Heights and Distances Questions in English

(D) Let the height of the building $AB$ be $H$ metres.
Let $Q$ be the top of the tower and $P$ be the base of the tower. Given $QP = 30 \ m$.
In $\Delta APQ$,the angle of elevation from $A$ to $Q$ is equal to the angle of depression from $Q$ to $A$,which is $60^{\circ}$.
$\tan 60^{\circ} = \frac{QP}{AP} = \frac{30}{AP} = \sqrt{3}$
$AP = \frac{30}{\sqrt{3}} = 10 \sqrt{3} \ m$.
Let $R$ be a point on $QP$ such that $BR$ is horizontal. Then $BR = AP = 10 \sqrt{3} \ m$.
The angle of elevation from $B$ to $Q$ is equal to the angle of depression from $Q$ to $B$,which is $45^{\circ}$.
In $\Delta BRQ$,$\tan 45^{\circ} = \frac{QR}{BR} = 1$.
$QR = BR = 10 \sqrt{3} \ m$.
The height of the building $H = AB = RP = QP - QR$.
$H = 30 - 10 \sqrt{3} = 30 - 10(1.732) = 30 - 17.32 = 12.68 \ m$.

(D) Let $AB$ and $PQ$ be the two towers of equal height $h$ metres. Let $T$ be a point on the road $BQ$ such that $BT = x$ and $TQ = 100 - x$.
In $\Delta ABT$, $\tan 60^{\circ} = \frac{AB}{BT} = \frac{h}{x}$.
Therefore, $x = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt{3}}$.
In $\Delta PQT$, $\tan 30^{\circ} = \frac{PQ}{TQ} = \frac{h}{100 - x}$.
Therefore, $100 - x = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}$.
Substituting $x = \frac{h}{\sqrt{3}}$ into the equation $100 - x = h\sqrt{3}$:
$100 = x + h\sqrt{3} = \frac{h}{\sqrt{3}} + h\sqrt{3}$.
$100 = h \left( \frac{1 + 3}{\sqrt{3}} \right) = h \left( \frac{4}{\sqrt{3}} \right)$.
$h = \frac{100 \sqrt{3}}{4} = 25 \sqrt{3} \, m$.

(C) Let the length of the ladder $AP$ be $l \ m$.
In the right-angled triangle $\Delta ABP$,the angle of elevation at the foot of the ladder $P$ is $\angle APB = 60^{\circ}$.
The distance from the foot of the ladder to the house is $BP = 6.5 \ m$.
Using the trigonometric ratio for cosine:
$\cos 60^{\circ} = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BP}{AP}$
$\frac{1}{2} = \frac{6.5}{l}$
$l = 6.5 \times 2 = 13 \ m$.
Thus,the length of the ladder is $13 \ m$.

(A) Let $H$ be the height of the aeroplane above the road in miles.
Let $P$ be the position of the aeroplane and $Q$ be the point on the road directly below it.
Let $R$ and $S$ be the positions of the two consecutive milestones on the road.
The distance between two consecutive milestones is $RS = 1 \text{ mile}$.
In $\Delta PQR$,$\tan 60^{\circ} = \frac{H}{QR} \Rightarrow QR = \frac{H}{\sqrt{3}}$.
In $\Delta PQS$,$\tan 30^{\circ} = \frac{H}{QS} \Rightarrow QS = H\sqrt{3}$.
Since $QS = QR + RS$,we have $H\sqrt{3} = \frac{H}{\sqrt{3}} + 1$.
Multiplying by $\sqrt{3}$,we get $3H = H + \sqrt{3}$.
$2H = \sqrt{3} \Rightarrow H = \frac{\sqrt{3}}{2} \text{ miles}$.

(A) Let $H$ be the height of the tower $PQ$. Let $S$ be the top of the cliff of height $60 \ m$. $R$ is the point on the ground directly below $S$.
In $\Delta S R Q$,$\tan 60^{\circ} = \frac{SR}{RQ} = \frac{60}{RQ}$.
Therefore,$RQ = \frac{60}{\tan 60^{\circ}} = \frac{60}{\sqrt{3}} = 20\sqrt{3} \ m$.
Since $RQ = TP$,we have $TP = 20\sqrt{3} \ m$.
In $\Delta STP$,$\tan 30^{\circ} = \frac{ST}{TP} = \frac{60 - H}{20\sqrt{3}}$.
$\frac{1}{\sqrt{3}} = \frac{60 - H}{20\sqrt{3}}$.
$20 = 60 - H$.
$H = 60 - 20 = 40 \ m$.

(D) Let $PQ$ be the height of the unfinished tower and $R$ be the point on the ground at a distance of $120 \ m$ from the base $Q$.
In $\Delta PQR$,$\tan 45^{\circ} = \frac{PQ}{RQ}$.
Since $\tan 45^{\circ} = 1$,we have $PQ = RQ = 120 \ m$.
Let the tower be raised to a height $P'$ such that the angle of elevation from $R$ becomes $60^{\circ}$.
In $\Delta P'QR$,$\tan 60^{\circ} = \frac{P'Q}{RQ}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $P'Q = RQ \cdot \sqrt{3} = 120 \sqrt{3} \ m$.
The additional height required is $P'P = P'Q - PQ = 120 \sqrt{3} - 120 = 120(\sqrt{3} - 1) \ m$.
Using $\sqrt{3} \approx 1.732$,we get $P'P = 120(1.732 - 1) = 120 \times 0.732 = 87.84 \ m$.

