Height and Distance Questions in English

(C) Let the height of the tower be $h$ and the distance from the tower to the first point be $x$.
From the right-angled triangle,we have:
$\tan(60^\circ) = \frac{h}{OA}$ $\Rightarrow \sqrt{3} = \frac{h}{OA}$ $\Rightarrow OA = \frac{h}{\sqrt{3}}$.
$\tan(30^\circ) = \frac{h}{OB}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{20 + OA}$ $\Rightarrow 20 + OA = h\sqrt{3}$.
Substituting $OA = \frac{h}{\sqrt{3}}$ into the equation:
$20 + \frac{h}{\sqrt{3}} = h\sqrt{3}$
$20 = h\sqrt{3} - \frac{h}{\sqrt{3}}$
$20 = h \left( \frac{3 - 1}{\sqrt{3}} \right) = h \left( \frac{2}{\sqrt{3}} \right)$
$h = \frac{20 \times \sqrt{3}}{2} = 10\sqrt{3} \, m$.

(C) Let the height of the tower be $h = 20 \, m$ and the distance from the base be $d$.
In the right-angled triangle formed by the tower and the ground,we have:
$\tan(30^\circ) = \frac{\text{height}}{\text{base}} = \frac{20}{d}$
Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$,we get:
$\frac{1}{\sqrt{3}} = \frac{20}{d}$
$d = 20\sqrt{3} \, m$
Thus,the correct option is $C$.

(C) Let $h$ be the height of the tower $OP$,where $O$ is the base of the tower.
In $\triangle OAP$,$\tan \alpha = \frac{h}{OA} \Rightarrow OA = h \cot \alpha$.
In $\triangle OBP$,$\tan \beta = \frac{h}{OB} \Rightarrow OB = h \cot \beta$.
Since $OA$ is south and $OB$ is east,$\triangle OAB$ is a right-angled triangle at $O$.
By Pythagoras theorem in $\triangle OAB$,$OA^2 + OB^2 = AB^2$.
$(h \cot \alpha)^2 + (h \cot \beta)^2 = d^2$.
$h^2 (\cot^2 \alpha + \cot^2 \beta) = d^2$.
$h^2 = \frac{d^2}{\cot^2 \alpha + \cot^2 \beta}$.
$h = \frac{d}{\sqrt{\cot^2 \alpha + \cot^2 \beta}}$.

(A) Let the height of the tree be $h$ and the breadth of the river be $b$.
From the given information:
In the first triangle,$\tan 60^\circ = \frac{h}{b} \Rightarrow h = b \tan 60^\circ = b\sqrt{3}$.
In the second triangle,$\tan 30^\circ = \frac{h}{b + 40} \Rightarrow h = (b + 40) \tan 30^\circ = \frac{b + 40}{\sqrt{3}}$.
Equating the two expressions for $h$:
$b\sqrt{3} = \frac{b + 40}{\sqrt{3}}$
$3b = b + 40$
$2b = 40$
$b = 20 \ m$.

(B) Let the total height of the pole be $H$. The lower part is $\frac{H}{3}$ and the upper part is $\frac{2H}{3}$. Let the point be at a distance $d = 20 \, m$ from the base. Let $\alpha$ be the angle subtended by the lower part and $\beta$ be the angle subtended by the whole pole at the point. The angle subtended by the upper part is $\theta = \beta - \alpha$,where $\tan \theta = \frac{1}{2}$.
We have $\tan \alpha = \frac{H/3}{d} = \frac{H}{3d}$ and $\tan \beta = \frac{H}{d}$.
Using the formula $\tan(\beta - \alpha) = \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha}$,we get:
$\frac{1}{2} = \frac{\frac{H}{d} - \frac{H}{3d}}{1 + (\frac{H}{d})(\frac{H}{3d})} = \frac{\frac{2H}{3d}}{1 + \frac{H^2}{3d^2}} = \frac{2Hd}{3d^2 + H^2}$.
Thus,$3d^2 + H^2 = 4Hd$,which simplifies to $H^2 - 4dH + 3d^2 = 0$.
Substituting $d = 20$,we get $H^2 - 4(20)H + 3(20^2) = 0$,so $H^2 - 80H + 1200 = 0$.
Factoring the quadratic equation: $(H - 20)(H - 60) = 0$.
Therefore,the possible heights of the pole are $H = 20 \, m$ and $H = 60 \, m$.

(D) Let $H = 60 \ m$ be the height of the tower and $h$ be the height of the house. Let $d$ be the distance between the tower and the house.
From the geometry of the problem:
$\tan \alpha = \frac{H - h}{d} \Rightarrow d = \frac{60 - h}{\tan \alpha}$
$\tan \beta = \frac{H}{d} \Rightarrow d = \frac{60}{\tan \beta}$
Equating the two expressions for $d$:
$\frac{60 - h}{\tan \alpha} = \frac{60}{\tan \beta}$
$(60 - h) \tan \beta = 60 \tan \alpha$
$60 \tan \beta - h \tan \beta = 60 \tan \alpha$
$h \tan \beta = 60(\tan \beta - \tan \alpha)$
$h = 60 \left( \frac{\sin \beta}{\cos \beta} - \frac{\sin \alpha}{\cos \alpha} \right) \cdot \frac{\cos \beta}{\sin \beta}$
$h = 60 \left( \frac{\sin \beta \cos \alpha - \cos \beta \sin \alpha}{\cos \beta \cos \alpha} \right) \cdot \frac{\cos \beta}{\sin \beta}$
$h = \frac{60 \sin(\beta - \alpha)}{\cos \alpha \sin \beta}$
Comparing this with the given expression $\frac{60 \sin(\beta - \alpha)}{x}$,we get $x = \cos \alpha \sin \beta$.

