\triangle PAB$.
We have $\tan 30^{\circ} = \frac{AB}{AP}$.
i.e.,$\frac{1}{\sqrt{3}} = \frac{10}{AP}$.
Therefore,$AP = 10\sqrt{3} \, m$.
i.e.,the distance of the building from $P$ is $10 \times 1.732 = 17.32 \, m$.
Next,let us suppose the length of the flagstaff $BD = x \, m$. Then the total height $AD = (10 + x) \, m$.
Now,in right $\triangle PAD$,$\tan 45^{\circ} = \frac{AD}{AP} = \frac{10 + x}{10\sqrt{3}}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{10 + x}{10\sqrt{3}}$.
Therefore,$10\sqrt{3} = 10 + x$.
$x = 10\sqrt{3} - 10 = 10(\sqrt{3} - 1) = 10(1.732 - 1) = 10(0.732) = 7.32 \, m$.
Thus,the length of the flagstaff is $7.32 \, m$ and the distance of the building from $P$ is $17.32 \, m$.

(N/A) Let $AB$ be the tower of height $h \, m$ and $BC$ be the length of the shadow when the Sun's altitude is $60^{\circ}$. Let $BC = x \, m$.
When the Sun's altitude is $30^{\circ}$,the length of the shadow is $DB = (x + 40) \, m$.
In $\triangle ABC$,$\tan 60^{\circ} = \frac{AB}{BC} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \dots (1)$.
In $\triangle ABD$,$\tan 30^{\circ} = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{h}{x + 40} \implies h = \frac{x + 40}{\sqrt{3}} \dots (2)$.
Equating $(1)$ and $(2)$,we get $x\sqrt{3} = \frac{x + 40}{\sqrt{3}}$.
$3x = x + 40 \implies 2x = 40 \implies x = 20 \, m$.
Substituting $x = 20$ in $(1)$,$h = 20\sqrt{3} \, m$.
Thus,the height of the tower is $20\sqrt{3} \, m$ (or approximately $34.64 \, m$).

(N/A) Let $PC$ be the multi-storeyed building and $AB$ be the $8 \, m$ tall building. We need to find the height of the multi-storeyed building $(PC)$ and the distance between the two buildings $(AC)$.
From the figure,$PQ$ is the horizontal line from the top of the building $P$. Since $PQ \parallel BD$,the alternate interior angles are equal.
Therefore,$\angle PBD = 30^{\circ}$ and $\angle PAC = 45^{\circ}$.
In right-angled $\triangle PBD$:
$\tan 30^{\circ} = \frac{PD}{BD} \implies \frac{1}{\sqrt{3}} = \frac{PD}{BD} \implies BD = PD\sqrt{3}$.
In right-angled $\triangle PAC$:
$\tan 45^{\circ} = \frac{PC}{AC} \implies 1 = \frac{PC}{AC} \implies PC = AC$.
Since $AC = BD$ and $PC = PD + DC$,where $DC = AB = 8 \, m$:
$PD + 8 = AC = BD = PD\sqrt{3}$.
Solving for $PD$:
$PD\sqrt{3} - PD = 8 \implies PD(\sqrt{3} - 1) = 8 \implies PD = \frac{8}{\sqrt{3} - 1}$.
Rationalizing the denominator:
$PD = \frac{8(\sqrt{3} + 1)}{3 - 1} = \frac{8(\sqrt{3} + 1)}{2} = 4(\sqrt{3} + 1) \, m$.
Height of the multi-storeyed building $PC = PD + DC = 4\sqrt{3} + 4 + 8 = 4\sqrt{3} + 12 = 4(3 + \sqrt{3}) \, m$.
Distance between the buildings $AC = PC = 4(3 + \sqrt{3}) \, m$.

(N/A) Let $A$ and $B$ represent points on the bank on opposite sides of the river,so that $AB$ is the width of the river. Let $P$ be a point on the bridge at a height of $3 \, m$ from the river level,i.e.,$PD = 3 \, m$,where $D$ is the point on the river surface directly below $P$.
In right-angled $\triangle APD$,$\angle PAD = 30^{\circ}$.
Using trigonometry,$\tan 30^{\circ} = \frac{PD}{AD}$.
$\frac{1}{\sqrt{3}} = \frac{3}{AD} \implies AD = 3\sqrt{3} \, m$.
In right-angled $\triangle PBD$,$\angle PBD = 45^{\circ}$.
Using trigonometry,$\tan 45^{\circ} = \frac{PD}{BD}$.
$1 = \frac{3}{BD} \implies BD = 3 \, m$.
The total width of the river is $AB = AD + BD = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1) \, m$.

(D) From the figure,let $AB$ be the vertical pole and $AC$ be the rope of length $20\, m$.
In the right-angled triangle $\triangle ABC$,the angle made by the rope with the ground is $\angle C = 30^{\circ}$.
We need to find the height of the pole,which is $AB$.
Using the trigonometric ratio $\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}}$,we have:
$\sin 30^{\circ} = \frac{AB}{AC}$
Since $\sin 30^{\circ} = \frac{1}{2}$ and $AC = 20\, m$,we get:
$\frac{1}{2} = \frac{AB}{20}$
$AB = \frac{20}{2} = 10\, m$.
Thus,the height of the pole is $10\, m$.