(D) Let the height of the cliff be $PS = 200 \, m$. Let the two ships be at points $Q$ and $R$ such that $S, Q, R$ are collinear.
In $\Delta PQS$,the angle of elevation is $45^{\circ}$.
$\tan 45^{\circ} = \frac{PS}{SQ} \implies 1 = \frac{200}{SQ} \implies SQ = 200 \, m$.
In $\Delta PRS$,the angle of elevation is $30^{\circ}$.
$\tan 30^{\circ} = \frac{PS}{SR} \implies \frac{1}{\sqrt{3}} = \frac{200}{SQ + QR}$.
$SQ + QR = 200\sqrt{3}$.
$200 + QR = 200(1.732) = 346.4$.
$QR = 346.4 - 200 = 146.4 \, m$.
Thus,the distance between the two ships is $146.4 \, m$.

(A) Let $PQ$ be the tower and $R$ and $S$ be the two positions of the boat. Given $RQ = 60 \ m$,$\angle PSQ = 30^{\circ}$,and $\angle PRQ = 45^{\circ}$.
From $\Delta PQR$,$\tan 45^{\circ} = \frac{PQ}{RQ} = 1$.
Thus,$PQ = RQ = 60 \ m$.
From $\Delta PQS$,$\tan 30^{\circ} = \frac{PQ}{QS} = \frac{60}{RQ + RS} = \frac{60}{60 + RS} = \frac{1}{\sqrt{3}}$.
This implies $60 + RS = 60\sqrt{3}$.
So,$RS = 60(\sqrt{3} - 1) \ m$.
Using $\sqrt{3} \approx 1.732$,$RS = 60(1.732 - 1) = 60 \times 0.732 = 43.92 \ m$.
Speed of the boat = $\frac{\text{Distance}}{\text{Time}} = \frac{43.92 \ m}{5 \ s} = 8.784 \ m/s$.
To convert to $km/hr$,multiply by $\frac{18}{5}$:
Speed = $8.784 \times \frac{18}{5} = 31.62 \ km/hr$.
Rounding to the nearest integer,the speed is approximately $32 \ km/hr$.

(C) Let the tower be represented by $QR$ and the point on the ground be $P$.
Given: Height of the tower $QR = 100 \ m$,and the angle of elevation $\angle QPR = 30^{\circ}$.
Let the distance of point $P$ from the foot of the tower $Q$ be $PQ = x$.
In the right-angled triangle $\Delta PQR$,we have:
$\tan(30^{\circ}) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{QR}{PQ}$
$\frac{1}{\sqrt{3}} = \frac{100}{x}$
$x = 100 \sqrt{3} \ m$
Using $\sqrt{3} \approx 1.732$,we get:
$x = 100 \times 1.732 = 173.2 \ m$
Rounding to the nearest integer,the distance is $173 \ m$.

(C) Let $PQ$ be the height of the tower $H$. Let the car be at point $S$ initially and at point $R$ after $12$ minutes. The angles of depression are $30^{\circ}$ and $45^{\circ}$ respectively,which are equal to the angles of elevation $\angle PSQ = 30^{\circ}$ and $\angle PRQ = 45^{\circ}$.
In $\Delta PQR$,$\tan 45^{\circ} = \frac{PQ}{RQ} = 1 \implies PQ = RQ = H$.
In $\Delta PQS$,$\tan 30^{\circ} = \frac{PQ}{SQ} = \frac{1}{\sqrt{3}} \implies SQ = H\sqrt{3}$.
The distance $SR = SQ - RQ = H\sqrt{3} - H = H(\sqrt{3} - 1)$.
The car covers distance $SR$ in $12$ minutes. Let the time taken to cover distance $RQ$ be $t$ minutes.
Since speed is uniform,$\frac{SR}{12} = \frac{RQ}{t}$.
$\frac{H(\sqrt{3} - 1)}{12} = \frac{H}{t} \implies t = \frac{12}{\sqrt{3} - 1} = \frac{12(\sqrt{3} + 1)}{3 - 1} = 6(\sqrt{3} + 1) \approx 6(1.732 + 1) = 6(2.732) = 16.392 \text{ minutes}$.
$0.392 \text{ minutes} = 0.392 \times 60 \approx 23.5 \text{ seconds}$.
Thus,the time is $16 \text{ minutes } 23 \text{ seconds}$.

(D) Let the height of the tower $NM = H$ and the distance $LM = x$.
In $\triangle NQM$,$\tan 60^{\circ} = \frac{NM}{QM} \implies \sqrt{3} = \frac{H}{QM} \implies QM = \frac{H}{\sqrt{3}}$.
In $\triangle NLM$,$\tan 30^{\circ} = \frac{NM}{LM} \implies \frac{1}{\sqrt{3}} = \frac{H}{x} \implies x = H\sqrt{3}$.
Since the height $H$ of the tower is not provided,the exact numerical value of $x$ cannot be determined.
Therefore,the data is inadequate.

