Mix Examples - Some Applications of Trigonometry Questions in English

(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).

(A) In a right-angled triangle,let the angles be $30^{\circ}$,$60^{\circ}$,and $90^{\circ}$.
According to trigonometric ratios,for an angle $\theta$,$\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}$.
Here,$\theta = 30^{\circ}$.
Therefore,$\sin(30^{\circ}) = \frac{\text{side opposite to } 30^{\circ}}{\text{hypotenuse}}$.
We know that $\sin(30^{\circ}) = \frac{1}{2}$.
Thus,$\frac{1}{2} = \frac{\text{side opposite to } 30^{\circ}}{\text{hypotenuse}}$.
This implies that the side opposite to the $30^{\circ}$ angle is $\frac{1}{2}$ times the hypotenuse.

(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 $N$.

(D) Let the height of the tower be $y$ and the distance from point $A$ to the base of the tower be $x$.
In the right-angled triangle formed by the tower and the ground,the angle of elevation is given as $\theta = 45^{\circ}$.
Using the trigonometric ratio $\tan(\theta) = \frac{\text{height}}{\text{base}}$,we have:
$\tan(45^{\circ}) = \frac{y}{x}$.
Since $\tan(45^{\circ}) = 1$,we get $1 = \frac{y}{x}$.
Therefore,$x = y$.

(A) Let the slope be represented by a right-angled triangle where the hypotenuse is the distance walked on the slope $(x \, m)$ and the side opposite to the angle of $30^{\circ}$ is the height from the ground $(y \, m)$.
Using the trigonometric ratio for sine:
$\sin(30^{\circ}) = \frac{\text{Opposite}}{\text{Hypotenuse}}$
$\sin(30^{\circ}) = \frac{y}{x}$
Since $\sin(30^{\circ}) = \frac{1}{2}$,we have:
$\frac{1}{2} = \frac{y}{x}$
By cross-multiplying,we get:
$x = 2y$.

(B) Let the height of the tower be $h$. Let $d_A$ and $d_B$ be the distances of houses $A$ and $B$ from the base of the tower,respectively.
From the geometry of the situation,the angle of depression is equal to the angle of elevation from the house to the top of the tower.
Thus,$\tan(25^{\circ}) = \frac{h}{d_A}$ and $\tan(40^{\circ}) = \frac{h}{d_B}$.
This implies $d_A = \frac{h}{\tan(25^{\circ})}$ and $d_B = \frac{h}{\tan(40^{\circ})}$.
Since $40^{\circ} > 25^{\circ}$,we know that $\tan(40^{\circ}) > \tan(25^{\circ})$.
Therefore,$\frac{h}{\tan(40^{\circ})} < \frac{h}{\tan(25^{\circ})}$,which means $d_B < d_A$.
Hence,house $B$ is closer to the tower than house $A$.

(C) Let the height of the tower be $AB = 30 \ m$. Let the stone be at point $C$ on the ground.
The angle of depression from the top $A$ to the stone $C$ is $45^{\circ}$.
Since the line of sight is parallel to the ground,the angle of elevation from the stone $C$ to the top $A$ is also $45^{\circ}$ (alternate interior angles).
In the right-angled triangle $ABC$,we have:
$\tan(45^{\circ}) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$.
Since $\tan(45^{\circ}) = 1$,we have $1 = \frac{30}{BC}$.
Therefore,$BC = 30 \ m$.
The distance of the stone from the base of the tower is $30 \ m$.

(D) Let the length of the ladder be $L\, m$.
In the right-angled triangle formed by the ladder,the wall,and the ground,the height of the wall is the side opposite to the angle of $30^{\circ}$,which is $3\, m$.
The ladder acts as the hypotenuse of the triangle.
Using the trigonometric ratio $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$:
$\sin(30^{\circ}) = \frac{3}{L}$
Since $\sin(30^{\circ}) = \frac{1}{2}$,we have:
$\frac{1}{2} = \frac{3}{L}$
$L = 3 \times 2 = 6\, m$.
Thus,the length of the ladder is $6\, m$.

(A) Let the height of the pole be $h \, m$.
Given that the distance from the base of the pole to the point of observation is $20 \, m$.
The angle of elevation is $\theta = 45^{\circ}$.
In the right-angled triangle formed by the pole,the ground,and the line of sight,we have:
$\tan(\theta) = \frac{\text{Height of the pole}}{\text{Distance from the base}}$
$\tan(45^{\circ}) = \frac{h}{20}$
Since $\tan(45^{\circ}) = 1$,we get:
$1 = \frac{h}{20}$
$h = 20 \, m$.
Therefore,the height of the pole is $20 \, m$.

(B) Let the length of the ladder be $L = 4 \, m$ (hypotenuse) and the distance from the wall be $d = 2 \, m$ (adjacent side).
Let $\theta$ be the angle the ladder makes with the ground.
Using the trigonometric ratio $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$.
$\cos \theta = \frac{2}{4} = \frac{1}{2}$.
Since $\cos 60^\circ = \frac{1}{2}$,we have $\theta = 60^\circ$.

