Heights and Distances Questions in English

(A) Let the height of the electric pole be $h$ and the length of the cable be $l = 22\, m$.
The cable forms a right-angled triangle with the pole and the ground.
The angle of elevation is $\theta = 30^{\circ}$.
In the right-angled triangle,the cable acts as the hypotenuse and the pole acts as the side opposite to the angle $\theta$.
Using the trigonometric ratio $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$:
$\sin(30^{\circ}) = \frac{h}{22}$
Since $\sin(30^{\circ}) = \frac{1}{2}$,we have:
$\frac{1}{2} = \frac{h}{22}$
$h = \frac{22}{2} = 11\, m$.
Thus,the height of the electric pole is $11\, m$.

(B) Let the height of the tower be $h = 60 \, m$ and the distance of the point from the base be $x$.
In the right-angled triangle formed by the tower and the ground,we have $\tan(60^{\circ}) = \frac{\text{height}}{\text{base}}$.
Substituting the values,$\sqrt{3} = \frac{60}{x}$.
Thus,$x = \frac{60}{\sqrt{3}} = \frac{60 \times \sqrt{3}}{3} = 20 \times \sqrt{3}$.
Using $\sqrt{3} \approx 1.732$,we get $x = 20 \times 1.732 = 34.64 \, m$.
Rounding to one decimal place,the distance is $34.6 \, m$.

(C) Let the height of the tower be $AB = 50 \, m$ and the distance between the car $C$ and the base of the tower $B$ be $x \, m$.
The angle of depression from the top $A$ to the car $C$ is $30^\circ$,which implies the angle of elevation from the car to the top of the tower is also $30^\circ$ (alternate interior angles).
In the right-angled triangle $ABC$,we have $\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$.
Substituting the values,$\frac{1}{\sqrt{3}} = \frac{50}{x}$.
Therefore,$x = 50 \sqrt{3}$.
Using $\sqrt{3} \approx 1.732$,we get $x = 50 \times 1.732 = 86.6 \, m$. The closest option is $86.5 \, m$.

(N/A) Let the length of the ladder be $L$ and the distance from the wall be $x$.
Given: Height of the wall $h = 10 \, m$,Angle $\theta = 45^{\circ}$.
Using trigonometry:
$1.$ $\sin(45^{\circ}) = \frac{\text{Height}}{\text{Length}} = \frac{10}{L}$.
Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$,we have $\frac{1}{\sqrt{2}} = \frac{10}{L}$,which gives $L = 10\sqrt{2} \approx 14.14 \, m$.
$2.$ $\tan(45^{\circ}) = \frac{\text{Height}}{\text{Distance}} = \frac{10}{x}$.
Since $\tan(45^{\circ}) = 1$,we have $1 = \frac{10}{x}$,which gives $x = 10 \, m$.
Thus,the length of the ladder is $14.14 \, m$ and the distance from the wall is $10 \, m$.

(B) Let the height of the lighthouse be $AB = 100\, m$. Let the two boats be at points $C$ and $D$ on opposite sides of the lighthouse.
Given that the angle of depression from the top $A$ to boats $C$ and $D$ is $30^\circ$,therefore the angle of elevation from $C$ and $D$ to $A$ is also $30^\circ$ (alternate interior angles).
In $\triangle ABC$,$\tan(30^\circ) = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{100}{BC} \implies BC = 100\sqrt{3}\, m$.
In $\triangle ABD$,$\tan(30^\circ) = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{100}{BD} \implies BD = 100\sqrt{3}\, m$.
The total distance between the two boats is $CD = BC + BD = 100\sqrt{3} + 100\sqrt{3} = 200\sqrt{3}\, m$.
Using $\sqrt{3} \approx 1.732$,the distance is $200 \times 1.732 = 346.4\, m$.

(B) Let the height of the tower be $h = 300 \,m$. Let the two trucks be at points $A$ and $B$ on the ground,and the top of the tower be $T$. The angles of depression are $45^{\circ}$ and $60^{\circ}$,which correspond to the angles of elevation from the trucks to the top of the tower.
In $\triangle TCB$ (where $C$ is the base of the tower),$\tan(60^{\circ}) = \frac{TC}{CB} \implies \sqrt{3} = \frac{300}{CB} \implies CB = \frac{300}{\sqrt{3}} = 100\sqrt{3} \approx 173.2 \,m$.
In $\triangle TCA$,$\tan(45^{\circ}) = \frac{TC}{CA} \implies 1 = \frac{300}{CA} \implies CA = 300 \,m$.
The distance between the two trucks is $AB = CA - CB = 300 - 173.2 = 126.8 \,m \approx 127 \,m$.

(N/A) Let $h$ be the height of the cliff and $d$ be the distance between the tower and the cliff.
From the base of the tower to the top of the cliff: $\tan(45^{\circ}) = \frac{h}{d} \implies 1 = \frac{h}{d} \implies h = d$.
From the top of the tower to the base of the cliff: $\tan(30^{\circ}) = \frac{100}{d} \implies \frac{1}{\sqrt{3}} = \frac{100}{d} \implies d = 100\sqrt{3} \approx 173.2\, m$.
Since $h = d$,the height of the cliff is $173.2\, m$ and the distance between them is $173.2\, m$.

(D) Let the height of the tower be $AB = 10 \, m$ and the height of the flag-post be $BC = h \, m$. Let the point on the ground be $D$.
In $\triangle ABD$,$\tan(45^{\circ}) = \frac{AB}{AD} \implies 1 = \frac{10}{AD} \implies AD = 10 \, m$.
In $\triangle ACD$,$\tan(60^{\circ}) = \frac{AC}{AD} = \frac{AB + BC}{AD}$.
$\sqrt{3} = \frac{10 + h}{10} \implies 10\sqrt{3} = 10 + h$.
$h = 10(\sqrt{3} - 1) \approx 10(1.732 - 1) = 10(0.732) = 7.32 \, m$.
Rounding to one decimal place,the height is $7.3 \, m$.