(A) Let $QR$ represent the tree and $PQ$ represent its shadow,as shown in the figure.
It is given that the length of the shadow is $\sqrt{3}$ times the height of the tree,i.e.,$PQ = \sqrt{3} QR$.
In the right-angled triangle $\triangle PQR$,the angle of elevation $\theta$ is given by:
$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{QR}{PQ}$
Substituting the given value of $PQ$:
$\tan \theta = \frac{QR}{\sqrt{3} QR} = \frac{1}{\sqrt{3}}$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have:
$\theta = 30^{\circ}$

(C) In the figure,the ladder is represented by $AC$,where $AC$ is the hypotenuse of the right-angled triangle $ABC$.
The distance from the foot of the ladder to the wall is $AB = 4.242 \ m$.
Using the trigonometric ratio for cosine:
$\cos 45^{\circ} = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC}$
$\frac{1}{\sqrt{2}} = \frac{4.242}{AC}$
$AC = 4.242 \times \sqrt{2}$
Given $\sqrt{2} \approx 1.414$,we have:
$AC = 4.242 \times 1.414 \approx 6 \ m$.
Thus,the length of the ladder is $6 \ m$.

(C) Let the tree be represented by $BC$ with height $H$ and the breadth of the river be $DB = x$.
From the given figure,in $\triangle DBC$,$\tan 60^{\circ} = \frac{H}{x}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $H = x\sqrt{3} \quad \dots(1)$.
In $\triangle ABC$,$\tan 30^{\circ} = \frac{H}{AB} = \frac{H}{36 + x}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $H = \frac{36 + x}{\sqrt{3}} \quad \dots(2)$.
Equating $(1)$ and $(2)$:
$x\sqrt{3} = \frac{36 + x}{\sqrt{3}}$
$3x = 36 + x$
$2x = 36$
$x = 18 \ m$.
Thus,the breadth of the river is $18 \ m$.

(A) Let $CD$ be the tower of height $60 \ m$ and $AB$ be the other tower of height $h$. Let the width of the river be $BD = x$.
From the diagram,$CD = 60 \ m$. The angle of depression of the top of tower $AB$ (point $A$) from $C$ is $30^{\circ}$,and the angle of depression of the bottom of tower $AB$ (point $B$) from $C$ is $60^{\circ}$.
In $\Delta CBD$,$\tan 60^{\circ} = \frac{CD}{BD} = \frac{60}{x}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $\sqrt{3} = \frac{60}{x}$,which gives $x = \frac{60}{\sqrt{3}} = 20 \sqrt{3} \ m$. Thus,the width of the river is $20 \sqrt{3} \ m$.
Now,draw a horizontal line from $A$ to $CD$,meeting $CD$ at point $E$. Then $AE = BD = 20 \sqrt{3} \ m$ and $ED = AB = h$.
In $\Delta CAE$,$\tan 30^{\circ} = \frac{CE}{AE} = \frac{CE}{20 \sqrt{3}}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{CE}{20 \sqrt{3}}$,which gives $CE = 20 \ m$.
Therefore,the height of the other tower $AB = ED = CD - CE = 60 - 20 = 40 \ m$.

(A) Let $AB$ be the rod and $AC$ be its shadow.
Let $\angle ACB = \theta$.
Given that the ratio of the length of the rod to its shadow is $AB : AC = 1 : \sqrt{3}$.
Let $AB = x$,then $AC = \sqrt{3}x$.
In the right-angled triangle $\triangle ABC$,we have:
$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AB}{AC}$
$\tan \theta = \frac{x}{\sqrt{3}x} = \frac{1}{\sqrt{3}}$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we get $\theta = 30^{\circ}$.
Therefore,the angle of elevation of the sun is $30^{\circ}$.

(B) Let the height of the pole be $AB = h$ and the length of its shadow be $AC = h$.
In the right-angled triangle $ABC$,where $\angle A = 90^{\circ}$ and $\angle C = \theta$ is the angle of elevation of the sun:
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{AC}$
$\tan \theta = \frac{h}{h} = 1$
Since $\tan 45^{\circ} = 1$,we have $\theta = 45^{\circ}$.

(C) Let the height of the tower be $h \ m$.
Given that the distance from the base of the tower is $100 \ m$ and the angle of elevation is $30^{\circ}$.
In the right-angled triangle formed,we have:
$\tan(30^{\circ}) = \frac{\text{Height of the tower}}{\text{Distance from the base}}$
$\tan(30^{\circ}) = \frac{h}{100}$
Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$,we get:
$\frac{1}{\sqrt{3}} = \frac{h}{100}$
$h = \frac{100}{\sqrt{3}} \ m$.

(B) Let $AB$ be the building and $AC$ be its shadow.
Given,height of the building $AB = 50 \ m$ and the angle of elevation of the sun $\theta = 30^{\circ}$.
In the right-angled triangle $\triangle ABC$,we have:
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{AC}$
$\tan 30^{\circ} = \frac{50}{AC}$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we get:
$\frac{1}{\sqrt{3}} = \frac{50}{AC}$
$AC = 50 \sqrt{3} \ m$
Therefore,the length of the shadow is $50 \sqrt{3} \ m$.