(C) Let $h$ be the height of the tree. Let the car be at point $A$ initially and at point $B$ after $3 \text{ minutes}$.
In the right-angled triangle formed with the $30^\circ$ angle,the distance from the tree base is $x_1 = h \cot(30^\circ) = h\sqrt{3}$.
In the right-angled triangle formed with the $60^\circ$ angle,the distance from the tree base is $x_2 = h \cot(60^\circ) = \frac{h}{\sqrt{3}}$.
The distance traveled in $3 \text{ minutes}$ is $d = x_1 - x_2 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h\left(\frac{3-1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}}$.
Since the car travels distance $\frac{2h}{\sqrt{3}}$ in $3 \text{ minutes}$,the speed $v = \frac{2h}{3\sqrt{3}} \text{ per minute}$.
The remaining distance to the tree is $x_2 = \frac{h}{\sqrt{3}}$.
Time taken to cover the remaining distance $t = \frac{\text{distance}}{\text{speed}} = \frac{h/\sqrt{3}}{2h/(3\sqrt{3})} = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{2h} = \frac{3}{2} = 1.5 \text{ minutes}$.

(A) Let the distance between the two houses be $d$. The window is at a height of $64 \ m$. The house of height $100 \ m$ is divided into two parts by the horizontal line from the window: one part of height $64 \ m$ (below the window) and another part of height $(100 - 64) = 36 \ m$ (above the window).
Let $\theta$ be the angle that the lower part makes with the horizontal line of distance $d$. Then $\tan \theta = \frac{64}{d}$.
The upper part makes an angle $(90^\circ - \theta)$ with the horizontal line. Then $\tan(90^\circ - \theta) = \frac{36}{d}$,which implies $\cot \theta = \frac{36}{d}$.
Multiplying the two equations: $\tan \theta \times \cot \theta = \frac{64}{d} \times \frac{36}{d}$.
$1 = \frac{64 \times 36}{d^2}$.
$d^2 = 64 \times 36$.
$d = \sqrt{64 \times 36} = 8 \times 6 = 48 \ m$.

(A) Let $OP = l$ be the length of the pole,inclined at $10^\circ$ to the vertical. Let $O$ be the base of the pole and $S$ be the tip of the shadow. The shadow length $OS = 2.05 \text{ m}$.
In $\triangle SPO$,the angle of elevation of the sun is $\angle OSP = 38^\circ$.
The angle between the pole and the vertical is $10^\circ$. The angle between the vertical and the ground is $90^\circ$.
Thus,the angle between the pole and the ground is $\angle SOP = 90^\circ + 10^\circ = 100^\circ$.
The sum of angles in $\triangle SPO$ is $180^\circ$,so $\angle SPO = 180^\circ - (38^\circ + 100^\circ) = 180^\circ - 138^\circ = 42^\circ$.
Using the Sine Rule in $\triangle SPO$:
$\frac{OP}{\sin \angle OSP} = \frac{OS}{\sin \angle SPO}$
$\frac{l}{\sin 38^\circ} = \frac{2.05}{\sin 42^\circ}$
$l = \frac{2.05 \sin 38^\circ}{\sin 42^\circ}$

(B) Let the height of the tower be $h \ m$.
Given that the distance from the base is $20 \ m$ and the angle of elevation is $45^\circ$.
Using the trigonometric ratio $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$:
$\tan(45^\circ) = \frac{h}{20}$
Since $\tan(45^\circ) = 1$,we have:
$1 = \frac{h}{20}$
$h = 20 \ m$.
Therefore,the height of the tower is $20 \ m$.

(C) Let the height of the first tower be $h \ m$ and the height of the second tower be $H = 150 \ m$.
The horizontal distance between them is $d = 60 \ m$.
From the top of the second tower,the angle of depression to the top of the first tower is $30^\circ$. This implies the angle of elevation from the top of the first tower to the top of the second tower is also $30^\circ$.
Considering the right-angled triangle formed by the difference in heights and the horizontal distance:
$\tan(30^\circ) = \frac{H - h}{d}$
$\frac{1}{\sqrt{3}} = \frac{150 - h}{60}$
$150 - h = \frac{60}{\sqrt{3}}$
$150 - h = 20\sqrt{3}$
$h = 150 - 20\sqrt{3} \ m$.
Thus,the height of the first tower is $(150 - 20\sqrt{3}) \ m$.

(B) Let the height of the lighthouse be $h = 60 \ m$ and the distance of the boat from the foot of the lighthouse be $x \ m$.
In the right-angled triangle formed by the lighthouse and the sea level,the angle of elevation of the top from the boat is equal to the angle of depression,which is $15^\circ$.
Thus,$\tan(15^\circ) = \frac{\text{height}}{\text{distance}} = \frac{60}{x}$.
Therefore,$x = \frac{60}{\tan(15^\circ)} = 60 \cot(15^\circ)$.
Using the value $\cot(15^\circ) = \frac{\cos(15^\circ)}{\sin(15^\circ)} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$.
So,$x = 60 \left( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \right) \ m$.
Thus,the correct option is $B$.

