Mix Examples - Some Applications of Trigonometry Questions in English

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

(A) True.
In the figure,point $B$ is moved closer to $C$ along the line segment $BC$. It is observed that:
$(i)$ The angle $\theta$ increases (as $\theta_{1} > \theta, \theta_{2} > \theta_{1}, \dots$) and
(ii) The length of the base $BC$ decreases (as $B_{1}C < BC, B_{2}C < B_{1}C, \dots$)
In a right-angled triangle,$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{BC}$.
Since the perpendicular $AC$ remains fixed and the base $BC$ decreases as $\theta$ increases,the ratio $\frac{AC}{BC}$ increases. Hence,$\tan \theta$ increases as $\theta$ increases.

(A) True.
We know that $\sin \theta$ increases as $\theta$ increases,but $\cos \theta$ decreases as $\theta$ increases in the interval $0^\circ < \theta < 90^\circ$.
We have $\tan \theta = \frac{\sin \theta}{\cos \theta}$.
As $\theta$ increases from $0^\circ$ to $90^\circ$,$\sin \theta$ increases (numerator increases) and $\cos \theta$ decreases (denominator decreases).
Since $\tan \theta$ is the ratio of an increasing value to a decreasing value,it grows at a significantly faster rate compared to $\sin \theta$,which is simply the ratio of an increasing value to a constant value of $1$.
Therefore,$\tan \theta$ increases faster than $\sin \theta$ as $\theta$ increases.

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

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

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

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

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

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

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

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

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

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

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

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

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

(B) Let $\overline{AB}$ be the building of height $40 \, m$. Let $C$ and $D$ be the two positions of the moving car.
From the figure,$AB = 40 \, m$ and $\angle B = 90^{\circ}$.
The angles of depression are $\angle XAC = 45^{\circ}$ and $\angle XAD = 30^{\circ}$.
Since the line $AX$ is parallel to $BD$,the alternate interior angles are equal:
$\angle ACB = \angle XAC = 45^{\circ}$ and $\angle ADB = \angle XAD = 30^{\circ}$.
In right-angled $\triangle ABC$:
$\tan(45^{\circ}) = \frac{AB}{BC} \implies 1 = \frac{40}{BC} \implies BC = 40 \, m$.
In right-angled $\triangle ABD$:
$\tan(30^{\circ}) = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{40}{BD} \implies BD = 40\sqrt{3} \, m$.
Using $\sqrt{3} \approx 1.73$,we get $BD = 40 \times 1.73 = 69.2 \, m$.
The distance covered by the car is $CD = BD - BC$.
$CD = 69.2 - 40 = 29.2 \, m$.
Thus,the distance covered by the car is $29.2 \, m$.

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

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

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

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

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

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

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

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

(C) Let $\overline{AB}$ and $\overline{ED}$ be two poles of equal height $h \, m$ and $C$ be the point of observation on the line segment $\overline{BD}$ joining the bases of the poles.
Given $BD = 200 \, m$. Let $BC = x \, m$,then $CD = (200 - x) \, m$.
The angles of elevation are $\angle ACB = 60^{\circ}$ and $\angle ECD = 30^{\circ}$.
In $\triangle ABC$,$\tan 60^{\circ} = \frac{AB}{BC} \implies \sqrt{3} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}} \quad \dots(1)$
In $\triangle EDC$,$\tan 30^{\circ} = \frac{ED}{CD} \implies \frac{1}{\sqrt{3}} = \frac{h}{200 - x} \implies 200 - x = \sqrt{3}h \implies x = 200 - \sqrt{3}h \quad \dots(2)$
Equating $(1)$ and $(2)$:
$\frac{h}{\sqrt{3}} = 200 - \sqrt{3}h$
$h = 200\sqrt{3} - 3h$
$4h = 200\sqrt{3}$
$h = 50\sqrt{3} \approx 50 \times 1.732 = 86.6 \, m$ (Using $\sqrt{3} \approx 1.732$,the closest option is $86.5 \, m$).

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

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

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

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

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

(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 $45^{\circ}$.