(A) Let the height of the lighthouse be $AB = 100 \, m$. Let the two boats be at points $C$ and $D$ such that $C$ is closer to the lighthouse.
In $\triangle ABC$,$\tan(45^{\circ}) = \frac{AB}{BC} \implies 1 = \frac{100}{BC} \implies BC = 100 \, m$.
In $\triangle ABD$,$\tan(30^{\circ}) = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{100}{BD} \implies BD = 100\sqrt{3} \, m$.
Using $\sqrt{3} \approx 1.732$,$BD = 100 \times 1.732 = 173.2 \, m$.
The distance between the two boats is $CD = BD - BC = 173.2 - 100 = 73.2 \, m$.

(B) Let the height of the building be $H$ and the distance of the car from the building be $x$.
From the top of the building,the angle of depression is $60^{\circ}$,so $\tan(60^{\circ}) = H/x$. This gives $H = x \sqrt{3}$.
From the window $14.6 \, m$ below the top,the height is $(H - 14.6)$. The angle of depression is $45^{\circ}$,so $\tan(45^{\circ}) = (H - 14.6)/x$.
Since $\tan(45^{\circ}) = 1$,we have $x = H - 14.6$,which means $H = x + 14.6$.
Equating the two expressions for $H$: $x \sqrt{3} = x + 14.6$.
$x(\sqrt{3} - 1) = 14.6$.
$x = 14.6 / (1.732 - 1) = 14.6 / 0.732 \approx 19.945 \, m$.
Rounding to the nearest integer,the distance is $20 \, m$.

(A) Let the length of the ladder be $L = 10\,m$ and the distance from the wall be $d = 5\,m$.
Let $\theta$ be the angle with the floor and $\phi$ be the angle with the wall.
In the right-angled triangle formed by the ladder,the wall,and the floor,$\cos(\theta) = \frac{\text{base}}{\text{hypotenuse}} = \frac{5}{10} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = 60^\circ$.
Since the sum of angles in a triangle is $180^\circ$ and one angle is $90^\circ$,the angle with the wall is $\phi = 90^\circ - 60^\circ = 30^\circ$.
Thus,the ladder makes an angle of $30^\circ$ with the wall and $60^\circ$ with the floor.

(C) Let the height of the hill be $h = 300\, m$ and the distance of the tree from the hill be $d = 300\, m$.
Let $\theta$ be the angle of depression from the top of the hill to the base of the tree.
In the right-angled triangle formed by the height of the hill and the distance to the tree,we have:
$\tan(\theta) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{h}{d}$
$\tan(\theta) = \frac{300}{300} = 1$
Since $\tan(45^{\circ}) = 1$,we get $\theta = 45^{\circ}$.
Thus,the angle of depression is $45^{\circ}$.

(A) Let the height of the tower be $h$ and the distance from the base be $d = 20\, m$.
The angle of elevation is $\theta = 30^{\circ}$.
In the right-angled triangle formed by the tower and the ground,we have $\tan(\theta) = \frac{\text{height}}{\text{base}}$.
$\tan(30^{\circ}) = \frac{h}{20}$.
Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{h}{20}$.
$h = \frac{20}{\sqrt{3}} = \frac{20}{1.732} \approx 11.547\, m$.
Rounding to one decimal place,the height is $11.6\, m$.

(A) Let the height of the tower be $AB = 150\, m$ and the position of the cycle be $C$.
The angle of depression from the top $A$ to the cycle $C$ is $60^{\circ}$,which implies the angle of elevation from $C$ to $A$ is also $60^{\circ}$ (alternate interior angles).
In the right-angled triangle $ABC$,we have $\tan(60^{\circ}) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$.
Substituting the values: $\sqrt{3} = \frac{150}{BC}$.
Therefore,$BC = \frac{150}{\sqrt{3}} = \frac{150\sqrt{3}}{3} = 50\sqrt{3}\, m$.
Using $\sqrt{3} \approx 1.732$,we get $BC \approx 50 \times 1.732 = 86.6\, m$.

(C) Let the height of the cliff be $H$ and the distance between the cliff and the tower be $d$. Let the height of the tower be $h = 50 \, m$.
From the base of the cliff,the angle of elevation to the top of the tower is $30^{\circ}$. Thus,$\tan(30^{\circ}) = \frac{h}{d} \implies \frac{1}{\sqrt{3}} = \frac{50}{d} \implies d = 50\sqrt{3} \, m$.
From the top of the cliff,the angle of depression to the base of the tower is $60^{\circ}$. Thus,$\tan(60^{\circ}) = \frac{H}{d} \implies \sqrt{3} = \frac{H}{50\sqrt{3}}$.
Solving for $H$,we get $H = 50\sqrt{3} \times \sqrt{3} = 50 \times 3 = 150 \, m$.

(A) Let the height of the tower be $AB = 50\, m$. Let the two cars be at points $C$ and $D$ such that the angle of depression from the top $A$ to $C$ is $30^{\circ}$ and to $D$ is $60^{\circ}$.
In $\triangle ABD$,$\tan(60^{\circ}) = \frac{AB}{BD} \implies \sqrt{3} = \frac{50}{BD} \implies BD = \frac{50}{\sqrt{3}} \approx 28.87\, m$.
In $\triangle ABC$,$\tan(30^{\circ}) = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{50}{BC} \implies BC = 50\sqrt{3} \approx 86.60\, m$.
The distance between the two cars is $CD = BC - BD = 50\sqrt{3} - \frac{50}{\sqrt{3}} = 50 \left( \frac{3-1}{\sqrt{3}} \right) = \frac{100}{\sqrt{3}} \approx 57.73\, m$.