(A) Let the height of the cliff be $h \ m$ and the distance of the observer from the base of the cliff be $x \ m$.
From the given information,the height of the tower is $60 \ m$.
In the right-angled triangle formed by the cliff and the observer,we have:
$\tan(30^\circ) = \frac{h}{x} \Rightarrow x = h \cot(30^\circ) = h\sqrt{3}$.
In the right-angled triangle formed by the tower and the cliff together,the total height is $(h + 60) \ m$:
$\tan(60^\circ) = \frac{h + 60}{x} \Rightarrow x = \frac{h + 60}{\tan(60^\circ)} = \frac{h + 60}{\sqrt{3}}$.
Equating the two expressions for $x$:
$h\sqrt{3} = \frac{h + 60}{\sqrt{3}}$
$3h = h + 60$
$2h = 60$
$h = 30 \ m$.

(B) Let the tower be $OT$ with height $H$,where $O$ is the foot of the tower. Let $A$ be a point on the ground such that $\angle OAT = \alpha$. Thus,in $\triangle AOT$,$\tan \alpha = \frac{H}{AO}$,which implies $AO = H \cot \alpha$.
Let $P$ be a point at height $l$ vertically above $A$,so $PA = l$. The angle of depression of the foot $O$ from $P$ is $\beta$,so $\angle APO = 90^\circ - \beta$ or $\angle POA = \beta$. In $\triangle PAO$,$\tan \beta = \frac{PA}{AO} = \frac{l}{AO}$,which implies $AO = l \cot \beta$.
Equating the two expressions for $AO$: $H \cot \alpha = l \cot \beta$.
Therefore,$H = l \frac{\cot \beta}{\cot \alpha} = l \tan \alpha \cot \beta$.

(B) Let $O$ be the base of the tower and $H = 100 \ m$ be its height.
In $\triangle AOT$,where $T$ is the top of the tower,$\tan 30^\circ = \frac{H}{OA} \implies OA = \frac{100}{\tan 30^\circ} = 100\sqrt{3} \ m$.
In $\triangle BOT$,$\tan 45^\circ = \frac{H}{OB} \implies OB = \frac{100}{\tan 45^\circ} = 100 \ m$.
Since $OA$ is due south and $OB$ is due west,$\angle AOB = 90^\circ$.
In right-angled $\triangle AOB$,$AB = \sqrt{OA^2 + OB^2} = \sqrt{(100\sqrt{3})^2 + 100^2} = \sqrt{30000 + 10000} = \sqrt{40000} = 200 \ m$.

$(B)$ Let the height of the aeroplane be $H = 1 \ km$. Let the distance covered by the aeroplane in $10$ seconds be $d$.
From the geometry of the problem,we have:
$d = H \cot(30^\circ) - H \cot(60^\circ)$
$d = 1 \times \sqrt{3} - 1 \times \frac{1}{\sqrt{3}} = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \ km$.
Time taken $t = 10 \ seconds = \frac{10}{3600} \ hours = \frac{1}{360} \ hours$.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{2/\sqrt{3}}{1/360} = \frac{2}{\sqrt{3}} \times 360 = \frac{720}{\sqrt{3}} = 240\sqrt{3} \ km/h$.

(B) Let the point of observation be $P$ at a height $a$ above the lake surface. Let the cloud be at $C$ at a height $H$ above the lake surface. The reflection of the cloud is at $C'$ at a depth $H$ below the lake surface.
In the right-angled triangle formed by the cloud,the point $P$,and the horizontal line,we have $\tan \alpha = \frac{H - a}{x}$,where $x$ is the horizontal distance.
So,$x = (H - a) \cot \alpha$.
In the right-angled triangle formed by the reflection $C'$,the point $P$,and the horizontal line,we have $\tan \beta = \frac{H + a}{x}$.
So,$x = (H + a) \cot \beta$.
Equating the two expressions for $x$: $(H - a) \cot \alpha = (H + a) \cot \beta$.
$H \cot \alpha - a \cot \alpha = H \cot \beta + a \cot \beta$.
$H(\cot \alpha - \cot \beta) = a(\cot \alpha + \cot \beta)$.
$H \left( \frac{\cos \alpha}{\sin \alpha} - \frac{\cos \beta}{\sin \beta} \right) = a \left( \frac{\cos \alpha}{\sin \alpha} + \frac{\cos \beta}{\sin \beta} \right)$.
$H \left( \frac{\sin \beta \cos \alpha - \cos \beta \sin \alpha}{\sin \alpha \sin \beta} \right) = a \left( \frac{\sin \beta \cos \alpha + \cos \beta \sin \alpha}{\sin \alpha \sin \beta} \right)$.
$H \sin(\beta - \alpha) = a \sin(\alpha + \beta)$.
$H = \frac{a \sin(\alpha + \beta)}{\sin(\beta - \alpha)}$.

(C) The angle of depression from the top of a tower to a point $A$ on the ground is the angle between the horizontal line of sight from the top and the line of sight to point $A$.
By the property of alternate interior angles,when two parallel lines (the horizontal line at the top and the ground) are intersected by a transversal (the line of sight),the angle of depression is equal to the angle of elevation.
Therefore,if the angle of depression is $30^\circ$,the angle of elevation of the top of the tower from point $A$ is also $30^\circ$.

(B) Let the height of each pole be $h \, m$. Let the positions of the two poles be $P_1$ and $P_2$. The distance between the poles is $120 \, m$.
Let $A$ and $B$ be points on the line joining the bottoms of the poles such that $AB = 30 \, m$.
From the geometry of the problem,the distance from the base of the first pole to $A$ is $h$ (since $\tan 45^\circ = \frac{h}{dist} = 1$).
Similarly,the distance from the base of the second pole to $B$ is $h$.
The total distance between the poles is the sum of the distance from the first pole to $A$,the distance $AB$,and the distance from $B$ to the second pole.
Therefore,$h + 30 + h = 120$.
$2h = 120 - 30 = 90$.
$h = 45 \, m$.