(N/A) Let $AB$ be the original height of the tree. Let the tree break at point $B$,and the top $A$ touches the ground at point $A^{\prime}$.
In the right-angled triangle $\triangle BCA^{\prime}$,where $\angle BCA^{\prime} = 90^{\circ}$ and $\angle BA^{\prime}C = 30^{\circ}$:
Given $CA^{\prime} = 8\, m$.
Using $\tan 30^{\circ} = \frac{BC}{CA^{\prime}}$:
$\frac{BC}{8} = \frac{1}{\sqrt{3}} \implies BC = \frac{8}{\sqrt{3}}\, m$.
Using $\cos 30^{\circ} = \frac{CA^{\prime}}{BA^{\prime}}$:
$\frac{8}{BA^{\prime}} = \frac{\sqrt{3}}{2} \implies BA^{\prime} = \frac{16}{\sqrt{3}}\, m$.
The total height of the tree is $AB = BC + BA^{\prime}$ (since $BA^{\prime} = BA$):
$AB = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}}\, m$.
Rationalizing the denominator:
$AB = \frac{24 \sqrt{3}}{3} = 8\sqrt{3}\, m$.
Thus,the height of the tree is $8\sqrt{3}\, m$.

(N/A) It can be observed that $AC$ and $PR$ are the slides for younger and elder children respectively.
In $\triangle ABC$:
$\sin 30^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC}$
$\frac{1}{2} = \frac{1.5}{AC}$
$AC = 3 \, m$
In $\triangle PQR$:
$\sin 60^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PQ}{PR}$
$\frac{\sqrt{3}}{2} = \frac{3}{PR}$
$PR = \frac{6}{\sqrt{3}} = 2\sqrt{3} \, m \approx 3.46 \, m$

(N/A) Let $AB$ be the tower and the angle of elevation from point $C$ (on the ground) is $30^{\circ}$.
In $\triangle ABC$,
$\frac{AB}{BC} = \tan 30^{\circ}$
$\frac{AB}{30} = \frac{1}{\sqrt{3}}$
$AB = \frac{30}{\sqrt{3}} = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3}\, m$
Therefore,the height of the tower is $10\sqrt{3}\, m$.

$(40\sqrt{3}\, m) $ Let $K$ be the position of the kite and $P$ be the point on the ground where the string is tied. Let $L$ be the point on the ground directly below the kite.
In the right-angled triangle $\triangle KLP$:
$KL = 60\, m$ (height of the kite)
$\angle KPL = 60^{\circ}$ (angle of inclination)
We need to find the length of the string,which is the hypotenuse $KP$.
Using the trigonometric ratio $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$:
$\sin 60^{\circ} = \frac{KL}{KP}$
$\frac{\sqrt{3}}{2} = \frac{60}{KP}$
$KP = \frac{60 \times 2}{\sqrt{3}}$
$KP = \frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$KP = \frac{120\sqrt{3}}{3} = 40\sqrt{3}\, m$
Thus,the length of the string is $40\sqrt{3}\, m$ (approximately $69.28\, m$).

(N/A) Let the boy be standing at point $A$ initially. He walks towards the building and reaches point $B$.
Let $PQ$ be the building of height $30 \, m$. The boy's height is $1.5 \, m$.
Therefore,the height of the triangle $PR = PQ - RQ = 30 - 1.5 = 28.5 \, m = \frac{57}{2} \, m$.
In $\triangle PAR$,$\tan 30^{\circ} = \frac{PR}{AR} \implies \frac{1}{\sqrt{3}} = \frac{57/2}{AR} \implies AR = \frac{57\sqrt{3}}{2} \, m$.
In $\triangle PBR$,$\tan 60^{\circ} = \frac{PR}{BR} \implies \sqrt{3} = \frac{57/2}{BR} \implies BR = \frac{57}{2\sqrt{3}} = \frac{19\sqrt{3}}{2} \, m$.
The distance walked towards the building is $AB = AR - BR$.
$AB = \frac{57\sqrt{3}}{2} - \frac{19\sqrt{3}}{2} = \frac{38\sqrt{3}}{2} = 19\sqrt{3} \, m$.
Thus,the boy walked $19\sqrt{3} \, m$ towards the building.

(N/A) Let $BC$ be the building,$AB$ be the transmission tower,and $D$ be the point on the ground from where the elevation angles are measured.
In $\triangle BCD$:
$\frac{BC}{CD} = \tan 45^{\circ}$
$\frac{20}{CD} = 1$
$CD = 20 \, m$
In $\triangle ACD$:
$\frac{AC}{CD} = \tan 60^{\circ}$
$\frac{AB + BC}{CD} = \sqrt{3}$
$\frac{AB + 20}{20} = \sqrt{3}$
$AB + 20 = 20\sqrt{3}$
$AB = 20\sqrt{3} - 20$
$AB = 20(\sqrt{3} - 1) \, m$
Therefore,the height of the transmission tower is $20(\sqrt{3} - 1) \, m$.

(N/A) Let $AB$ be the statue,$BC$ be the pedestal,and $D$ be the point on the ground from where the angles of elevation are measured.
In $\triangle BCD$,we have:
$\frac{BC}{CD} = \tan 45^{\circ}$
$\frac{BC}{CD} = 1$
$BC = CD$
In $\triangle ACD$,we have:
$\frac{AC}{CD} = \tan 60^{\circ}$
$\frac{AB + BC}{CD} = \sqrt{3}$
Since $CD = BC$,we can write:
$\frac{1.6 + BC}{BC} = \sqrt{3}$
$1.6 + BC = BC \sqrt{3}$
$1.6 = BC(\sqrt{3} - 1)$
$BC = \frac{1.6}{\sqrt{3} - 1}$
Rationalizing the denominator:
$BC = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$
$BC = \frac{1.6(\sqrt{3} + 1)}{3 - 1}$
$BC = \frac{1.6(\sqrt{3} + 1)}{2}$
$BC = 0.8(\sqrt{3} + 1) \, m$
Thus,the height of the pedestal is $0.8(\sqrt{3} + 1) \, m$.