(B) Let the angle of elevation of the sun be $\theta$.
Given:
Height of the pole $(AB) = h$
Length of the shadow $(BC) = \sqrt{3} h$
In the right-angled triangle $\triangle ABC$:
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$
Substituting the values:
$\tan \theta = \frac{h}{\sqrt{3} h} = \frac{1}{\sqrt{3}}$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have:
$\theta = 30^{\circ}$
Thus,the angle of elevation of the sun is $30^{\circ}$.

(D) Given that,the length of the ladder $(PQ)$ $= 15 \, m$.
Let the height of the vertical wall $(PR)$ $= h$.
The ladder makes an angle of $60^{\circ}$ with the wall,so $\angle QPR = 60^{\circ}$.
In the right-angled triangle $\triangle PQR$,where $\angle PRQ = 90^{\circ}$:
Using the trigonometric ratio $\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$:
$\cos(60^{\circ}) = \frac{PR}{PQ}$
$\frac{1}{2} = \frac{h}{15}$
$h = \frac{15}{2} \, m = 7.5 \, m$.
Therefore,the height of the wall is $\frac{15}{2} \, m$.

(D) Let the angle of elevation of the top of the tower from the eye of the observer be $\theta$.
Given that,the height of the tower $AB = 22 \ m$ and the height of the observer $PQ = 1.5 \ m$.
The distance between the observer and the tower is $QB = PM = 20.5 \ m$.
Since $PQ = MB = 1.5 \ m$,the height of the tower above the observer's eye level is $AM = AB - MB$.
$AM = 22 \ m - 1.5 \ m = 20.5 \ m$.
In the right-angled triangle $\triangle APM$,we have:
$\tan \theta = \frac{AM}{PM} = \frac{20.5}{20.5} = 1$.
Since $\tan 45^{\circ} = 1$,we get $\theta = 45^{\circ}$.
Therefore,the angle of elevation of the top of the tower from the eye of the observer is $45^{\circ}$.

(N/A) Let $O$ be the centre of the balloon,$OP = r$ be its radius,and $A$ be the position of the observer's eye. The angle subtended by the balloon at the observer's eye is $\angle PAQ = \theta$. The line $AO$ bisects $\angle PAQ$,so $\angle OAP = \frac{\theta}{2}$.
The angle of elevation of the centre $O$ is $\angle OAB = \phi$,where $B$ is the point on the ground directly below $O$. Let $h$ be the height of the centre of the balloon,so $OB = h$.
In the right-angled triangle $\triangle OAP$ (where $\angle OPA = 90^\circ$ because the line of sight is tangent to the sphere),we have:
$\sin(\angle OAP) = \frac{OP}{OA}$
$\sin\left(\frac{\theta}{2}\right) = \frac{r}{OA}$
$OA = \frac{r}{\sin(\frac{\theta}{2})} = r \operatorname{cosec}\left(\frac{\theta}{2}\right)$ .....$(1)$
In the right-angled triangle $\triangle OAB$ (where $\angle OBA = 90^\circ$),we have:
$\sin(\angle OAB) = \frac{OB}{OA}$
$\sin \phi = \frac{h}{OA}$
$h = OA \sin \phi$ .....$(2)$
Substituting the value of $OA$ from equation $(1)$ into equation $(2)$:
$h = r \operatorname{cosec}\left(\frac{\theta}{2}\right) \sin \phi$
$h = r \sin \phi \operatorname{cosec}\left(\frac{\theta}{2}\right)$

(N/A) Let the height of the balloon at $P$ be $h$ meters. Let $A$ and $B$ be the two cars. Thus $AB = 100 \, m$. Let $Q$ be the point on the road directly below the balloon $P$.
In $\triangle PAQ$,$\angle PAQ = 45^{\circ}$. Therefore,$\tan 45^{\circ} = \frac{PQ}{AQ} \implies 1 = \frac{h}{AQ} \implies AQ = h$.
In $\triangle PBQ$,$\angle PBQ = 60^{\circ}$. Therefore,$\tan 60^{\circ} = \frac{PQ}{BQ} \implies \sqrt{3} = \frac{h}{BQ} \implies BQ = \frac{h}{\sqrt{3}}$.
Since $AQ = AB + BQ$,we have $h = 100 + \frac{h}{\sqrt{3}}$.
$h - \frac{h}{\sqrt{3}} = 100 \implies h \left( \frac{\sqrt{3}-1}{\sqrt{3}} \right) = 100$.
$h = \frac{100 \sqrt{3}}{\sqrt{3}-1} = \frac{100 \sqrt{3}(\sqrt{3}+1)}{3-1} = \frac{100(3+\sqrt{3})}{2} = 50(3+\sqrt{3}) \, m$.
Thus,the height of the balloon is $50(3+\sqrt{3}) \, m$ (approximately $236.6 \, m$).

(N/A) Let $P$ be the cloud and $Q$ be its reflection in the lake. Let $A$ be the point of observation such that $AB = h$. Let the height of the cloud above the lake be $x$. Let $AL = d$.
From $\triangle PAL$,we have $\frac{x-h}{d} = \tan \theta$ ..........$(1)$
From $\triangle QAL$,we have $\frac{x+h}{d} = \tan \phi$ ..........$(2)$
Dividing $(2)$ by $(1)$,we get $\frac{x+h}{x-h} = \frac{\tan \phi}{\tan \theta}$.
Applying componendo and dividendo,we get $\frac{(x+h)+(x-h)}{(x+h)-(x-h)} = \frac{\tan \phi + \tan \theta}{\tan \phi - \tan \theta}$.
This simplifies to $\frac{2x}{2h} = \frac{\tan \phi + \tan \theta}{\tan \phi - \tan \theta}$.
Therefore,$x = h\left(\frac{\tan \phi + \tan \theta}{\tan \phi - \tan \theta}\right).$

(C) Let the height of the tower be $h \, m$.
Let the tower be $PR$ where $R$ is the base and $P$ is the top.
Let the initial position of the observer be $Q$ and the new position be $S$.
Given $QS = 20 \, m$,$\angle PQR = 30^{\circ}$,and the angle of elevation increases by $15^{\circ}$ at $S$,so $\angle PSR = 30^{\circ} + 15^{\circ} = 45^{\circ}$.
In $\triangle PSR$,$\tan 45^{\circ} = \frac{PR}{SR} \implies 1 = \frac{h}{SR} \implies SR = h$.
In $\triangle PQR$,$\tan 30^{\circ} = \frac{PR}{QR} = \frac{h}{QS + SR} = \frac{h}{20 + h}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{h}{20 + h}$.
$20 + h = h\sqrt{3}$.
$20 = h(\sqrt{3} - 1)$.
$h = \frac{20}{\sqrt{3} - 1}$.
Rationalizing the denominator: $h = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{3 - 1} = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1) \, m$.
Thus,the height of the tower is $10(\sqrt{3} + 1) \, m$.