(A) Let the height of the pole be $p$. The tower has height $h$ and is at a distance $2h$ from the observer.
Let the angle subtended by the tower be $\alpha$. Then,$\tan \alpha = \frac{h}{2h} = \frac{1}{2}$.
Let the pole be on top of the tower. The angle subtended by the pole is also $\alpha$.
The total angle subtended by the tower and the pole together is $\alpha + \alpha = 2\alpha$.
The total height is $h + p$.
Thus,$\tan(2\alpha) = \frac{h + p}{2h}$.
Using the formula $\tan(2\alpha) = \frac{2\tan \alpha}{1 - \tan^2 \alpha}$,we get:
$\frac{h + p}{2h} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
Therefore,$\frac{h + p}{2h} = \frac{4}{3}$ $\Rightarrow h + p = \frac{8h}{3}$ $\Rightarrow p = \frac{8h}{3} - h = \frac{5h}{3}$.

(B) Let the height of the first house be $H$. Let the window be at height $y$ from the ground. The distance between the houses is $d = 6 \ m$.
From the triangle formed by the bottom of the first house,the base of the second house,and the window,we have $\tan(60^\circ) = \frac{y}{6}$.
Thus,$y = 6 \tan(60^\circ) = 6\sqrt{3} \ m$.
Let the window divide the house of height $H$ into two parts $h_1$ and $h_2$ such that $H = h_1 + h_2$. The window is at height $y$ from the ground,so $h_2 = y = 6\sqrt{3} \ m$.
The angle subtended by the house at the window is $90^\circ$. Let the angle of elevation of the top of the house from the window be $\alpha$. Then the angle of depression of the bottom of the house from the window is $90^\circ - \alpha$.
In the upper triangle,$\tan(\alpha) = \frac{h_1}{6}$. In the lower triangle,$\tan(90^\circ - \alpha) = \cot(\alpha) = \frac{y}{6} = \frac{6\sqrt{3}}{6} = \sqrt{3}$.
Since $\cot(\alpha) = \sqrt{3}$,we have $\tan(\alpha) = \frac{1}{\sqrt{3}}$.
Therefore,$h_1 = 6 \tan(\alpha) = 6 \times \frac{1}{\sqrt{3}} = 2\sqrt{3} \ m$.
The total height $H = h_1 + h_2 = 2\sqrt{3} + 6\sqrt{3} = 8\sqrt{3} \ m$.

(B) Let the height of the pole be $h$ and the length of the shadow be $s$.
Given that $s = \sqrt{3}h$.
Let $\alpha$ be the angle of elevation of the sun.
In the right-angled triangle formed by the pole and its shadow,we have:
$\tan \alpha = \frac{\text{height}}{\text{shadow}} = \frac{h}{\sqrt{3}h} = \frac{1}{\sqrt{3}}$.
Since $\tan 30^o = \frac{1}{\sqrt{3}}$,we get $\alpha = 30^o$.

(B) Let the length of the ladder be $L$ and the height of the wall be $h = 6\sqrt{3} \text{ m}$.
Given that the ladder makes an angle of $\theta = 60^{\circ}$ with the horizontal.
In the right-angled triangle formed by the ladder,the wall,and the ground,we have:
$\sin(\theta) = \frac{\text{Height of the wall}}{\text{Length of the ladder}}$
$\sin(60^{\circ}) = \frac{6\sqrt{3}}{L}$
$\frac{\sqrt{3}}{2} = \frac{6\sqrt{3}}{L}$
$L = \frac{6\sqrt{3} \times 2}{\sqrt{3}}$
$L = 12 \text{ m}$.
Thus,the length of the ladder is $12 \text{ m}$.

(C) Let the heights of the two towers be $H_1$ and $H_2$,and the distance from the middle point to the foot of each tower be $d$.
From the given information:
$H_1 = d \tan 60^\circ = d \sqrt{3}$
$H_2 = d \tan 30^\circ = d \frac{1}{\sqrt{3}}$
Therefore,the ratio of their heights is:
$\frac{H_1}{H_2} = \frac{d \sqrt{3}}{d / \sqrt{3}} = \sqrt{3} \times \sqrt{3} = \frac{3}{1}$
Thus,the ratio is $3 : 1$.

(A) Let the height of the tower be $h$ and the angles of elevation be $\alpha$ and $\beta$ respectively.
Given: $\cot \alpha = 3/5$ and $\cot \beta = 2/5$.
Let the distance from the first point to the base of the tower be $x_1 = h \cot \alpha = 3h/5$.
Let the distance from the second point to the base of the tower be $x_2 = h \cot \beta = 2h/5$.
The distance walked towards the tower is $x_1 - x_2 = 32 \, m$.
Substituting the values: $3h/5 - 2h/5 = 32$.
$h/5 = 32$.
$h = 32 \times 5 = 160 \, m$.

(C) Let the total height of the tree be $H = 20 \ m$. Let the tree break at a height $h$ from the ground. The broken part forms the hypotenuse $l$ of a right-angled triangle,where $l = 20 - h$.
In the right-angled triangle,$\sin 30^\circ = \frac{h}{l}$.
Since $\sin 30^\circ = \frac{1}{2}$,we have $\frac{1}{2} = \frac{h}{20 - h}$.
$20 - h = 2h \implies 3h = 20 \implies h = \frac{20}{3} \ m$.