(N/A) Let $AB$ be the building and $CD$ be the tower.
In $\triangle CDB$,
$\frac{CD}{BD} = \tan 60^{\circ}$
$\frac{50}{BD} = \sqrt{3}$
$BD = \frac{50}{\sqrt{3}}$
In $\triangle ABD$,
$\frac{AB}{BD} = \tan 30^{\circ}$
$AB = BD \times \tan 30^{\circ} = \frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{50}{3} = 16 \frac{2}{3} \, m$
Therefore,the height of the building is $16 \frac{2}{3} \, m$.

(N/A) Let $AB$ and $CD$ be the poles of equal height $h$,and $O$ be the point on the road $BD$ such that $BD = 80\, m$.
Let $BO = x\, m$,then $OD = (80 - x)\, m$.
In $\triangle ABO$,$\tan 60^{\circ} = \frac{AB}{BO} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \dots(1)$.
In $\triangle CDO$,$\tan 30^{\circ} = \frac{CD}{OD} \implies \frac{1}{\sqrt{3}} = \frac{h}{80 - x} \implies h = \frac{80 - x}{\sqrt{3}} \quad \dots(2)$.
Equating $(1)$ and $(2)$:
$x\sqrt{3} = \frac{80 - x}{\sqrt{3}}$
$3x = 80 - x$
$4x = 80 \implies x = 20\, m$.
So,$BO = 20\, m$ and $OD = 80 - 20 = 60\, m$.
Substituting $x$ in $(1)$:
$h = 20\sqrt{3}\, m$.
Thus,the height of the poles is $20\sqrt{3}\, m$ and the distances of the point from the poles are $20\, m$ and $60\, m$.

(N/A) Let the height of the $TV$ tower be $AB = h$ and the width of the canal be $BC = x$.
In $\triangle ABC$,we have:
$\frac{AB}{BC} = \tan 60^{\circ}$
$\frac{h}{x} = \sqrt{3} \implies h = x\sqrt{3} \quad \dots(1)$
In $\triangle ABD$,we have:
$\frac{AB}{BD} = \tan 30^{\circ}$
$\frac{h}{x + 20} = \frac{1}{\sqrt{3}} \implies h\sqrt{3} = x + 20 \quad \dots(2)$
Substituting the value of $h$ from equation $(1)$ into equation $(2)$:
$(x\sqrt{3})\sqrt{3} = x + 20$
$3x = x + 20$
$2x = 20 \implies x = 10 \, m$
Now,substituting $x = 10$ into equation $(1)$:
$h = 10\sqrt{3} \, m$
Therefore,the height of the tower is $10\sqrt{3} \, m$ and the width of the canal is $10 \, m$.

(N/A) Let $AB$ be the building of height $7 \,m$ and $CD$ be the cable tower.
In $\triangle ABD$,$\angle ABD = 90^{\circ}$ and $\angle ADB = 45^{\circ}$.
$\tan 45^{\circ} = \frac{AB}{BD}$
$1 = \frac{7}{BD}$
$BD = 7 \,m$.
Since $ABED$ is a rectangle,$AE = BD = 7 \,m$ and $ED = AB = 7 \,m$.
In $\triangle ACE$,$\angle AEC = 90^{\circ}$ and $\angle CAE = 60^{\circ}$.
$\tan 60^{\circ} = \frac{CE}{AE}$
$\sqrt{3} = \frac{CE}{7}$
$CE = 7\sqrt{3} \,m$.
The total height of the tower $CD = CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \,m$.
Thus,the height of the tower is $7(\sqrt{3} + 1) \,m$.

(N/A) Let $AB$ be the lighthouse and the two ships be at points $C$ and $D$ respectively.
In $\triangle ABC$,
$\frac{AB}{BC} = \tan 45^{\circ}$
$\frac{75}{BC} = 1$
$BC = 75 \, m$
In $\triangle ABD$,
$\frac{AB}{BD} = \tan 30^{\circ}$
$\frac{75}{BC + CD} = \frac{1}{\sqrt{3}}$
$\frac{75}{75 + CD} = \frac{1}{\sqrt{3}}$
$75\sqrt{3} = 75 + CD$
$CD = 75(\sqrt{3} - 1) \, m$
Therefore,the distance between the two ships is $75(\sqrt{3} - 1) \, m$.

(N/A) Let the initial position of the balloon be $A$ and its final position be $B$. Let $CD$ be the girl,where $CD = 1.2\, m$.
The height of the balloon from the ground is $88.2\, m$. The height of the balloon from the eye level of the girl is $88.2\, m - 1.2\, m = 87\, m$.
Let the eye level of the girl be the horizontal line $CG$. Thus,$AE = BG = 87\, m$.
In $\triangle ACE$,$\tan 60^{\circ} = \frac{AE}{CE} \implies \sqrt{3} = \frac{87}{CE} \implies CE = \frac{87}{\sqrt{3}} = 29\sqrt{3}\, m$.
In $\triangle BCG$,$\tan 30^{\circ} = \frac{BG}{CG} \implies \frac{1}{\sqrt{3}} = \frac{87}{CG} \implies CG = 87\sqrt{3}\, m$.
The distance travelled by the balloon is $EG = CG - CE = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}\, m$.