(B) Let $OP$ be the original pole,which breaks at point $A$. Let the top $P$ strike the ground at $P^{\prime}$.
Given: $OP^{\prime} = 21 \ m$ and $\angle OP^{\prime}A = 30^{\circ}$.
In $\triangle AOP^{\prime}$,$\tan 30^{\circ} = \frac{OA}{OP^{\prime}}$
$\frac{1}{\sqrt{3}} = \frac{OA}{21} \Rightarrow OA = \frac{21}{\sqrt{3}} = 7\sqrt{3} \ m$.
Also,$\cos 30^{\circ} = \frac{OP^{\prime}}{AP^{\prime}}$
$\frac{\sqrt{3}}{2} = \frac{21}{AP^{\prime}} \Rightarrow AP^{\prime} = \frac{42}{\sqrt{3}} = 14\sqrt{3} \ m$.
Since the broken part $AP$ is equal to $AP^{\prime}$,the total height of the pole is $OA + AP = OA + AP^{\prime}$.
Total height $= 7\sqrt{3} + 14\sqrt{3} = 21\sqrt{3} \ m$.

(A) Let the tree be represented by $PQ$,where $Q$ is the root and $M$ is the point where it breaks. When the top $P$ touches the ground at $A$,we have a right-angled triangle $\triangle MQA$ with $\angle MAQ = 30^{\circ}$ and $AQ = 10 \ m$.
In $\triangle MQA$:
$\tan 30^{\circ} = \frac{MQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{MQ}{10} \Rightarrow MQ = \frac{10}{\sqrt{3}} \ m$.
Also,$\cos 30^{\circ} = \frac{AQ}{AM} \Rightarrow \frac{\sqrt{3}}{2} = \frac{10}{AM} \Rightarrow AM = \frac{20}{\sqrt{3}} \ m$.
Since the broken part $MP$ is equal to $AM$,the total height of the tree is $MQ + MP = MQ + AM$.
Height $= \frac{10}{\sqrt{3}} + \frac{20}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3} \ m$.

(D) Let the height of the tower $AB$ be $h$. Let the two points on the road be $C$ and $D$ such that $CD = 500 \ m$. Let $AD = x$.
In $\triangle ABD$,$\tan 60^{\circ} = \frac{AB}{AD} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}}$.
In $\triangle ABC$,$\tan 45^{\circ} = \frac{AB}{AC} = \frac{h}{x+500} \implies 1 = \frac{h}{\frac{h}{\sqrt{3}} + 500}$.
$h = \frac{h}{\sqrt{3}} + 500 \implies h(1 - \frac{1}{\sqrt{3}}) = 500 \implies h(\frac{\sqrt{3}-1}{\sqrt{3}}) = 500$.
$h = \frac{500 \sqrt{3}}{\sqrt{3}-1} \ m$.
Rationalizing the denominator: $h = \frac{500 \sqrt{3}(\sqrt{3}+1)}{3-1} = \frac{500(3+\sqrt{3})}{2} = 250(3+\sqrt{3}) \ m$.

(C) Let the height of the tower be $h \ m$. Let $AB$ be the shadow when the sun's altitude is $60^{\circ}$ and $AC$ be the shadow when the sun's altitude is $30^{\circ}$.
Given,$BC = 50 \ m$.
In $\triangle TAB$,$\tan 60^{\circ} = \frac{h}{AB} \Rightarrow \sqrt{3} = \frac{h}{AB} \Rightarrow AB = \frac{h}{\sqrt{3}}$.
In $\triangle TAC$,$\tan 30^{\circ} = \frac{h}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + 50} \Rightarrow AB + 50 = h\sqrt{3}$.
Substituting $AB = \frac{h}{\sqrt{3}}$ in the second equation:
$\frac{h}{\sqrt{3}} + 50 = h\sqrt{3}$
$50 = h\sqrt{3} - \frac{h}{\sqrt{3}}$
$50 = h \left( \frac{3 - 1}{\sqrt{3}} \right) = h \left( \frac{2}{\sqrt{3}} \right)$
$h = \frac{50 \sqrt{3}}{2} = 25 \sqrt{3} \ m$.

(B) Let $ABCD$ be the rectangle where $\angle BAC = 30^{\circ}$ and the diagonal $AC = 6 \text{ cm}$.
In the right-angled triangle $\triangle ABC$:
$\cos 30^{\circ} = \frac{AB}{AC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{AB}{6} \Rightarrow AB = 3\sqrt{3} \text{ cm}$.
$\sin 30^{\circ} = \frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{BC}{6} \Rightarrow BC = 3 \text{ cm}$.
The area of the rectangle is $AB \times BC = (3\sqrt{3}) \times 3 = 9\sqrt{3} \text{ cm}^2$.