(B) Let the height of the cliff be $h = 500 \ m$.
Let the diameter of the base be $D = d_1 + d_2$.
From the right-angled triangles formed:
$d_1 = h \cot(60^\circ) = 500 \times \frac{1}{\sqrt{3}} = \frac{500}{\sqrt{3}} \ m$.
$d_2 = h \cot(30^\circ) = 500 \times \sqrt{3} = 500\sqrt{3} \ m$.
The diameter $D = d_1 + d_2 = \frac{500}{\sqrt{3}} + 500\sqrt{3}$.
$D = \frac{500 + 500(3)}{\sqrt{3}} = \frac{500 + 1500}{\sqrt{3}} = \frac{2000}{\sqrt{3}} \ m$.

(B) Let $h$ be the height of the house and $H$ be the height of the tower above the top of the house.
From the triangle formed by the angle of depression,we have $\tan(30^\circ) = \frac{h}{12}$.
Thus,$h = 12 \tan(30^\circ) = 12 \times \frac{1}{\sqrt{3}} = 4\sqrt{3} \ m$.
From the triangle formed by the angle of elevation,we have $\tan(60^\circ) = \frac{H}{12}$.
Thus,$H = 12 \tan(60^\circ) = 12 \sqrt{3} \ m$.
The total height of the tower is $h + H = 4\sqrt{3} + 12\sqrt{3} = 16\sqrt{3} \ m$.

(A) Let the height of the tower be $H$. The eye level of the man is $h = 1.5 \ m$ above the ground.
The distance of the man from the tower is $d = 10 \ m$.
The angle of elevation is $\theta = 60^\circ$.
Considering the right-angled triangle formed by the line of sight,the horizontal distance,and the height of the tower above the eye level,we have:
$\tan(60^\circ) = \frac{\text{Height above eye level}}{d}$
$\sqrt{3} = \frac{H - 1.5}{10}$
$H - 1.5 = 10\sqrt{3}$
$H = (10\sqrt{3} + 1.5) \ m$.

(A) Let the height of the tower be $H$ and the distance from the foot be $d$.
From the first point,we have $\tan(30^\circ) = H/d$,so $H = d \tan(30^\circ) = d/\sqrt{3}$.
From the second point at height $h$ above the first,the angle of depression to the foot of the tower is $60^\circ$.
This implies $\tan(60^\circ) = h/d$,so $d = h/\tan(60^\circ) = h/\sqrt{3}$.
Substituting $d$ into the expression for $H$:
$H = (h/\sqrt{3}) / \sqrt{3} = h/3$.
Thus,the height of the tower is $h/3$.

(B) Let the tower be $AB$ with height $b$ and the pole be $BP$ with height $p$. The point $O$ is at distance $a$ from $A$.
Given $\angle AOB = \alpha$ and $\angle POB = \alpha$.
In $\triangle OAB$,$\tan \alpha = \frac{AB}{OA} = \frac{b}{a}$.
In $\triangle OAP$,$\angle POA = 2\alpha$.
So,$\tan 2\alpha = \frac{AP}{OA} = \frac{p + b}{a}$.
Using the formula $\tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}$,we get:
$\frac{2(b/a)}{1 - (b/a)^2} = \frac{p + b}{a}$
$\frac{2b/a}{(a^2 - b^2)/a^2} = \frac{p + b}{a}$
$\frac{2ba}{a^2 - b^2} = \frac{p + b}{a}$
$p + b = \frac{2ba^2}{a^2 - b^2}$
$p = \frac{2ba^2}{a^2 - b^2} - b$
$p = \frac{2ba^2 - b(a^2 - b^2)}{a^2 - b^2}$
$p = \frac{2ba^2 - ba^2 + b^3}{a^2 - b^2}$
$p = \frac{ba^2 + b^3}{a^2 - b^2} = \frac{b(a^2 + b^2)}{a^2 - b^2}$.

(C) Let the height of the tree be $H$. Let the broken part be $x$ and the remaining part be $h$.
The distance from the foot of the tree to the point where the top touches the ground is $d = 10 \ m$.
In the right-angled triangle formed,$\tan(45^\circ) = \frac{h}{10} \implies 1 = \frac{h}{10} \implies h = 10 \ m$.
Also,$\cos(45^\circ) = \frac{10}{x} \implies \frac{1}{\sqrt{2}} = \frac{10}{x} \implies x = 10\sqrt{2} \ m$.
The total length of the tree is $H = h + x = 10 + 10\sqrt{2} = 10(1 + \sqrt{2}) \ m$.

(C) Let $h = 30 \, m$ be the height of the tower and $x$ be the distance of the ship from the base of the tower.
Since the angle of depression is $60^\circ$,the angle of elevation from the ship to the top of the tower is also $60^\circ$.
In the right-angled triangle formed by the tower and the distance on the ground,we have:
$\tan 60^\circ = \frac{\text{Height}}{\text{Distance}} = \frac{30}{x}$
Since $\tan 60^\circ = \sqrt{3}$,we get:
$\sqrt{3} = \frac{30}{x}$
$x = \frac{30}{\sqrt{3}} = \frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{30 \sqrt{3}}{3} = 10 \sqrt{3} \, m$.
Thus,the distance of the ship from the base of the tower is $10 \sqrt{3} \, m$.