(B) Let $AB$ be the tower of height $h$ and $C$ be the initial position of the car. After $6$ seconds,the car reaches point $D$.
In $\triangle ADB$,$\tan 60^{\circ} = \frac{AB}{DB} \implies \sqrt{3} = \frac{h}{DB} \implies DB = \frac{h}{\sqrt{3}}$.
In $\triangle ABC$,$\tan 30^{\circ} = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{h}{BC} \implies BC = h\sqrt{3}$.
The distance covered in $6$ seconds is $CD = BC - DB = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$.
Since the speed is uniform,the time taken to cover distance $CD$ is $6$ seconds.
Therefore,time taken to cover distance $DB = \frac{h}{\sqrt{3}}$ is $\frac{6 \times (h/\sqrt{3})}{2h/\sqrt{3}} = \frac{6}{2} = 3$ seconds.

(N/A) Let $AQ$ be the tower and $R, S$ be the points at a distance of $4 \,m$ and $9 \,m$ from the base of the tower $(Q)$ respectively.
The angles of elevation are complementary. Therefore,if one angle is $\theta$,the other will be $(90^\circ - \theta)$.
In $\triangle AQR$,
$\tan \theta = \frac{AQ}{QR} = \frac{AQ}{4} \quad \dots(i)$
In $\triangle AQS$,
$\tan(90^\circ - \theta) = \frac{AQ}{SQ}$
$\cot \theta = \frac{AQ}{9} \quad \dots(ii)$
Multiplying equations $(i)$ and $(ii)$:
$\tan \theta \cdot \cot \theta = \left(\frac{AQ}{4}\right) \cdot \left(\frac{AQ}{9}\right)$
$1 = \frac{AQ^2}{36}$
$AQ^2 = 36$
$AQ = \sqrt{36} = 6 \,m$ (Since height cannot be negative).
Hence,the height of the tower is $6 \,m$.

(C) Let $BC = 6\, m$ be the height of the pole and $AB = 2\sqrt{3}\, m$ be the length of the shadow on the ground. Let the Sun's elevation angle be $\theta$.
In the right-angled $\triangle ABC$,we have:
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB}$
Substituting the given values:
$\tan \theta = \frac{6}{2\sqrt{3}}$
$\tan \theta = \frac{3}{\sqrt{3}}$
Rationalizing the denominator:
$\tan \theta = \frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}$
Since $\tan 60^{\circ} = \sqrt{3}$,we get:
$\theta = 60^{\circ}$
Thus,the Sun's elevation is $60^{\circ}$.

(B) False.
Let the height of the tower be $h$ and the length of the shadow be $x$. The angle of elevation of the sun is $\theta$. In the right-angled triangle formed by the tower and its shadow,we have $\tan(\theta) = \frac{h}{x}$.
As the length of the shadow $x$ increases,the value of the fraction $\frac{h}{x}$ decreases. Since $\tan(\theta)$ is an increasing function for $0^\circ < \theta < 90^\circ$,a decrease in $\tan(\theta)$ implies a decrease in the angle $\theta$. Therefore,as the shadow length increases,the angle of elevation of the sun decreases.

(B) False.
Let the height of the cloud above the surface of the lake be $H$. The reflection of the cloud in the lake will be at a depth of $H$ below the surface of the lake.
The man is standing at a height of $3 \ m$ above the surface of the lake. Let the observer's eye level be $O$.
The angle of elevation $\theta_1$ of the cloud is given by $\tan(\theta_1) = \frac{H - 3}{x}$,where $x$ is the horizontal distance from the man to the cloud.
The angle of depression $\theta_2$ of the reflection is given by $\tan(\theta_2) = \frac{H + 3}{x}$.
Since $H - 3 \neq H + 3$,the angle of elevation $\theta_1$ is not equal to the angle of depression $\theta_2$.

(FALSE) False.
$Case-I$: Let the height of the tower be $h$ and the distance from the base be $BC = x \ m$.
In $\triangle ABC$,$\tan 30^{\circ} = \frac{AC}{BC} = \frac{h}{x}$.
$\frac{1}{\sqrt{3}} = \frac{h}{x} \implies h = \frac{x}{\sqrt{3}}$ ... $(i)$
$Case-II$: By the given condition,the height of the tower is doubled,i.e.,$h' = 2h$.
Let the new angle of elevation be $\theta$.
In the new triangle,$\tan \theta = \frac{h'}{x} = \frac{2h}{x}$.
Substituting $h = \frac{x}{\sqrt{3}}$ from $(i)$:
$\tan \theta = \frac{2(x/\sqrt{3})}{x} = \frac{2}{\sqrt{3}} \approx 1.1547$.
Since $\tan 60^{\circ} = \sqrt{3} \approx 1.732$,and $\tan \theta \approx 1.1547 < 1.732$,it follows that $\theta < 60^{\circ}$.
Thus,the angle of elevation is not doubled.