(C) Let $AB$ be the tower of height $100 \text{ m}$. Let $AC$ and $AD$ be the lengths of the shadows when the angle of elevation is $30^{\circ}$ and $45^{\circ}$ respectively.
In $\triangle ABD$,$\tan 45^{\circ} = \frac{AB}{AD} \Rightarrow 1 = \frac{100}{AD} \Rightarrow AD = 100 \text{ m}$.
In $\triangle ABC$,$\tan 30^{\circ} = \frac{AB}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{AC} \Rightarrow AC = 100\sqrt{3} \text{ m}$.
The decrease in the length of the shadow is $x = AC - AD$.
$x = 100\sqrt{3} - 100 = 100(\sqrt{3} - 1) \text{ m}$.

(A) Let the height of the electric pole be $PQ = 20 \, m$ and the length of the steel wire be $OP$. The pole makes a right angle with the ground,so $\triangle OPQ$ is a right-angled triangle at $Q$.
Using the trigonometric ratio $\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$:
$\sin 60^{\circ} = \frac{PQ}{OP}$
$\frac{\sqrt{3}}{2} = \frac{20}{OP}$
$OP = \frac{20 \times 2}{\sqrt{3}} = \frac{40}{\sqrt{3}} \, m$.
Thus,the length of the steel wire is $\frac{40}{\sqrt{3}} \, m$.

(C) Let $T$ be the top of the lighthouse and $L$ be the base of the lighthouse at sea level. Let $B$ be the position of the boat.
Given,height of the lighthouse $TL = 50 \ m$.
The angle of depression from the top $T$ to the boat $B$ is $30^{\circ}$.
Since the line of sight is parallel to the sea level,the angle of elevation from the boat $B$ to the top $T$ is also $30^{\circ}$ (alternate interior angles).
In the right-angled triangle $\triangle TBL$,we have:
$\tan(30^{\circ}) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{TL}{BL}$
$\frac{1}{\sqrt{3}} = \frac{50}{BL}$
$BL = 50 \sqrt{3} \ m$
Therefore,the boat is $50 \sqrt{3} \ m$ away from the lighthouse.

(C) Let $AB$ be the cliff of height $25 \ m$ and $CD$ be the tower.
Let the angle of elevation of the top of the tower $D$ from the top of the cliff $B$ be $\alpha$.
Let the angle of depression of the foot of the tower $C$ from the top of the cliff $B$ be $\alpha$.
Draw $BE \perp CD$,where $E$ is on $CD$.
In $\triangle ABC$,$\tan \alpha = \frac{AB}{AC} = \frac{25}{AC}$.
In $\triangle BDE$,$\tan \alpha = \frac{DE}{BE}$.
Since $BE = AC$,we have $\frac{DE}{BE} = \frac{25}{BE}$,which implies $DE = 25 \ m$.
Since $CE = AB = 25 \ m$,the total height of the tower $CD = CE + DE = 25 + 25 = 50 \ m$.

(B) Let the height of the pole be $h$ and the length of the shadow be $x$.
Given that the length of the shadow is equal to the height of the pole,we have $h = x$.
In the right-angled triangle formed by the pole and its shadow,the angle of elevation $\theta$ is given by:
$\tan \theta = \frac{\text{Height of the pole}}{\text{Length of the shadow}} = \frac{h}{x}$
Since $h = x$,we get:
$\tan \theta = \frac{h}{h} = 1$
We know that $\tan 45^{\circ} = 1$.
Therefore,$\theta = 45^{\circ}$.

(B) Let $AC$ be the lighthouse of height $60 \ m$ and $B$ be the position of the boat.
Let $x$ be the distance of the boat from the base of the lighthouse $(AB = x)$.
The angle of depression from the top $C$ to the boat $B$ is $15^{\circ}$,so the angle of elevation from the boat to the top is also $15^{\circ}$.
In $\triangle ABC$,$\tan 15^{\circ} = \frac{AC}{AB} = \frac{60}{x}$.
Therefore,$x = \frac{60}{\tan 15^{\circ}}$.
We know that $\tan 15^{\circ} = \tan(45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.
Substituting this into the equation for $x$:
$x = 60 \div \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right) = 60 \left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) \ m$.

(B) Let the height of the tower be $h$ and the distance from the tower to the second point be $x$.
In $\triangle BAQ$,$\tan 60^{\circ} = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \dots(1)$
In $\triangle BAP$,$\tan 30^{\circ} = \frac{h}{x+20} \implies \frac{1}{\sqrt{3}} = \frac{h}{x+20} \implies x+20 = h\sqrt{3} \quad \dots(2)$
Substitute $x = \frac{h}{\sqrt{3}}$ from $(1)$ into $(2)$:
$\frac{h}{\sqrt{3}} + 20 = h\sqrt{3}$
$20 = h\sqrt{3} - \frac{h}{\sqrt{3}}$
$20 = h \left( \frac{3-1}{\sqrt{3}} \right)$
$20 = h \left( \frac{2}{\sqrt{3}} \right)$
$h = \frac{20 \times \sqrt{3}}{2} = 10\sqrt{3} \ m$.

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

(A) Let $MP$ represent the house with height $15 \, m$.
Let $O$ be the point on the ground at a distance of $15 \, m$ from the base $M$ of the house.
In the right-angled triangle $\triangle OMP$,the angle of elevation of the top $P$ from point $O$ is $\theta$.
Using the trigonometric ratio for tangent:
$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{MP}{OM}$
$\tan \theta = \frac{15 \, m}{15 \, m} = 1$
Since $\tan 45^{\circ} = 1$,we have $\theta = 45^{\circ}$.