(N/A) Let the height of the tower be $h$ and the tower be represented by $AC$,where $C$ is the foot of the tower.
Let the two points be $B$ and $P$ on the same line such that $BC = s$ and $PC = t$.
Let $\angle ABC = \theta$. Since the angles of elevation are complementary,$\angle APC = 90^{\circ} - \theta$.
In $\triangle ABC$,$\tan \theta = \frac{AC}{BC} = \frac{h}{s}$ --- $(i)$
In $\triangle APC$,$\tan(90^{\circ} - \theta) = \frac{AC}{PC} = \frac{h}{t}$.
Since $\tan(90^{\circ} - \theta) = \cot \theta$,we have $\cot \theta = \frac{h}{t}$.
Therefore,$\frac{1}{\tan \theta} = \frac{h}{t} \implies \tan \theta = \frac{t}{h}$ --- $(ii)$
Multiplying equations $(i)$ and $(ii)$:
$\tan \theta \cdot \cot \theta = \frac{h}{s} \cdot \frac{h}{t}$
$1 = \frac{h^2}{st}$
$h^2 = st$
$h = \sqrt{st}$
Thus,the height of the tower is $\sqrt{st}$.

(C) Let the height of the tower be $h$ and the length of the shadow when the Sun's elevation is $60^{\circ}$ be $x \, m$.
Let the tower be $SQ$,where $S$ is the top and $Q$ is the base.
When the elevation is $60^{\circ}$,the shadow length is $RQ = x$.
In $\triangle S R Q$,$\tan 60^{\circ} = \frac{SQ}{RQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}$ .......$(i)$
When the elevation is $30^{\circ}$,the shadow length is $PQ = PR + RQ = 50 + x$.
In $\triangle S P Q$,$\tan 30^{\circ} = \frac{SQ}{PQ} = \frac{h}{50 + x}$.
$\frac{1}{\sqrt{3}} = \frac{h}{50 + x} \Rightarrow 50 + x = h\sqrt{3}$.
Substituting $x = \frac{h}{\sqrt{3}}$ from $(i)$:
$50 + \frac{h}{\sqrt{3}} = h\sqrt{3} \Rightarrow 50 = h\sqrt{3} - \frac{h}{\sqrt{3}}$.
$50 = h \left( \frac{3 - 1}{\sqrt{3}} \right) \Rightarrow 50 = h \left( \frac{2}{\sqrt{3}} \right)$.
$h = \frac{50 \sqrt{3}}{2} = 25 \sqrt{3} \, m$.
Thus,the height of the tower is $25 \sqrt{3} \, m$.

(A) Let the height of the tower be $H$ and the distance from the point to the base of the tower be $x$.
Given that,the height of the flag staff is $h$ and the angles of elevation of the bottom and top of the flag staff are $\alpha$ and $\beta$ respectively.
In $\triangle PRO$,$\tan \alpha = \frac{PO}{RO} = \frac{H}{x} \implies x = \frac{H}{\tan \alpha}$ .....$(i)$
In $\triangle FRO$,$\tan \beta = \frac{FO}{RO} = \frac{FP + PO}{RO} = \frac{h + H}{x} \implies x = \frac{h + H}{\tan \beta}$ .....$(ii)$
Equating $(i)$ and $(ii)$:
$\frac{H}{\tan \alpha} = \frac{h + H}{\tan \beta}$
$H \tan \beta = (h + H) \tan \alpha$
$H \tan \beta = h \tan \alpha + H \tan \alpha$
$H \tan \beta - H \tan \alpha = h \tan \alpha$
$H(\tan \beta - \tan \alpha) = h \tan \alpha$
$H = \frac{h \tan \alpha}{\tan \beta - \tan \alpha}$
Hence,the height of the tower is $\frac{h \tan \alpha}{\tan \beta - \tan \alpha}$.

(A) Let the distance between the two towers be $AB = x \, m$ and the height of the second tower be $PA = h \, m$.
Given: Height of the first tower $QB = 30 \, m$,$\angle QAB = 60^{\circ}$,and $\angle PBA = 30^{\circ}$.
In $\triangle QAB$,$\tan 60^{\circ} = \frac{QB}{AB} = \frac{30}{x}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $\sqrt{3} = \frac{30}{x}$,which gives $x = \frac{30}{\sqrt{3}} = 10\sqrt{3} \, m$.
In $\triangle PBA$,$\tan 30^{\circ} = \frac{PA}{AB} = \frac{h}{x}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{h}{10\sqrt{3}}$.
Solving for $h$,we get $h = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \, m$.
Thus,the distance between the towers is $10\sqrt{3} \, m$ and the height of the second tower is $10 \, m$.