(A) True.
$Case-I$: Let the height of a tower be $h$ and the distance of the point of observation from its foot be $x$.
In $\triangle ABC$,$\tan \theta_1 = \frac{h}{x}$,which implies $\theta_1 = \tan^{-1}(\frac{h}{x})$ ... $(i)$
$Case-II$: Now,the height of the tower is increased by $10 \%$,so the new height $h' = h + 0.1h = 1.1h = \frac{11h}{10}$.
The distance of the point of observation from its foot is also increased by $10 \%$,so the new distance $x' = x + 0.1x = 1.1x = \frac{11x}{10}$.
In the new triangle,$\tan \theta_2 = \frac{h'}{x'} = \frac{1.1h}{1.1x} = \frac{h}{x}$.
Thus,$\theta_2 = \tan^{-1}(\frac{h}{x})$ ... $(ii)$
From equations $(i)$ and $(ii)$,we get $\theta_1 = \theta_2$.
Therefore,the angle of elevation remains unchanged.

(A) Let $\overline{AC}$ be the ladder and $\overline{AB}$ be the wall.
Given that $AB = 3 \, m$,$\angle B = 90^{\circ}$,and $\angle C = 30^{\circ}$.
In $\triangle ABC$,we have:
$\sin C = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{AB}{AC}$
$\sin 30^{\circ} = \frac{3}{AC}$
Since $\sin 30^{\circ} = \frac{1}{2}$,we have:
$\frac{1}{2} = \frac{3}{AC}$
$AC = 3 \times 2 = 6 \, m$.
Thus,the length of the ladder is $6 \, m$.

(B) Let $\overline{AB}$ be the building and $C$ be the position of the man on the ground.
Given that the distance $BC = 10 \, m$.
The angle of depression from the top $A$ to the man at $C$ is $45^{\circ}$.
Since the line of sight is parallel to the ground,the angle of elevation from $C$ to $A$ is equal to the angle of depression,so $\angle ACB = 45^{\circ}$.
In the right-angled triangle $\Delta ABC$,we have:
$\tan(C) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AB}{BC}$
$\tan(45^{\circ}) = \frac{AB}{10}$
Since $\tan(45^{\circ}) = 1$,we have:
$1 = \frac{AB}{10}$
$AB = 10 \, m$.
Thus,the height of the building is $10 \, m$.

(C) Let $\overline{AB}$ be the pole and $\overline{AC}$ be the wire fixed on the ground at point $C$.
Given: $AC = 7 \, m$,$m\angle B = 90^{\circ}$,and $m\angle C = 30^{\circ}$.
In $\Delta ABC$,we have:
$\sin C = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{AB}{AC}$
$\sin 30^{\circ} = \frac{AB}{7}$
Since $\sin 30^{\circ} = \frac{1}{2}$,we have:
$\frac{1}{2} = \frac{AB}{7}$
$AB = \frac{7}{2} = 3.5 \, m$.
Thus,the height of the pole is $3.5 \, m$.

(D) Let $AB$ be the tower and $C$ be the stone on the ground.
Given that the height of the tower $AB = 30 \, m$.
The angle of depression from the top $A$ to the stone $C$ is $45^{\circ}$.
Since the horizontal line through $A$ is parallel to the ground $BC$,the angle of elevation from $C$ to $A$ is equal to the angle of depression,so $\angle ACB = 45^{\circ}$.
In the right-angled triangle $\triangle ABC$,we have:
$\tan(45^{\circ}) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$
Since $\tan(45^{\circ}) = 1$,we have:
$1 = \frac{30}{BC}$
$BC = 30 \, m$.
Thus,the distance between the stone and the tower is $30 \, m$.

(A) From the figure,we can see that the angle of elevation of the sun is the angle of elevation of the top of the tower from the end of its shadow.
Here,$AB$ is the tower and $BC$ is its shadow.
Given,$AB = 30 \ m$,$\angle B = 90^\circ$ and $\angle C = 45^\circ$.
In $\Delta ABC$,$\tan C = \frac{AB}{BC}$.
$\tan 45^\circ = \frac{30}{BC}$.
$1 = \frac{30}{BC}$.
$BC = 30 \ m$.
Thus,the length of the shadow of the tower is $30 \ m$.

(B) Let $\overline{AB}$ be the vertical pole and $C$ be the point of observation.
Given that the distance from the base of the pole is $BC = 20\, m$ and the angle of elevation is $\angle C = 60^\circ$.
In the right-angled triangle $\Delta ABC$,where $\angle B = 90^\circ$:
$\tan C = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AB}{BC}$
$\tan 60^\circ = \frac{AB}{20}$
Since $\tan 60^\circ = \sqrt{3} \approx 1.732$:
$\sqrt{3} = \frac{AB}{20}$
$AB = 20 \times \sqrt{3}$
$AB = 20 \times 1.732 = 34.64\, m$.
Rounding to one decimal place,the height of the pole is $34.6\, m$.

(C) Let $\overline{AB}$ be the tower and $C$ be the object.
Given that the distance of the object from the base of the tower is $BC = 30 \ m$.
The angle of depression from the top $A$ to the object $C$ is $45^\circ$,so $\angle XAC = 45^\circ$.
Since the line of sight is parallel to the ground,$\angle ACB = \angle XAC = 45^\circ$ (alternate interior angles).
In the right-angled triangle $\Delta ABC$,where $\angle B = 90^\circ$:
$\tan(C) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AB}{BC}$
$\tan(45^\circ) = \frac{AB}{30}$
Since $\tan(45^\circ) = 1$,we have:
$1 = \frac{AB}{30}$
$AB = 30 \ m$.
Thus,the height of the tower is $30 \ m$.