(C) Let $MA$ be the lighthouse of height $h$ meters. Let $P$ and $Q$ be the positions of the two boats such that $PQ = 60 \, m$.
In $\triangle AMP$,$\tan 45^{\circ} = \frac{MA}{MP} \implies 1 = \frac{h}{MP} \implies MP = h$.
In $\triangle AMQ$,$\tan 30^{\circ} = \frac{MA}{MQ} \implies \frac{1}{\sqrt{3}} = \frac{h}{MP + PQ} \implies \frac{1}{\sqrt{3}} = \frac{h}{h + 60}$.
Cross-multiplying,we get $h + 60 = h\sqrt{3}$.
Rearranging,$h(\sqrt{3} - 1) = 60$.
$h = \frac{60}{\sqrt{3} - 1}$.
Rationalizing the denominator,$h = \frac{60(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{60(\sqrt{3} + 1)}{3 - 1} = \frac{60(\sqrt{3} + 1)}{2} = 30(\sqrt{3} + 1) \, m$.

(B) Let $h$ be the height of the hill and $O$ be the foot of the hill on the horizontal plane.
Since the angle of elevation from each vertex $A, B, C$ is $\alpha$,we have $OA = OB = OC = h \cot \alpha$.
This implies that $O$ is the circumcentre of $\triangle ABC$ with circumradius $R = h \cot \alpha$.
In any triangle $ABC$,the circumradius $R$ is given by $R = \frac{a}{2 \sin A}$.
Equating the two expressions for $R$: $h \cot \alpha = \frac{a}{2 \sin A}$.
Therefore,$h = \frac{a \tan \alpha}{2 \sin A} = \frac{a}{2} \tan \alpha \operatorname{cosec} A$.

(A) Let $PQ = x$ be the height of the tower and $A$ be the point on the same level as the foot of the tower $Q$. Given $\angle PAQ = 30^{\circ}$.
Let $B$ be a point $h \ m$ vertically above $A$,so $AB = h$. The angle of depression of the foot of the tower $Q$ from $B$ is $60^{\circ}$,which implies $\angle BQA = 60^{\circ}$.
In the right-angled triangle $\triangle ABQ$,we have:
$\tan 60^{\circ} = \frac{AB}{AQ}$
$\sqrt{3} = \frac{h}{AQ}$
$AQ = \frac{h}{\sqrt{3}} = h \cot 60^{\circ}$.
Thus,the horizontal distance of the tower from the point $A$ is $h \cot 60^{\circ}$.

(A) Let $OP$ be the tower of height $h \, m$ and $PQ$ be the flagstaff of height $6 \, m$. Let the sun make an angle $\theta$ with the ground. Let $OA = x$ be the shadow of the tower and $AB = 2 \sqrt{3} \, m$ be the shadow of the flagstaff.
In $\triangle OAP$,$\tan \theta = \frac{OP}{OA} = \frac{h}{x}$.
In $\triangle OBQ$,$\tan \theta = \frac{OQ}{OB} = \frac{h + 6}{x + 2 \sqrt{3}}$.
Equating the two expressions for $\tan \theta$:
$\frac{h}{x} = \frac{h + 6}{x + 2 \sqrt{3}}$
$h(x + 2 \sqrt{3}) = x(h + 6)$
$hx + 2 \sqrt{3}h = hx + 6x$
$2 \sqrt{3}h = 6x$
$\frac{h}{x} = \frac{6}{2 \sqrt{3}} = \sqrt{3}$.
Since $\tan \theta = \frac{h}{x}$,we have $\tan \theta = \sqrt{3}$.
Therefore,$\theta = 60^{\circ}$.

(C) Let $OP$ be the height of the tree and $OA$ be the breadth of the river.
In $\triangle OAP$,$\tan 60^{\circ} = \frac{OP}{OA} = \sqrt{3} \implies OP = OA \sqrt{3}$.
When the person moves $40 \ m$ away,the new distance from the bank is $OB = OA + 40$.
In $\triangle OBP$,$\tan 30^{\circ} = \frac{OP}{OB} = \frac{1}{\sqrt{3}}$.
Substituting $OP = OA \sqrt{3}$ into the second equation:
$\frac{OA \sqrt{3}}{OA + 40} = \frac{1}{\sqrt{3}}$
$3 OA = OA + 40$
$2 OA = 40$
$OA = 20 \ m$.

(A) Let the height of the pole be $AB = h$ and the length of the shadow be $AC = \sqrt{3}h$.
In the right-angled triangle $\triangle ABC$,the angle of elevation is $\theta$.
We know that $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{AC}$.
Substituting the values,we get $\tan \theta = \frac{h}{\sqrt{3}h} = \frac{1}{\sqrt{3}}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,therefore $\theta = 30^{\circ}$.