(C) In $\Delta ABC$,$\angle B = 90^{\circ}$.
We are given the hypotenuse $AC = 20$ and the adjacent side to $\angle ACB$,which is $BC = 10$.
Using the trigonometric ratio for cosine:
$\cos(\angle ACB) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC}$
$\cos(\angle ACB) = \frac{10}{20} = \frac{1}{2}$
Since $\cos(60^{\circ}) = \frac{1}{2}$,we have $m\angle ACB = 60^{\circ}$.

(D) In a right-angled triangle $\Delta ABC$ where $m \angle B = 90^{\circ}$,the side opposite to $\angle C$ is $AB$ (perpendicular) and the side adjacent to $\angle C$ is $BC$ (base).
By the definition of the tangent ratio:
$\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}$
$\tan \theta = \frac{AB}{BC}$
Therefore,the correct option is $D$.

(A) The angle of depression is defined as the angle formed by the line of sight with the horizontal level when the object is below the horizontal level. Therefore,if an observer looks down at an object,the object is situated below the horizontal level.

(B) The square root of $3$ is an irrational number. Its decimal expansion is $1.7320508\ldots$
Rounding this value to $2$ decimal places,we look at the third decimal digit,which is $2$. Since $2 < 5$,we keep the second decimal digit as it is.
Therefore,the approximate value of $\sqrt{3}$ up to $2$ decimal places is $1.73$.

(C) We know that the value of $\sqrt{3} \approx 1.732$.
Therefore,$\frac{1}{\sqrt{3}} = \frac{1}{1.732}$.
By performing the division,we get $\frac{1}{1.732} \approx 0.5773$.
Rounding this value to $2$ decimal places,we get $0.58$.

(D) The square root of $2$ is an irrational number,which is approximately $1.41421356...$
Rounding this value to $2$ decimal places,we look at the third decimal digit,which is $4$. Since $4 < 5$,we keep the second decimal digit as it is.
Therefore,the approximate value of $\sqrt{2}$ up to $2$ decimal places is $1.41$.

(C) Let the height of the building be $h$ and the distance of the object from the foot of the building be $l$.
According to the problem,the angle of depression from the top of the building to the object is $\theta$.
Since the line of sight is parallel to the ground,the angle of elevation from the object to the top of the building is also $\theta$ (alternate interior angles).
In the right-angled triangle formed by the building and the ground,we have:
$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{h}{l}$
Rearranging the formula to solve for $l$:
$l = \frac{h}{\tan \theta}$
Since $\frac{1}{\tan \theta} = \cot \theta$,we get:
$l = h \cot \theta$
Therefore,the distance of the object from the foot of the building is $h \cot \theta$.

(C) Let the height of the lighthouse be $h$. Let the distance of ship $Q$ from the foot of the lighthouse be $x$ and the distance of ship $P$ be $y$.
From the geometry of the problem,the angle of elevation of the top of the lighthouse from the ships is equal to the angle of depression.
For ship $Q$: $\tan(50^{\circ}) = \frac{h}{x} \implies x = \frac{h}{\tan(50^{\circ})}$.
For ship $P$: $\tan(35^{\circ}) = \frac{h}{y} \implies y = \frac{h}{\tan(35^{\circ})}$.
Since $35^{\circ} < 50^{\circ}$,we have $\tan(35^{\circ}) < \tan(50^{\circ})$.
Therefore,$\frac{h}{\tan(35^{\circ})} > \frac{h}{\tan(50^{\circ})}$,which means $y > x$.
Thus,the distance of ship $P$ from the lighthouse is greater than the distance of ship $Q$.

(C) Let $AB$ be the wall and $AC$ be the ladder.
Given that the height of the wall $AB = 3 \, m$ and the angle of elevation $\angle C = 30^{\circ}$.
In the right-angled triangle $\triangle ABC$,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 get:
$\frac{1}{2} = \frac{3}{AC}$
$AC = 3 \times 2 = 6 \, m$.
Therefore,the length of the ladder is $6 \, m$.

(B) Let the distance between the base of the tower and the midpoint be $a$,and the distance between the base of the building and the midpoint be $a$.
In the right-angled triangle formed by the tower,we have $\tan \alpha = \frac{50}{a}$.
In the right-angled triangle formed by the building,we have $\tan \beta = \frac{30}{a}$.
Taking the ratio,we get $\frac{\tan \alpha}{\tan \beta} = \frac{50/a}{30/a} = \frac{50}{30} = \frac{5}{3}$.
Since $\frac{5}{3} > 1$,it follows that $\tan \alpha > \tan \beta$.
Since the tangent function is strictly increasing in the interval $(0, 90^\circ)$,$\tan \alpha > \tan \beta$ implies $\alpha > \beta$.

(C) Let $AB$ be the tower of height $30 \, m$ and $C$ be the position of the stone.
In the right-angled triangle $\Delta ABC$,the angle of elevation of the top of the tower from the stone is equal to the angle of depression,which is $45^{\circ}$.
Therefore,$\angle ACB = 45^{\circ}$.
Using the trigonometric ratio $\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}$:
$\tan 45^{\circ} = \frac{AB}{BC}$
Since $\tan 45^{\circ} = 1$ and $AB = 30 \, m$:
$1 = \frac{30}{BC}$
$BC = 30 \, m$.
Thus,the distance between the tower and the stone is $30 \, m$.