(D) Let $B$ be the base of the building,$C$ be the first floor,and $A$ be the fifteenth floor. Let $P$ be the point of observation such that $BP = 50 \ m$.
In $\Delta CBP$,$\angle B = 90^{\circ}$ and $\angle CPB = 30^{\circ}$.
$\tan 30^{\circ} = \frac{CB}{BP}$
$\frac{1}{\sqrt{3}} = \frac{CB}{50}$
$CB = \frac{50}{\sqrt{3}} = \frac{50 \times 1.732}{3} \approx 28.87 \ m$.
In $\Delta ABP$,$\angle B = 90^{\circ}$ and $\angle APB = 60^{\circ}$.
$\tan 60^{\circ} = \frac{AB}{BP}$
$\sqrt{3} = \frac{AB}{50}$
$AB = 50 \times \sqrt{3} = 50 \times 1.732 = 86.6 \ m$.
The distance between the first floor and the fifteenth floor is $AC = AB - CB$.
$AC = 86.6 - 28.87 = 57.73 \ m$.
Rounding to the nearest provided option,the correct answer is $57.5 \ m$.

(N/A) Let $\overline{PM}$ be the tree and $\overline{OM}$ be the width of the river. $O$ and $A$ are the points of observation.
Given: $OA = 40 \ m$,$m\angle A = 30^{\circ}$,$m\angle O = 60^{\circ}$,and $m\angle M = 90^{\circ}$.
Let $OM = x \ m$ and $PM = h \ m$.
Then,$MA = OM + OA = (x + 40) \ m$.
In $\Delta PMO$,$m\angle M = 90^{\circ}$.
$\tan O = \frac{PM}{OM} \implies \tan 60^{\circ} = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies h = \sqrt{3}x \quad ...(1)$
In $\Delta PMA$,$m\angle M = 90^{\circ}$.
$\tan A = \frac{PM}{AM} \implies \tan 30^{\circ} = \frac{h}{x + 40} \implies \frac{1}{\sqrt{3}} = \frac{h}{x + 40} \implies h = \frac{x + 40}{\sqrt{3}} \quad ...(2)$
From $(1)$ and $(2)$:
$\sqrt{3}x = \frac{x + 40}{\sqrt{3}} \implies 3x = x + 40 \implies 2x = 40 \implies x = 20 \ m$.
Now,$h = \sqrt{3} \times 20 = 20\sqrt{3} \approx 20 \times 1.732 = 34.64 \ m$.
Thus,the width of the river is $20 \ m$ and the height of the tree is $34.64 \ m$.

(C) Let $CD$ be the tower of height $150\, m$. Let $A$ and $B$ be the two houses situated to the west and east of the tower respectively.
In $\Delta CDA$,$\angle D = 90^{\circ}$ and $\angle CAD = 45^{\circ}$.
$\tan(45^{\circ}) = \frac{CD}{AD} \implies 1 = \frac{150}{AD} \implies AD = 150\, m$.
In $\Delta CDB$,$\angle D = 90^{\circ}$ and $\angle CBD = 30^{\circ}$.
$\tan(30^{\circ}) = \frac{CD}{BD} \implies \frac{1}{\sqrt{3}} = \frac{150}{BD} \implies BD = 150\sqrt{3} \approx 150 \times 1.73 = 259.5\, m$.
The total distance between the houses is $AB = AD + BD = 150 + 259.5 = 409.5\, m$.

(D) Let $\overline{AB}$ be the hill and $C$ and $D$ be two houses situated to the west and east of the hill respectively. Given,$AB = 340 \ m$.
The angle of depression from the top of the hill to house $C$ is $60^{\circ}$ and to house $D$ is $30^{\circ}$.
Therefore,the angle of elevation $\angle ACB = 60^{\circ}$ and $\angle ADB = 30^{\circ}$ (alternate interior angles).
In right-angled $\Delta ABC$,$\tan(60^{\circ}) = \frac{AB}{BC}$.
$\sqrt{3} = \frac{340}{BC} \implies BC = \frac{340}{\sqrt{3}} = \frac{340 \times 1.732}{3} \approx 196.3 \ m$.
In right-angled $\Delta ABD$,$\tan(30^{\circ}) = \frac{AB}{BD}$.
$\frac{1}{\sqrt{3}} = \frac{340}{BD} \implies BD = 340 \times \sqrt{3} = 340 \times 1.732 \approx 588.9 \ m$.
The total distance between the two houses $CD = BC + BD$.
$CD = 196.3 + 588.9 = 785.2 \ m$.
Rounding to the nearest provided option,the distance is $785.4 \ m$.

(A) Let the total height of the tree be $AC = 15\ m$. The tree breaks at point $B$ at a height of $BC = 5\ m$ from the ground.
The broken part $AB$ bends to touch the ground at point $D$. Thus,the length of the broken part is $BD = AB = AC - BC = 15\ m - 5\ m = 10\ m$.
In the right-angled triangle $\Delta BCD$ (where $\angle C = 90^{\circ}$),we have:
$\sin D = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{BD}$
Substituting the values:
$\sin D = \frac{5}{10} = \frac{1}{2}$
Since $\sin 30^{\circ} = \frac{1}{2}$,we get:
$D = 30^{\circ}$
Therefore,the angle made by the broken part of the tree with the ground is $30^{\circ}$.