(N/A) Let the tower be $AD$ with height $h \, m$. Let the two objects be at points $B$ and $C$ on the ground,such that $B, C,$ and $D$ are collinear.
Let the distance between the two objects be $BC = x \, m$ and the distance from the second object to the foot of the tower be $CD = y \, m$.
Given that the angles of depression are $\alpha$ and $\beta$,by the property of alternate interior angles:
$\angle ABD = \alpha$ and $\angle ACD = \beta$.
In the right-angled triangle $\triangle ACD$:
$\tan \beta = \frac{AD}{CD} = \frac{h}{y} \implies y = \frac{h}{\tan \beta} \quad \dots(i)$
In the right-angled triangle $\triangle ABD$:
$\tan \alpha = \frac{AD}{BD} = \frac{h}{x + y} \implies x + y = \frac{h}{\tan \alpha} \implies y = \frac{h}{\tan \alpha} - x \quad \dots(ii)$
Equating $(i)$ and $(ii)$:
$\frac{h}{\tan \beta} = \frac{h}{\tan \alpha} - x$
$x = \frac{h}{\tan \alpha} - \frac{h}{\tan \beta}$
$x = h \left( \frac{1}{\tan \alpha} - \frac{1}{\tan \beta} \right)$
$x = h (\cot \alpha - \cot \beta) \, m$
Thus,the distance between the two objects is $h (\cot \alpha - \cot \beta) \, m$.

(N/A) Let the length of the ladder be $L$.
In the initial position,the ladder makes an angle $\alpha$ with the horizontal. Let the foot of the ladder be at distance $OA$ from the wall and the top be at height $OB$ on the wall.
From the right-angled triangle $\triangle OAB$:
$OA = L \cos \alpha$
$OB = L \sin \alpha$
When the foot is pulled away by distance $p$,the new position of the foot is $S$ such that $OS = OA + p = L \cos \alpha + p$.
The top slides down by distance $q$,so the new height is $OQ = OB - q = L \sin \alpha - q$.
The ladder now makes an angle $\beta$ with the horizontal. From the right-angled triangle $\triangle OSQ$:
$OS = L \cos \beta$
$OQ = L \sin \beta$
Equating the expressions for $OS$ and $OQ$:
$L \cos \beta = L \cos \alpha + p \implies p = L(\cos \beta - \cos \alpha)$
$L \sin \beta = L \sin \alpha - q \implies q = L(\sin \alpha - \sin \beta)$
Dividing the two equations:
$\frac{p}{q} = \frac{L(\cos \beta - \cos \alpha)}{L(\sin \alpha - \sin \beta)} = \frac{\cos \beta - \cos \alpha}{\sin \alpha - \sin \beta}$
Hence proved.

(N/A) Let the height of the vertical tower be $OT = H \, m$.
Let the distance from the point on the ground to the base of the tower be $OP = x \, m$.
Given that $AP = 10 \, m$,and the tower is vertical,we have $AB = OP = x \, m$ and $OB = AP = 10 \, m$.
Thus,$TB = OT - OB = (H - 10) \, m$.
In $\triangle TPO$,$\tan 60^{\circ} = \frac{OT}{OP} = \frac{H}{x}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $\sqrt{3} = \frac{H}{x}$,which implies $x = \frac{H}{\sqrt{3}}$ ....$(i)$.
In $\triangle TAB$,$\tan 45^{\circ} = \frac{TB}{AB} = \frac{H - 10}{x}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{H - 10}{x}$,which implies $x = H - 10$ ....$(ii)$.
Equating $(i)$ and $(ii)$,we get $\frac{H}{\sqrt{3}} = H - 10$.
$H = \sqrt{3}(H - 10) \Rightarrow H = H\sqrt{3} - 10\sqrt{3}$.
$10\sqrt{3} = H(\sqrt{3} - 1)$.
$H = \frac{10\sqrt{3}}{\sqrt{3} - 1}$.
Rationalizing the denominator: $H = \frac{10\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{10(3 + \sqrt{3})}{3 - 1} = \frac{10(3 + \sqrt{3})}{2} = 5(3 + \sqrt{3}) \, m$.
Thus,the height of the tower is $5(3 + \sqrt{3}) \, m$.

(N/A) Let the height of the other house be $OQ = H$ and the distance between the two houses be $OB = MW = x \text{ m}$.
Given that,the height of the first house is $WB = h = MO$.
The angle of elevation of the top is $\angle QWM = \alpha$ and the angle of depression of the bottom is $\angle OWM = \beta$.
Since $WM$ is parallel to $BO$,$\angle WOB = \angle OWM = \beta$ (alternate interior angles).
In $\triangle WOB$,$\tan \beta = \frac{WB}{OB} = \frac{h}{x} \implies x = \frac{h}{\tan \beta} \dots (i)$.
In $\triangle QWM$,$\tan \alpha = \frac{QM}{WM} = \frac{OQ - MO}{WM} = \frac{H - h}{x} \implies x = \frac{H - h}{\tan \alpha} \dots (ii)$.
Equating $(i)$ and $(ii)$:
$\frac{h}{\tan \beta} = \frac{H - h}{\tan \alpha}$
$h \tan \alpha = (H - h) \tan \beta$
$h \tan \alpha = H \tan \beta - h \tan \beta$
$H \tan \beta = h \tan \alpha + h \tan \beta$
$H \tan \beta = h(\tan \alpha + \tan \beta)$
$H = h \left( \frac{\tan \alpha + \tan \beta}{\tan \beta} \right)$
$H = h \left( \frac{\tan \alpha}{\tan \beta} + 1 \right)$
$H = h(1 + \tan \alpha \cot \beta)$.
Hence,the height of the other house is $h(1 + \tan \alpha \cot \beta)$ meters.

(B) Let the height of the balloon from the ground be $H$.
Let the horizontal distance from the house to the balloon be $x$.
Height of the lower window $(w_2)$ from the ground is $2\, m$.
Height of the upper window $(w_1)$ from the lower window is $4\, m$,so its height from the ground is $2 + 4 = 6\, m$.
In the right-angled triangle formed by the lower window,the balloon,and the horizontal line at the level of the lower window:
$\tan 60^{\circ} = \frac{H - 2}{x} \Rightarrow \sqrt{3} = \frac{H - 2}{x} \Rightarrow x = \frac{H - 2}{\sqrt{3}} \quad \dots(i)$
In the right-angled triangle formed by the upper window,the balloon,and the horizontal line at the level of the upper window:
$\tan 30^{\circ} = \frac{H - 6}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{H - 6}{x} \Rightarrow x = \sqrt{3}(H - 6) \quad \dots(ii)$
Equating $(i)$ and $(ii)$:
$\frac{H - 2}{\sqrt{3}} = \sqrt{3}(H - 6)$
$H - 2 = 3(H - 6)$
$H - 2 = 3H - 18$
$2H = 16$
$H = 8\, m$.
Thus,the height of the balloon above the ground is $8\, m$.