(C) Let $OP = h$ be the height of the tower,where $O$ is the foot of the tower.
In $\triangle OAP$,$\angle OAP = \alpha$. Thus,$OA = h \cot \alpha$.
In $\triangle OBP$,$\angle OBP = 2 \alpha$. Thus,$OB = h \cot 2 \alpha$.
Given $AB = a$,we have $OA - OB = a$.
Therefore,$h \cot \alpha - h \cot 2 \alpha = a$.
$h (\frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2 \alpha}{\sin 2 \alpha}) = a$.
$h (\frac{\sin 2 \alpha \cos \alpha - \cos 2 \alpha \sin \alpha}{\sin \alpha \sin 2 \alpha}) = a$.
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:
$h (\frac{\sin(2 \alpha - \alpha)}{\sin \alpha \sin 2 \alpha}) = a$.
$h (\frac{\sin \alpha}{\sin \alpha \sin 2 \alpha}) = a$.
$h (\frac{1}{\sin 2 \alpha}) = a$.
$h = a \sin 2 \alpha$.
Thus,the height of the tower is $a \sin 2 \alpha$.

(B) Let the initial height of the tower be $h$ and the additional height required be $x$. The distance from the base is $120 \ m$.
From the first condition,the angle of elevation is $45^{\circ}$:
$\tan 45^{\circ} = \frac{h}{120}$
$1 = \frac{h}{120} \implies h = 120 \ m$.
From the second condition,the angle of elevation of the new top is $60^{\circ}$:
$\tan 60^{\circ} = \frac{h+x}{120}$
$\sqrt{3} = \frac{120+x}{120}$
$120\sqrt{3} = 120 + x$
$x = 120\sqrt{3} - 120$
$x = 120(\sqrt{3}-1) \ m$.
Thus,the tower must be raised by $120(\sqrt{3}-1) \ m$.

(A) Let $h$ be the height of the hill $QP$ and $x$ be the distance from the second point $A$ to the base $P$ of the hill.
In $\triangle QPA$,$\tan 60^{\circ} = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies h = \sqrt{3}x$.
In $\triangle QPO$,the distance $OA = \sqrt{3} \text{ km}$,so $OP = OA + AP = \sqrt{3} + x$.
$\tan 30^{\circ} = \frac{h}{\sqrt{3} + x} \implies \frac{1}{\sqrt{3}} = \frac{h}{\sqrt{3} + x} \implies \sqrt{3}h = \sqrt{3} + x$.
Substitute $h = \sqrt{3}x$ into the equation: $\sqrt{3}(\sqrt{3}x) = \sqrt{3} + x \implies 3x = \sqrt{3} + x \implies 2x = \sqrt{3} \implies x = \frac{\sqrt{3}}{2} \text{ km}$.
Therefore,$h = \sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{3}{2} \text{ km}$.

(A) Let the heights of the two poles be $h_1 = 20 \ m$ and $h_2 = 14 \ m$.
The difference in the heights of the poles is $h_1 - h_2 = 20 \ m - 14 \ m = 6 \ m$.
Let the length of the wire be $l$.
In the right-angled triangle formed by the wire,the horizontal line,and the vertical difference between the poles,the angle of elevation is $30^{\circ}$.
Using the sine ratio: $\sin 30^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{6}{l}$.
Since $\sin 30^{\circ} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{6}{l}$.
Therefore,$l = 6 \times 2 = 12 \ m$.

(C) Let the height of the pole be $H = 6 \, m$ and the length of its shadow be $S = 8 \, m$.
Let the height of the man be $h$ and the length of his shadow be $s = 2.4 \, m$.
Since the angle of elevation of the sun is the same for both the pole and the man,the ratio of height to shadow length is constant.
$\frac{H}{S} = \frac{h}{s}$
Substituting the given values:
$\frac{6}{8} = \frac{h}{2.4}$
$\frac{3}{4} = \frac{h}{2.4}$
$h = \frac{3}{4} \times 2.4$
$h = 3 \times 0.6 = 1.8 \, m$.
Thus,the height of the man is $1.8 \, m$.

(D) Let $AB = h$ be the height of the tower and $A$ be the foot of the tower.
Let $GA = x$ be the initial distance from the tower.
In $\triangle GAB$,$\tan 30^{\circ} = \frac{AB}{GA} = \frac{h}{x}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{h}{x} = \frac{1}{\sqrt{3}}$,so $x = h\sqrt{3}$.
After walking $20 \ m$ towards the tower,the new distance is $HA = x - 20$.
In $\triangle HAB$,$\tan 60^{\circ} = \frac{AB}{HA} = \frac{h}{x - 20}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $\frac{h}{x - 20} = \sqrt{3}$,so $h = \sqrt{3}(x - 20)$.
Substituting $x = h\sqrt{3}$ into the equation: $h = \sqrt{3}(h\sqrt{3} - 20)$.
$h = 3h - 20\sqrt{3}$.
$2h = 20\sqrt{3}$.
$h = 10\sqrt{3} \ m$.