(A) Let $H$ be the height of the cloud above the lake level and $h = 2500 \, m$ be the height of the observation point above the lake.
Let $x$ be the horizontal distance from the observation point to the cloud.
From the angle of elevation,we have $\tan(15^\circ) = \frac{H - h}{x}$,so $x = \frac{H - h}{\tan(15^\circ)} = (H - h) \cot(15^\circ)$.
From the angle of depression of the reflection,the depth of the reflection below the lake is $H$. The total vertical distance from the observation point to the reflection is $H + h$.
Thus,$\tan(45^\circ) = \frac{H + h}{x}$,so $x = \frac{H + h}{\tan(45^\circ)} = (H + h) \cot(45^\circ)$.
Equating the two expressions for $x$: $(H - h) \cot(15^\circ) = (H + h) \cot(45^\circ)$.
Since $\cot(45^\circ) = 1$ and $\cot(15^\circ) = 2 + \sqrt{3}$,we have $(H - 2500)(2 + \sqrt{3}) = H + 2500$.
$H(2 + \sqrt{3}) - 2500(2 + \sqrt{3}) = H + 2500$.
$H(1 + \sqrt{3}) = 2500(3 + \sqrt{3}) = 2500 \sqrt{3}(\sqrt{3} + 1)$.
$H = 2500 \sqrt{3} \, m$.

(D) Let $h$ be the height of the aeroplane above the road.
Let the two milestones be at points $A$ and $B$,and the point on the road directly below the aeroplane be $P$.
Given that the distance between the two consecutive milestones is $1 \text{ mile}$,so $AP + PB = 1$.
In the right-angled triangles formed,we have $AP = h \cot \alpha$ and $PB = h \cot \beta$.
Thus,$h \cot \alpha + h \cot \beta = 1$.
$h(\cot \alpha + \cot \beta) = 1$.
$h = \frac{1}{\cot \alpha + \cot \beta}$.
Since $\cot \theta = \frac{1}{\tan \theta}$,we have $h = \frac{1}{\frac{1}{\tan \alpha} + \frac{1}{\tan \beta}} = \frac{1}{\frac{\tan \beta + \tan \alpha}{\tan \alpha \cdot \tan \beta}} = \frac{\tan \alpha \cdot \tan \beta}{\tan \alpha + \tan \beta}$.

(D) Let the height of the balloon be $h$ and the distance from the point directly under the balloon to $C$ be $x$. From the geometry of the problem:
$x = h \cot(3\alpha) \dots (i)$
$x + 100 = h \cot(2\alpha) \dots (ii)$
$x + 300 = h \cot(\alpha) \dots (iii)$
Subtracting $(i)$ from $(ii)$:
$100 = h(\cot(2\alpha) - \cot(3\alpha)) = h \left( \frac{\sin(3\alpha - 2\alpha)}{\sin(2\alpha)\sin(3\alpha)} \right) = h \frac{\sin(\alpha)}{\sin(2\alpha)\sin(3\alpha)} \dots (iv)$
Subtracting $(ii)$ from $(iii)$:
$200 = h(\cot(\alpha) - \cot(2\alpha)) = h \left( \frac{\sin(2\alpha - \alpha)}{\sin(\alpha)\sin(2\alpha)} \right) = h \frac{1}{\sin(2\alpha)} \dots (v)$
Dividing $(iv)$ by $(v)$:
$\frac{100}{200} = \frac{\sin(\alpha)}{\sin(3\alpha)} \Rightarrow \frac{\sin(3\alpha)}{\sin(\alpha)} = 2$
Using $\sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha)$:
$3 - 4\sin^2(\alpha) = 2$ $\Rightarrow 4\sin^2(\alpha) = 1$ $\Rightarrow \sin(\alpha) = \frac{1}{2}$ $\Rightarrow \alpha = 30^\circ$
Substituting $\alpha = 30^\circ$ in $(v)$:
$200 = h \frac{1}{\sin(60^\circ)} = h \frac{1}{\sqrt{3}/2} = \frac{2h}{\sqrt{3}}$
$h = 100\sqrt{3}$ metres.

(B) Let the total height of the pole be $h \, ft$. The lower portion is $\frac{h}{3}$ and the upper portion is $\frac{2h}{3}$.
Let the point be $P$ at a distance of $40 \, ft$ from the foot of the pole.
Let $\alpha$ be the angle subtended by the lower portion at $P$,and $\beta$ be the angle subtended by the whole pole at $P$.
Then $\tan \alpha = \frac{h/3}{40} = \frac{h}{120}$ and $\tan \beta = \frac{h}{40} = \frac{3h}{120}$.
The upper portion subtends an angle $\theta = \tan^{-1}(\frac{1}{2})$ at $P$,so $\tan \theta = \frac{1}{2}$.
Since $\theta = \beta - \alpha$,we have $\tan \theta = \tan(\beta - \alpha) = \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha}$.
Substituting the values: $\frac{1}{2} = \frac{\frac{3h}{120} - \frac{h}{120}}{1 + (\frac{3h}{120})(\frac{h}{120})} = \frac{\frac{2h}{120}}{1 + \frac{3h^2}{14400}} = \frac{h/60}{1 + \frac{h^2}{4800}}$.
$\frac{1}{2} = \frac{h}{60} \times \frac{4800}{4800 + h^2} = \frac{80h}{4800 + h^2}$.
$4800 + h^2 = 160h \Rightarrow h^2 - 160h + 4800 = 0$.
$(h - 120)(h - 40) = 0$.
So $h = 120$ or $h = 40$.
Since the pole is more than $100 \, ft$ high,$h = 120 \, ft$.