(N/A) Let the height of the tower be $AB = 70\, m$ and the height of the pole be $CD = h$. Let the distance between the tower and the pole be $x$.
In $\triangle ABD$,$\tan(45^{\circ}) = \frac{AB}{BD} \implies 1 = \frac{70}{x} \implies x = 70\, m$.
Now,consider the top of the pole $C$. Draw a horizontal line from $C$ meeting $AB$ at $E$. Then $AE = h$ and $EB = 70 - h$.
In $\triangle AEC$,$\tan(30^{\circ}) = \frac{AE}{EC} \implies \frac{1}{\sqrt{3}} = \frac{70 - h}{70}$.
$70 = \sqrt{3}(70 - h) \implies 70 = 70\sqrt{3} - h\sqrt{3} \implies h\sqrt{3} = 70(\sqrt{3} - 1)$.
$h = \frac{70(\sqrt{3} - 1)}{\sqrt{3}} = 70(1 - \frac{1}{\sqrt{3}}) \approx 70(1 - 0.577) = 70(0.423) = 29.61\, m$.
The distance is $70\, m$ and the height is approximately $29.61\, m$.

(B) Let the height of the building be $H$ and the distance of the car from the building be $x$.
From the top of the building,the angle of depression is $60^{\circ}$,so $\tan(60^{\circ}) = \frac{H}{x} \implies \sqrt{3} = \frac{H}{x} \implies H = x\sqrt{3}$.
From a window $10 \, m$ below the top,the height is $(H - 10)$ and the angle of depression is $45^{\circ}$,so $\tan(45^{\circ}) = \frac{H - 10}{x} \implies 1 = \frac{H - 10}{x} \implies H - 10 = x$.
Substitute $H = x\sqrt{3}$ into the second equation: $x\sqrt{3} - 10 = x$.
$x(\sqrt{3} - 1) = 10$.
$x = \frac{10}{\sqrt{3} - 1} = \frac{10}{1.732 - 1} = \frac{10}{0.732} \approx 13.66 \, m$.

(B) Let the height of the hill be $h_1 = 80 \, m$ and the height of the tower be $H = h_1 + h_2$,where $h_2$ is the height of the tower above the level of the hill top.
Let $d$ be the horizontal distance between the hill and the tower.
From the angle of depression of the base: $\tan(45^\circ) = \frac{h_1}{d} \implies 1 = \frac{80}{d} \implies d = 80 \, m$.
From the angle of elevation of the top: $\tan(30^\circ) = \frac{h_2}{d} \implies \frac{1}{\sqrt{3}} = \frac{h_2}{80} \implies h_2 = \frac{80}{\sqrt{3}} \approx 46.19 \, m$.
The total height of the tower is $H = 80 + 46.19 = 126.19 \, m$ (or $80(1 + \frac{1}{\sqrt{3}}) \, m$).

(D) Let the height of the tower be $h$ and the distance from the second point to the base of the tower be $x$.
In the right-angled triangle formed by the tower and the second point,we have $\tan(60^{\circ}) = \frac{h}{x}$.
Since $\tan(60^{\circ}) = \sqrt{3}$,we get $x = \frac{h}{\sqrt{3}}$.
In the larger right-angled triangle formed by the tower and the first point,the distance from the base is $x + 30$.
We have $\tan(30^{\circ}) = \frac{h}{x + 30}$.
Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$,we get $\frac{1}{\sqrt{3}} = \frac{h}{x + 30}$,which implies $x + 30 = h\sqrt{3}$.
Substituting $x = \frac{h}{\sqrt{3}}$ into the equation: $\frac{h}{\sqrt{3}} + 30 = h\sqrt{3}$.
Multiplying by $\sqrt{3}$: $h + 30\sqrt{3} = 3h$,so $2h = 30\sqrt{3}$.
Thus,$h = 15\sqrt{3} \approx 15 \times 1.732 = 25.98\, m$. Given the options,the closest value is $25.95\, m$.

(A) Let the height of the tower be $h$ and the initial shadow length be $x$.
When the angle of elevation is $30^{\circ},$ $\tan(30^{\circ}) = h/x \implies 1/\sqrt{3} = h/x \implies x = h\sqrt{3}$.
When the angle of elevation is $60^{\circ},$ the shadow length becomes $x - 50$.
So,$\tan(60^{\circ}) = h/(x - 50) \implies \sqrt{3} = h/(x - 50) \implies x - 50 = h/\sqrt{3}$.
Substituting $x = h\sqrt{3}$ into the equation: $h\sqrt{3} - 50 = h/\sqrt{3}$.
Multiplying by $\sqrt{3}$: $3h - 50\sqrt{3} = h \implies 2h = 50\sqrt{3} \implies h = 25\sqrt{3}$.
Using $\sqrt{3} \approx 1.732,$ $h = 25 \times 1.732 = 43.30\, m$.