(B) Let the height of the aeroplane be $h = 4500 \text{ m}$.
Let the initial position be $A$ and the final position be $B$.
At point $A$, the angle of elevation is $60^{\circ}$. Let the distance from the point on the ground to the point directly below $A$ be $x$.
$\tan(60^{\circ}) = \frac{4500}{x} \implies \sqrt{3} = \frac{4500}{x} \implies x = \frac{4500}{\sqrt{3}} = 1500\sqrt{3} \text{ m}$.
At point $B$, the angle of elevation is $30^{\circ}$. Let the total distance from the point on the ground to the point directly below $B$ be $y$.
$\tan(30^{\circ}) = \frac{4500}{y} \implies \frac{1}{\sqrt{3}} = \frac{4500}{y} \implies y = 4500\sqrt{3} \text{ m}$.
The distance covered by the aeroplane is $d = y - x = 4500\sqrt{3} - 1500\sqrt{3} = 3000\sqrt{3} \text{ m}$.
The time taken is $t = 30 \text{ s}$.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{3000\sqrt{3}}{30} = 100\sqrt{3} \text{ m/s}$.

(D) Let the two poles be $AB$ and $CD$ with heights $60 \ m$ and $35 \ m$ respectively.
Let $DE$ be the horizontal line from $D$ to the pole $AB$,meeting $AB$ at $E$.
Then $BE = CD = 35 \ m$.
The height $AE = AB - BE = 60 - 35 = 25 \ m$.
In the right-angled triangle $\triangle AED$,the angle of depression/elevation is $\phi$.
Given $\tan \phi = \frac{5}{9}$.
In $\triangle AED$,$\tan \phi = \frac{AE}{DE}$.
Substituting the values,$\frac{5}{9} = \frac{25}{DE}$.
$DE = \frac{25 \times 9}{5} = 5 \times 9 = 45 \ m$.
Thus,the distance between the two poles is $45 \ m$.

(D) Let the height of the shorter pillar $CD = h$ and the height of the taller pillar $AB = 3h$.
The distance between the pillars $BD = 120 \text{ m}$.
Let $E$ be the midpoint of $BD$,so $BE = ED = 60 \text{ m}$.
In $\triangle ABE$,$\tan \theta = \frac{AB}{BE} = \frac{3h}{60} = \frac{h}{20} \quad ... (i)$
In $\triangle CDE$,$\tan(90^\circ - \theta) = \frac{CD}{ED} = \frac{h}{60} \implies \cot \theta = \frac{h}{60} \quad ... (ii)$
Multiplying $(i)$ and $(ii)$:
$\tan \theta \cdot \cot \theta = \frac{h}{20} \cdot \frac{h}{60}$
$1 = \frac{h^2}{1200}$
$h^2 = 1200$
$h = \sqrt{1200} = 20\sqrt{3} \text{ m}$.
The height of the taller pillar is $3h = 3 \times 20\sqrt{3} = 60\sqrt{3} \text{ m}$.
Using $\sqrt{3} = 1.732$,height $= 60 \times 1.732 = 103.92 \text{ m}$.

(A) Let $H$ be the height of the aeroplane from the water surface of the lake.
Let the point of observation be $P$,which is $30 \ m$ above the water surface.
The height of the aeroplane above the point $P$ is $(H - 30) \ m$.
The depth of the image of the aeroplane in the lake is $H$ below the water surface.
Therefore,the total depth of the image from the point $P$ is $(H + 30) \ m$.
Let $x$ be the horizontal distance of the aeroplane from the point of observation.
From the angle of elevation: $\tan(30^{\circ}) = \frac{H - 30}{x} \implies x = \frac{H - 30}{\tan(30^{\circ})} = (H - 30)\sqrt{3}$.
From the angle of depression of the image: $\tan(60^{\circ}) = \frac{H + 30}{x} \implies x = \frac{H + 30}{\tan(60^{\circ})} = \frac{H + 30}{\sqrt{3}}$.
Equating the two expressions for $x$: $(H - 30)\sqrt{3} = \frac{H + 30}{\sqrt{3}}$.
$3(H - 30) = H + 30$.
$3H - 90 = H + 30$.
$2H = 120$.
$H = 60 \ m$.

(B) Let $AD$ be the height of the tree,$h$. Let $C$ and $B$ be the positions of the two posts on the ground such that $CB = 2 \text{ m}$. Let $DC = x \text{ m}$.
In $\Delta ADC$,the angle of elevation is $60^{\circ}$:
$\tan 60^{\circ} = \frac{AD}{DC} = \frac{h}{x}$
$\sqrt{3} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}} \quad \dots(i)$
In $\Delta ADB$,the angle of elevation is $45^{\circ}$:
$\tan 45^{\circ} = \frac{AD}{DB} = \frac{h}{x+2}$
$1 = \frac{h}{x+2} \implies x+2 = h \implies x = h-2 \quad \dots(ii)$
Equating $(i)$ and $(ii)$:
$\frac{h}{\sqrt{3}} = h-2$
$2 = h - \frac{h}{\sqrt{3}} = h \left(1 - \frac{1}{\sqrt{3}}\right) = h \left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)$
$h = \frac{2\sqrt{3}}{\sqrt{3}-1}$
Rationalizing the denominator:
$h = \frac{2\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{2(3+\sqrt{3})}{3-1} = \frac{2(3+\sqrt{3})}{2} = (3+\sqrt{3}) \text{ m}$.