(C) Let the height of the house be $h \ m$. The distance from the foot of the house is $d = 70 \ m$. The height of the flag-post is $20 \ m$.
Let $\theta$ be the angle subtended by the house at the point,and $\phi$ be the angle subtended by the house plus the flag-post.
Then $\tan \theta = \frac{h}{70}$ and $\tan \phi = \frac{h + 20}{70}$.
The angle subtended by the flag-post is $\alpha = \phi - \theta$,where $\tan \alpha = \frac{1}{6}$.
Using the formula $\tan \alpha = \frac{\tan \phi - \tan \theta}{1 + \tan \phi \tan \theta}$:
$\frac{1}{6} = \frac{\frac{h + 20}{70} - \frac{h}{70}}{1 + (\frac{h + 20}{70})(\frac{h}{70})}$
$\frac{1}{6} = \frac{\frac{20}{70}}{1 + \frac{h(h + 20)}{4900}}$
$1 + \frac{h^2 + 20h}{4900} = 6 \times \frac{20}{70} = \frac{120}{70} = \frac{12}{7}$
$\frac{h^2 + 20h}{4900} = \frac{12}{7} - 1 = \frac{5}{7}$
$h^2 + 20h = \frac{5}{7} \times 4900 = 5 \times 700 = 3500$
$h^2 + 20h - 3500 = 0$
$(h + 70)(h - 50) = 0$
Since $h > 0$,we have $h = 50 \ m$.

(B) Let the distance of the balloon from the observer be $d$. Let the initial height be $h_1$ and the final height be $h_2$.
From the geometry,$h_1 = d \tan 45^\circ = d$ and $h_2 = d \tan 30^\circ = \frac{d}{\sqrt{3}}$.
The distance descended in $10 \, minutes$ is $h_1 - h_2 = d - \frac{d}{\sqrt{3}} = d \left( \frac{\sqrt{3} - 1}{\sqrt{3}} \right)$.
Given the rate of descent is $4 \, m/min$,the distance descended in $10 \, minutes$ is $4 \times 10 = 40 \, m$.
Equating the two: $d \left( \frac{\sqrt{3} - 1}{\sqrt{3}} \right) = 40$.
$d = \frac{40 \sqrt{3}}{\sqrt{3} - 1} = \frac{40 \sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{40(3 + \sqrt{3})}{2} = 20(3 + \sqrt{3}) \, m$.

(B) Let the height of the tower be $h$ and the breadth of the river be $x$.
According to the problem,the angle of elevation is $45^\circ$.
Using the trigonometric ratio,$\tan(45^\circ) = \frac{\text{height of tower}}{\text{breadth of river}} = \frac{h}{x}$.
Since $\tan(45^\circ) = 1$,we have $1 = \frac{h}{x}$,which implies $x = h$.
Therefore,the breadth of the river and the height of the tower are the same.

(B) Let $AC = x = CB$. Then $AB = AC + CB = 2x$.
Given $AP = 3 \, AB = 3(2x) = 6x$.
Let $\angle APC = \alpha$. In $\Delta ACP$,$\tan \alpha = \frac{AC}{AP} = \frac{x}{6x} = \frac{1}{6}$.
In $\Delta ABP$,$\angle APB = \alpha + \beta$. Thus,$\tan(\alpha + \beta) = \frac{AB}{AP} = \frac{2x}{6x} = \frac{1}{3}$.
Using the formula $\tan \beta = \tan((\alpha + \beta) - \alpha) = \frac{\tan(\alpha + \beta) - \tan \alpha}{1 + \tan(\alpha + \beta)\tan \alpha}$.
Substituting the values: $\tan \beta = \frac{\frac{1}{3} - \frac{1}{6}}{1 + (\frac{1}{3})(\frac{1}{6})} = \frac{\frac{2-1}{6}}{1 + \frac{1}{18}} = \frac{\frac{1}{6}}{\frac{19}{18}} = \frac{1}{6} \times \frac{18}{19} = \frac{3}{19}$.

(A) Let $h$ be the height of the tower and $x$ be the length of its shadow. The flagstaff of height $6 \ m$ is placed on top of the tower.
From the geometry of the problem,we have two similar triangles $AEC$ and $BDC$ as shown in the figure.
For the tower of height $h$ and shadow $x$,we have $\tan \theta = \frac{h}{x}$.
For the combined height $(h + 6)$ and total shadow $(x + 2\sqrt{3})$,we have $\tan \theta = \frac{h + 6}{x + 2\sqrt{3}}$.
Equating the two expressions for $\tan \theta$:
$\frac{h}{x} = \frac{h + 6}{x + 2\sqrt{3}}$
$h(x + 2\sqrt{3}) = x(h + 6)$
$hx + 2\sqrt{3}h = xh + 6x$
$2\sqrt{3}h = 6x$
$\frac{h}{x} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$
Since $\tan \theta = \frac{h}{x} = \sqrt{3}$,we have $\theta = 60^o$.

(C) Let the height of the cliff be $OP = h$. Let $O$ be the base of the cliff on the ground.
From the given information,$AB = 100 \ m$ and $AB$ is vertical,so $OC = 100 \ m$.
Then $CP = OP - OC = h - 100$.
In $\triangle AOP$,$\tan \alpha = \frac{OP}{OA} = \frac{h}{OA} \implies OA = h \cot \alpha$.
In $\triangle BCP$,$\tan \beta = \frac{CP}{BC} = \frac{h - 100}{BC}$. Since $BC = OA$,we have $BC = h \cot \alpha$.
Substituting $BC$ in the second equation: $\tan \beta = \frac{h - 100}{h \cot \alpha}$.
$h \cot \alpha \tan \beta = h - 100$.
$100 = h - h \cot \alpha \tan \beta = h(1 - \cot \alpha \tan \beta)$.
$100 = h \left(1 - \frac{\cot \alpha}{\cot \beta}\right) = h \left(\frac{\cot \beta - \cot \alpha}{\cot \beta}\right)$.
Therefore,$h = \frac{100 \cot \beta}{\cot \beta - \cot \alpha}$.