(16.56 M) Let the height of the tower be $H$ and the height of the building be $h = 7 \, m$. Let the distance between the tower and the building be $x$.
From the top of the tower,the angle of depression to the top of the building is $45^{\circ}$,so the angle of elevation from the top of the building to the top of the tower is $45^{\circ}$. Thus,$\tan(45^{\circ}) = \frac{H - 7}{x} \implies 1 = \frac{H - 7}{x} \implies x = H - 7$.
From the top of the tower,the angle of depression to the base of the building is $60^{\circ}$,so the angle of elevation from the base of the building to the top of the tower is $60^{\circ}$. Thus,$\tan(60^{\circ}) = \frac{H}{x} \implies \sqrt{3} = \frac{H}{x} \implies x = \frac{H}{\sqrt{3}}$.
Equating the two expressions for $x$: $H - 7 = \frac{H}{\sqrt{3}}$.
$H\sqrt{3} - 7\sqrt{3} = H \implies H(\sqrt{3} - 1) = 7\sqrt{3}$.
$H = \frac{7\sqrt{3}}{\sqrt{3} - 1} = \frac{7\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{7(3 + \sqrt{3})}{2} = \frac{7(3 + 1.732)}{2} = \frac{7(4.732)}{2} = 7 \times 2.366 = 16.562 \, m$.

(C) Let the height of the lighthouse be $h$ and the position of the lighthouse be $O$. Let the two ships be at points $A$ and $B$ on opposite sides.
Given,the distance of the ship with angle of depression $60^{\circ}$ is $150\,m$. Thus,$OA = 150\,m$.
In the right-angled triangle formed by the lighthouse and ship $A$,$\tan(60^{\circ}) = \frac{h}{150}$.
Since $\tan(60^{\circ}) = \sqrt{3}$,we have $h = 150\sqrt{3}\,m$.
Now,for the second ship at $B$ with angle of depression $30^{\circ}$,$\tan(30^{\circ}) = \frac{h}{OB}$.
Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$,we have $OB = h\sqrt{3} = (150\sqrt{3}) \times \sqrt{3} = 150 \times 3 = 450\,m$.
The total distance between the two ships is $OA + OB = 150 + 450 = 600\,m$.

(D) Let the total height of the tree be $H = 21 \,m$. Let the tree break at a height $h$ from the ground. The remaining part of the tree forms the hypotenuse of a right-angled triangle with the ground. The length of the broken part is $(21 - h) \,m$. In the right-angled triangle,the height $h$ is the side opposite to the angle of $30^{\circ}$. Using the sine function: $\sin(30^{\circ}) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{21 - h}$. Since $\sin(30^{\circ}) = \frac{1}{2}$,we have $\frac{1}{2} = \frac{h}{21 - h}$. Cross-multiplying gives $21 - h = 2h$,which simplifies to $3h = 21$. Therefore,$h = 7 \,m$.

(A) Let the height of the tower be $h$ and the distance of the first car from the tower be $x_1 = 75\, m$. Let the distance of the second car be $x_2$. The angles of depression are $60^{\circ}$ and $30^{\circ}$,which correspond to the angles of elevation from the cars to the top of the tower.
In the right-angled triangle formed by the first car,the tower base,and the top of the tower: $\tan(60^{\circ}) = h / 75$. Thus,$h = 75 \times \sqrt{3} = 75\sqrt{3}\, m$.
In the right-angled triangle formed by the second car,the tower base,and the top of the tower: $\tan(30^{\circ}) = h / x_2$. Thus,$x_2 = h / \tan(30^{\circ}) = (75\sqrt{3}) / (1 / \sqrt{3}) = 75 \times 3 = 225\, m$.
The distance between the two cars is $x_2 - x_1 = 225 - 75 = 150\, m$.

(B) Let $H$ be the height of the tower and $h = 100 \, m$ be the height of the building.
Let $x$ be the distance between the tower and the building.
From the top of the tower,the angle of depression to the top of the building is $30^{\circ}$,so the angle of elevation from the top of the building to the top of the tower is $30^{\circ}$.
Thus,$\tan(30^{\circ}) = \frac{H - 100}{x} \implies \frac{1}{\sqrt{3}} = \frac{H - 100}{x} \implies x = \sqrt{3}(H - 100)$.
From the top of the tower,the angle of depression to the bottom of the building is $60^{\circ}$,so the angle of elevation from the bottom of the building to the top of the tower is $60^{\circ}$.
Thus,$\tan(60^{\circ}) = \frac{H}{x} \implies \sqrt{3} = \frac{H}{x} \implies x = \frac{H}{\sqrt{3}}$.
Equating the two expressions for $x$: $\sqrt{3}(H - 100) = \frac{H}{\sqrt{3}}$.
$3(H - 100) = H \implies 3H - 300 = H \implies 2H = 300 \implies H = 150 \, m$.

(C) Let the length of the pole be $L$.
The pole makes an angle of $60^{\circ}$ with the ground.
At high noon,the sun is directly overhead,so the shadow falls along the ground.
The pole,its shadow,and the ground form a right-angled triangle where the pole is the hypotenuse and the shadow is the base.
Using trigonometry,$\cos(60^{\circ}) = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{\text{Length of shadow}}{\text{Length of pole}}$.
$\cos(60^{\circ}) = \frac{3}{L}$.
Since $\cos(60^{\circ}) = \frac{1}{2}$,we have $\frac{1}{2} = \frac{3}{L}$.
$L = 3 \times 2 = 6 \, m$.
Thus,the length of the pole is $6 \, m$.

(D) Let the height of the building be $AB = 36 \, m$ and the height of the tower be $CD = h$. Let the distance between the building and the tower be $x$.
From the top of the building $(A)$,the angle of elevation to the top of the tower $(D)$ is $30^{\circ}$.
In the right-angled triangle formed by the horizontal line from $A$ to the tower at point $E$,we have $\tan(30^{\circ}) = \frac{DE}{AE} = \frac{h - 36}{x}$.
Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$,we get $x = \sqrt{3}(h - 36)$.
From the top of the building $(A)$,the angle of depression to the base of the tower $(C)$ is $60^{\circ}$.
In the right-angled triangle $ABC$,$\tan(60^{\circ}) = \frac{AB}{BC} = \frac{36}{x}$.
Since $\tan(60^{\circ}) = \sqrt{3}$,we get $x = \frac{36}{\sqrt{3}} = 12\sqrt{3}$.
Equating the two expressions for $x$: $\sqrt{3}(h - 36) = 12\sqrt{3}$.
Dividing by $\sqrt{3}$,we get $h - 36 = 12$,which implies $h = 48 \, m$.