(B) In the figure,$\overline{AB}$ is a tree and $C$ and $D$ represent two positions of the boat.
Suppose the constant velocity of the boat is $v \text{ m/min}$ and the boat takes time $t \text{ min}$ to reach from $D$ to $B$.
Let $AB = h$.
Now,according to the formula,$\text{Distance} = \text{Speed} \times \text{Time}$,
$CD = 10v$ and $BD = vt$.
Now,in $\Delta ABD$,$\tan 60^{\circ} = \frac{AB}{BD}$.
$\therefore \sqrt{3} = \frac{h}{vt} \implies h = vt\sqrt{3} \quad ....(1)$
In $\Delta ABC$,$\tan 30^{\circ} = \frac{AB}{BC}$.
$\therefore \frac{1}{\sqrt{3}} = \frac{h}{vt + 10v} \implies h = \frac{vt + 10v}{\sqrt{3}} \quad ....(2)$
Equating $(1)$ and $(2)$,
$vt\sqrt{3} = \frac{vt + 10v}{\sqrt{3}}$
$3vt = vt + 10v$
$2vt = 10v$
$t = 5$.
Thus,the boat takes $5$ minutes to reach the bank.

(C) Let $AC$ be the hill of height $100 \, m$ and $ED$ be the tower.
Let $EB$ be the horizontal line from $E$ to the hill $AC$,where $B$ is on $AC$.
Given: $AC = 100 \, m$,$\angle XAE = 30^{\circ}$,and $\angle XAD = 45^{\circ}$.
Since $AX \parallel ED$,we have $\angle AEB = 30^{\circ}$ and $\angle ADC = 45^{\circ}$ (alternate interior angles).
In right-angled $\Delta ACD$,$\tan(45^{\circ}) = \frac{AC}{DC}$.
$1 = \frac{100}{DC} \implies DC = 100 \, m$.
Since $EDCB$ is a rectangle,$EB = DC = 100 \, m$ and $BC = ED$.
In right-angled $\Delta ABE$,$\tan(30^{\circ}) = \frac{AB}{EB}$.
$\frac{1}{\sqrt{3}} = \frac{AB}{100} \implies AB = \frac{100}{\sqrt{3}} \approx 57.74 \, m$.
Height of the tower $ED = BC = AC - AB = 100 - 57.74 = 42.26 \, m$.
Rounding to the nearest integer,the height is $42 \, m$.

(D) Let $\overline{AB}$ be the flag-staff on top of the building $\overline{BC}$ and $D$ be the point of observation.
Given,$CD = 15\, m$.
In $\Delta BCD$,$\angle C = 90^{\circ}$ and $\angle BDC = 45^{\circ}$.
$\tan(45^{\circ}) = \frac{BC}{CD} \implies 1 = \frac{BC}{15} \implies BC = 15\, m$.
In $\Delta ACD$,$\angle C = 90^{\circ}$ and $\angle ADC = 60^{\circ}$.
$\tan(60^{\circ}) = \frac{AC}{CD} \implies \sqrt{3} = \frac{AC}{15} \implies AC = 15\sqrt{3} \approx 15 \times 1.732 = 25.98\, m$.
The height of the flag-staff $AB = AC - BC = 25.98 - 15 = 10.98\, m$. Using $\sqrt{3} \approx 1.73$,$AC = 25.95\, m$,so $AB = 25.95 - 15 = 10.95\, m$.

(A) Let $AB$ be the tower of height $510 \ m$. Let $C$ and $D$ be the two houses on the west and east sides of the tower respectively.
In $\Delta ABC$,$\tan 60^{\circ} = \frac{AB}{BC} \implies \sqrt{3} = \frac{510}{BC} \implies BC = \frac{510}{\sqrt{3}} = 170\sqrt{3} \ m$.
In $\Delta ABD$,$\tan 30^{\circ} = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{510}{BD} \implies BD = 510\sqrt{3} \ m$.
The distance between the two houses is $CD = BC + BD = 170\sqrt{3} + 510\sqrt{3} = 680\sqrt{3} \ m$.
Using $\sqrt{3} \approx 1.732$,$CD = 680 \times 1.732 = 1177.76 \ m \approx 1178.1 \ m$ (using the approximation $1.7325$ or similar).
Thus,the distance is $1178.1 \ m$.

(B) Let $\overline{AB}$ be the hill of height $h \, m$. Let $D$ be the first point of observation and $C$ be the second point of observation.
Given: $DC = 30 \, m$,$\angle ADB = 30^\circ$,and $\angle ACB = 45^\circ$.
Let $BC = x \, m$.
In $\Delta ABC$,$\tan 45^\circ = \frac{AB}{BC} = \frac{h}{x}$.
Since $\tan 45^\circ = 1$,we have $1 = \frac{h}{x}$,so $x = h$ $(1)$.
In $\Delta ABD$,$\tan 30^\circ = \frac{AB}{BD} = \frac{h}{x + 30}$.
Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{h}{x + 30}$.
$x + 30 = h\sqrt{3}$.
Substituting $x = h$ from $(1)$ into this equation:
$h + 30 = h\sqrt{3}$.
$30 = h(\sqrt{3} - 1)$.
$h = \frac{30}{\sqrt{3} - 1} = \frac{30}{1.732 - 1} = \frac{30}{0.732} \approx 40.98 \, m$.
Rounding to the nearest option,the height is approximately $41.10 \, m$.