(B) Let the observer be at point $O$ at a height of $30 \, m$ from the ground. The building height is $25 \, m$ and the flag-staff is $5 \, m$ on top of it,so the total height is $30 \, m$.
Let the horizontal distance between the observer and the building be $x$.
The observer is at the same level as the top of the flag-staff.
Let the angle subtended by the flag-staff at the observer be $\alpha$.
Let the angle subtended by the building at the observer be $\alpha$.
From the geometry,$\tan \alpha = \frac{5}{x}$ and $\tan 2\alpha = \frac{30}{x}$.
Using the identity $\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^2 \alpha}$,we have:
$\frac{30}{x} = \frac{2(5/x)}{1 - (5/x)^2} = \frac{10/x}{(x^2 - 25)/x^2} = \frac{10x}{x^2 - 25}$.
$30(x^2 - 25) = 10x^2$ $\Rightarrow 3x^2 - 75 = x^2$ $\Rightarrow 2x^2 = 75$ $\Rightarrow x^2 = \frac{75}{2}$ $\Rightarrow x = \sqrt{\frac{75}{2}} = 5\sqrt{\frac{3}{2}}$.
The distance of the observer from the top of the flag-staff is the horizontal distance $x$ because the observer and the top of the flag-staff are at the same height $(30 \, m)$.
Thus,the distance is $5\sqrt{\frac{3}{2}} \, m$.

(C) Let the tree be $AC$ before breaking,where $C$ is the root and $A$ is the top. Let it break at point $D$ such that the top $A$ touches the ground at point $B$.
In $\triangle BCD$,$\angle DBC = 30^\circ$ and $BC = 10 \, m$.
Using trigonometry:
$\tan(30^\circ) = \frac{CD}{BC} \implies \frac{1}{\sqrt{3}} = \frac{CD}{10} \implies CD = \frac{10}{\sqrt{3}} \, m$.
$\cos(30^\circ) = \frac{BC}{BD} \implies \frac{\sqrt{3}}{2} = \frac{10}{BD} \implies BD = \frac{20}{\sqrt{3}} \, m$.
The original height of the tree is $CD + AD = CD + BD$ (since $AD = BD$).
Height $= \frac{10}{\sqrt{3}} + \frac{20}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3} \, m$.
Using $\sqrt{3} \approx 1.732$,the height is $10 \times 1.732 = 17.32 \, m$.

(A) Let the height of the tower be $h \ m$.
Given that the distance from the base of the tower is $70 \ m$.
In the right-angled triangle,we have:
$\tan(45^\circ) = \frac{\text{Height}}{\text{Base}} = \frac{h}{70}$
Since $\tan(45^\circ) = 1$,we get:
$1 = \frac{h}{70}$
$h = 70 \ m$.
Thus,the height of the tower is $70 \ m$.

(D) Let the height of the tower $CD$ be $h$.
From $\Delta CDA$,we have $x = h \cot 60^\circ = \frac{h}{\sqrt{3}}$.
From $\Delta CDB$,we have $y = h \cot 30^\circ = \sqrt{3}h$.
In $\Delta ABC$,since $A$ is south of $C$ and $B$ is west of $A$,$\angle CAB = 90^\circ$.
By Pythagoras theorem,$AB^2 + AC^2 = BC^2$,which is $3^2 + x^2 = y^2$.
Substituting the values,$9 + \frac{h^2}{3} = 3h^2$.
$9 = 3h^2 - \frac{h^2}{3} = \frac{8h^2}{3}$.
$h^2 = \frac{27}{8} = \frac{54}{16}$.
$h = \sqrt{\frac{54}{16}} = \frac{3\sqrt{6}}{4} \, km$.

(A) Let the height of the tower $AB = h = 15(\sqrt{3} + 1) \ m$.
Let the car be at point $C$ initially and move to point $D$ in $3 \ s$.
In $\triangle ABD$,$\tan(45^\circ) = \frac{AB}{BD} \implies 1 = \frac{h}{BD} \implies BD = h$.
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 by the car is $CD = BC - BD = h\sqrt{3} - h = h(\sqrt{3} - 1)$.
Substituting $h = 15(\sqrt{3} + 1)$:
$CD = 15(\sqrt{3} + 1)(\sqrt{3} - 1) = 15(3 - 1) = 15 \times 2 = 30 \ m$.
The time taken is $t = 3 \ s$.
Speed $v = \frac{\text{Distance}}{\text{Time}} = \frac{30 \ m}{3 \ s} = 10 \ m/s$.
To convert $m/s$ to $km/hr$,multiply by $\frac{18}{5}$:
$v = 10 \times \frac{18}{5} = 36 \ km/hr$.

(D) Let the tower be $AB$ with height $40 \, m$. Let the two men be at points $O_1$ and $O_2$ on opposite sides of the tower.
In $\Delta O_1BA$,$\tan(45^{\circ}) = \frac{AB}{O_1B}$ $\Rightarrow 1 = \frac{40}{x}$ $\Rightarrow x = 40 \, m$.
In $\Delta O_2BA$,$\tan(30^{\circ}) = \frac{AB}{O_2B}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{40}{y}$ $\Rightarrow y = 40\sqrt{3} \, m$.
Using $\sqrt{3} \approx 1.732$,$y = 40 \times 1.732 = 69.28 \, m$.
The distance between the men is $x + y = 40 + 69.28 = 109.28 \, m$.