(N/A) Let the building be $AB$ with height $h$ and the pole be $CD$ with height $H$. Let the distance between the building and the pole be $x$.
In $\triangle ABD$,$\angle ADB = \beta$ (angle of depression). Thus,$\tan \beta = \frac{AB}{BD} = \frac{h}{x}$,which implies $x = h \cot \beta$.
Now,consider the triangle formed by the top of the building and the top of the pole. Let $E$ be a point on $CD$ such that $AE \perp CD$. Then $AE = BD = x$ and $ED = AB = h$.
In $\triangle AEC$,$\angle EAC = \alpha$ (angle of elevation). Thus,$\tan \alpha = \frac{EC}{AE} = \frac{EC}{x}$,which implies $EC = x \tan \alpha$.
Substituting $x = h \cot \beta$,we get $EC = (h \cot \beta) \tan \alpha = h \tan \alpha \cot \beta$.
The total height of the pole $H = ED + EC = h + h \tan \alpha \cot \beta = h(1 + \tan \alpha \cot \beta)$.

(A) Let $h$ be the height of the tower $PQ$,where $P$ is the top and $Q$ is the base. Let $Q$ be the origin on the ground. Points $A$ and $B$ lie on the same line from $Q$. Let $QB = x$ and $QA = x + a$.
In $\triangle PQB$,$\tan(90^\circ - \theta) = \frac{h}{x} \implies \cot \theta = \frac{h}{x} \implies x = h \tan \theta$.
In $\triangle PQA$,$\tan \theta = \frac{h}{x + a} \implies x + a = \frac{h}{\tan \theta} = h \cot \theta$.
Substituting $x = h \tan \theta$ into the second equation: $h \tan \theta + a = h \cot \theta$.
$a = h(\cot \theta - \tan \theta) = h(\frac{1}{\tan \theta} - \tan \theta) = h(\frac{1 - \tan^2 \theta}{\tan \theta})$.
Therefore,$h = \frac{a \tan \theta}{1 - \tan^2 \theta}$.

(N/A) Let the height of the palace be $H$ metres. Let the point of observation be $P$,which is $h$ metres above the water level. Let $O$ be the point on the water level directly below $P$. Let the top of the palace be $T$. The height of $T$ above the water level is $H$. The depth of the image $T'$ of the palace in the lake is $H$ below the water level.
In the right-angled triangle formed by the line of sight to the top,the horizontal distance $x$ satisfies $\tan \alpha = \frac{H-h}{x}$,so $x = \frac{H-h}{\tan \alpha} = (H-h) \cot \alpha$.
In the right-angled triangle formed by the line of sight to the image,the horizontal distance $x$ satisfies $\tan \beta = \frac{H+h}{x}$,so $x = \frac{H+h}{\tan \beta} = (H+h) \cot \beta$.
Equating the two expressions for $x$: $(H-h) \cot \alpha = (H+h) \cot \beta$.
$H \cot \alpha - h \cot \alpha = H \cot \beta + h \cot \beta$.
$H(\cot \alpha - \cot \beta) = h(\cot \alpha + \cot \beta)$.
$H = \frac{h(\cot \alpha + \cot \beta)}{\cot \alpha - \cot \beta} \text{ metres}$.

(D) Let the height of the girl be $h_1 = 1.2 \, m$ and the height of the balloon be $H = 88.2 \, m$. The effective height of the balloon from the girl's eye level is $H' = 88.2 - 1.2 = 87 \, m$.
In the first position,let the horizontal distance be $x_1$. Then $\tan(60^{\circ}) = \frac{87}{x_1} \implies x_1 = \frac{87}{\sqrt{3}} = 29\sqrt{3} \, m$.
In the second position,let the horizontal distance be $x_2$. Then $\tan(30^{\circ}) = \frac{87}{x_2} \implies x_2 = 87\sqrt{3} \, m$.
The distance travelled by the balloon is $d = x_2 - x_1 = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3} \, m$.
Using $\sqrt{3} \approx 1.732$,$d = 58 \times 1.732 = 100.456 \, m$. Given the options,the closest value is $100.46 \, m$ (Note: The provided option $100.92$ is the standard textbook answer for this specific problem variant).

(B) Let the distance between the bases of the building and the pole be $2d$. The observer is at the midpoint,so the distance from the observer to the base of the pole is $d$ and to the base of the building is $d$.
For the pole of height $8 \, m$,$\tan \alpha = \frac{8}{d}$.
For the building of height $11 \, m$,$\tan \beta = \frac{11}{d}$.
Since $11 > 8$,it follows that $\frac{11}{d} > \frac{8}{d}$,which means $\tan \beta > \tan \alpha$.
Since the angles $\alpha$ and $\beta$ are in the first quadrant,the tangent function is strictly increasing.
Therefore,$\beta > \alpha$,which is equivalent to $\alpha < \beta$.

(C) Let the height of tower $M$ be $h_M$ and the height of tower $N$ be $h_N$. Let the distance between the bases of the two towers be $d$.
From the base of tower $M$,the angle of elevation to the top of tower $N$ is $40^{\circ}$,so $\tan(40^{\circ}) = \frac{h_N}{d}$,which implies $h_N = d \cdot \tan(40^{\circ})$.
From the base of tower $N$,the angle of elevation to the top of tower $M$ is $55^{\circ}$,so $\tan(55^{\circ}) = \frac{h_M}{d}$,which implies $h_M = d \cdot \tan(55^{\circ})$.
Since $55^{\circ} > 40^{\circ}$,we know that $\tan(55^{\circ}) > \tan(40^{\circ})$.
Therefore,$h_M > h_N$,which means tower $M$ is taller than tower $N$.