(N/A) Let $\overline{CD}$ be the tower and $A$ be the point of observation $h \text{ m}$ above the ground level.
Let $\overline{AE} \perp \overline{CD}$,where $E$ lies on $\overline{CD}$.
Then,$\angle DAE = \alpha$,$\angle EAC = \beta$,and $AB = h \text{ m}$.
Let $CD = x \text{ m}$ and $BC = y \text{ m}$.
Then $AE = BC = y \text{ m}$ and $CE = AB = h \text{ m}$.
Also,$DE = DC - CE = (x - h) \text{ m}$.
In $\Delta ABC$,$\angle B = 90^{\circ}$.
$\therefore \tan \beta = \frac{AB}{BC} = \frac{h}{y} \implies y = \frac{h}{\tan \beta} \quad \dots(1)$
In $\Delta DEA$,$\angle E = 90^{\circ}$.
$\therefore \tan \alpha = \frac{DE}{AE} = \frac{x - h}{y} \implies y = \frac{x - h}{\tan \alpha} \quad \dots(2)$
From $(1)$ and $(2)$:
$\frac{h}{\tan \beta} = \frac{x - h}{\tan \alpha}$
$h \tan \alpha = (x - h) \tan \beta$
$h \tan \alpha = x \tan \beta - h \tan \beta$
$x \tan \beta = h \tan \alpha + h \tan \beta$
$x \tan \beta = h(\tan \alpha + \tan \beta)$
$x = \frac{h(\tan \alpha + \tan \beta)}{\tan \beta}$
Thus,the height of the tower is $\frac{h(\tan \alpha + \tan \beta)}{\tan \beta} \text{ m}$.

(N/A) Let $\overline{AB}$ be the electric pole and $\overline{AC}$ be the cable fixed at point $C$ on the ground.
In the right-angled $\Delta ABC$,$\angle B = 90^{\circ}$,$\angle C = \theta$,and the base $BC = a \text{ m}$.
To find the height of the pole $(AB)$:
Using the trigonometric ratio $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$.
$\tan \theta = \frac{AB}{a}$
$AB = a \tan \theta \text{ m}$.
To find the length of the cable $(AC)$:
Using the trigonometric ratio $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{AC}{BC}$.
$\sec \theta = \frac{AC}{a}$
$AC = a \sec \theta \text{ m}$.
Thus,the height of the pole is $a \tan \theta \text{ m}$ and the length of the cable is $a \sec \theta \text{ m}$.

(N/A) Let $AB$ be the tower of height $h$ and $C$ and $D$ be the two vehicles on the same side of the tower,where $C$ is closer to the tower.
Given $CD = b$. Let $BC = x$.
Then $BD = BC + CD = x + b$.
The angles of depression from the top $A$ to $C$ and $D$ are $\alpha$ and $\beta$ respectively.
Thus,$\angle ACB = \alpha$ and $\angle ADB = \beta$ (alternate interior angles).
In right-angled $\Delta ABC$,$\tan \alpha = \frac{AB}{BC} = \frac{h}{x} \implies x = \frac{h}{\tan \alpha} \quad (1)$.
In right-angled $\Delta ABD$,$\tan \beta = \frac{AB}{BD} = \frac{h}{x + b} \implies x + b = \frac{h}{\tan \beta} \implies x = \frac{h}{\tan \beta} - b \quad (2)$.
Equating $(1)$ and $(2)$,we get $\frac{h}{\tan \alpha} = \frac{h}{\tan \beta} - b$.
Rearranging the terms: $b = \frac{h}{\tan \beta} - \frac{h}{\tan \alpha} = h \left( \frac{\tan \alpha - \tan \beta}{\tan \alpha \tan \beta} \right)$.
Therefore,$h = \frac{b \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}$.

(A) Let the height of the hill be $H = 100\, m$. Let the height of the tower be $h$. Let the distance between the hill and the tower be $x$.
From the geometry of the problem,for the bottom of the tower (angle of depression $45^{\circ}$):
$\tan(45^{\circ}) = \frac{H}{x} \implies 1 = \frac{100}{x} \implies x = 100\, m$.
For the top of the tower (angle of depression $30^{\circ}$),the height of the hill above the level of the tower top is $H - h$:
$\tan(30^{\circ}) = \frac{H - h}{x} \implies \frac{1}{\sqrt{3}} = \frac{100 - h}{100}$.
$100 - h = \frac{100}{\sqrt{3}} \implies h = 100 - \frac{100}{\sqrt{3}} = 100(1 - \frac{1}{\sqrt{3}})$.
Using $\sqrt{3} \approx 1.732$,$h = 100(1 - 0.577) = 100(0.423) = 42.3\, m$ (approx).

(D) Let the height of the tower be $h$ and the length of the shadow be $s = 90\, m$.
The angle of elevation $\theta = 30^{\circ}$.
In the right-angled triangle formed by the tower and its shadow,we have $\tan(\theta) = \frac{\text{height}}{\text{shadow length}}$.
$\tan(30^{\circ}) = \frac{h}{90}$.
Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{h}{90}$.
$h = \frac{90}{\sqrt{3}} = \frac{90 \times \sqrt{3}}{3} = 30\sqrt{3}$.
Using $\sqrt{3} \approx 1.732$,we get $h = 30 \times 1.732 = 51.96\, m$.
Rounding to one decimal place,the height is $51.9\, m$.