Heights and Distances Questions in English

(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 $20 \, m$.

(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.

(A) Let the length of the ladder be the hypotenuse $c = 6 \,m$.
Let the height reached on the wall be one leg $a = 3.6 \,m$.
Let the distance from the base of the wall to the lower end of the ladder be $b$.
According to the Pythagorean theorem for a right-angled triangle formed by the wall,the ground,and the ladder:
$a^2 + b^2 = c^2$
$(3.6)^2 + b^2 = 6^2$
$12.96 + b^2 = 36$
$b^2 = 36 - 12.96$
$b^2 = 23.04$
$b = \sqrt{23.04} = 4.8 \,m$.
Thus,the distance is $4.8 \,m$.

(B) Let the length of the ladder be the hypotenuse $h = 4 \,m$.
Let the height reached on the wall be the perpendicular $p = 3.2 \,m$.
Let the distance between the base of the wall and the lower end of the ladder be the base $b$.
According to the Pythagorean theorem,$h^2 = p^2 + b^2$.
Substituting the values: $4^2 = (3.2)^2 + b^2$.
$16 = 10.24 + b^2$.
$b^2 = 16 - 10.24 = 5.76$.
$b = \sqrt{5.76} = 2.4 \,m$.

(C) Let the two poles be $AB = 6 \, m$ and $CD = 11 \, m$ standing on a flat ground.
The distance between their bases is $BD = 12 \, m$.
Draw a line segment $AE$ parallel to $BD$ such that $E$ lies on $CD$.
Then $AE = BD = 12 \, m$ and $ED = AB = 6 \, m$.
Now,$CE = CD - ED = 11 \, m - 6 \, m = 5 \, m$.
In the right-angled triangle $\triangle AEC$,by the Pythagorean theorem:
$AC^2 = AE^2 + CE^2$
$AC^2 = (12)^2 + (5)^2$
$AC^2 = 144 + 25 = 169$
$AC = \sqrt{169} = 13 \, m$.
Thus,the distance between their tops is $13 \, m$.

(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 $6 \, m$.

(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 $30 \, m$.

(B) 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.
Since the observer is looking down at the object,the object must be situated below the horizontal line passing through the observer's eye.
Therefore,the correct option is $B$.

(C) Let the height of the building be $y$ and the distance from the base be $x$. The angle of elevation is $\theta = 25^\circ$.
In the right-angled triangle formed by the building,the ground,and the line of sight,we have:
$\tan(\theta) = \frac{\text{height}}{\text{base}} = \frac{y}{x}$.
Therefore,$\tan(25^\circ) = \frac{y}{x}$,which implies $y = x \cdot \tan(25^\circ)$.
Since the value of $\tan(25^\circ)$ is approximately $0.466$,which is less than $1$ $(0 < \tan(25^\circ) < 1)$,
we have $y = x \cdot (0.466)$.
This means $y < x$,or $x > y$.

(B) Let the wall be $AB$ and the ground be $BC$. The ladder is $AC$.
Given that the distance of the lower end of the ladder from the base of the wall is $BC = a \ m$.
The angle made by the ladder with the ground is $\angle C = \theta$.
In the right-angled triangle $\triangle ABC$,we have:
$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AB}{BC}$
$\tan \theta = \frac{AB}{a}$
Therefore,the height at which the ladder reaches the wall is $AB = a \tan \theta \ m$.

(D) Let $AB$ be the building of height $x \ m$ and $C$ be the position of the child on the ground.
The angle of depression from the top $A$ to the child at $C$ is $\theta$.
Since the horizontal line at $A$ is parallel to the ground $BC$,the angle of elevation from $C$ to $A$ is also $\theta$ (alternate interior angles).
In the right-angled triangle $\triangle ABC$,we have:
$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$
$\tan \theta = \frac{x}{BC}$
Therefore,$BC = \frac{x}{\tan \theta} = x \cot \theta \ m$.
Thus,the distance of the child from the base of the building is $x \cot \theta \ m$.

(A) Let $AB$ be the height of the electric pole and $AC$ be the length of the cable.
In the right-angled triangle $ABC$,the angle of elevation is $\angle C = 30^{\circ}$.
The length of the hypotenuse $AC = 20 \, m$.
Using the trigonometric ratio $\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}}$:
$\sin 30^{\circ} = \frac{AB}{AC}$
$\frac{1}{2} = \frac{AB}{20}$
$AB = \frac{20}{2} = 10 \, m$.
Thus,the height of the electric pole is $10 \, m$.

(B) Let $AB$ be the height of the pole and $BC$ be the distance of the stone from the base of the pole.
Given,$BC = 10 \, m$ and the angle of depression is $60^{\circ}$.
Since the angle of depression equals the angle of elevation,the angle of elevation of the top of the pole from the stone is $\angle C = 60^{\circ}$.
In $\triangle ABC$,$\tan 60^{\circ} = \frac{AB}{BC}$.
$\sqrt{3} = \frac{AB}{10}$.
$AB = 10 \times \sqrt{3}$.
Given $\sqrt{3} = 1.73$,so $AB = 10 \times 1.73 = 17.3 \, m$.
Thus,the height of the pole is $17.3 \, m$.

(C) Let $h$ be the height of the lighthouse. Let $d_A$ and $d_B$ be the distances of boats $A$ and $B$ from the base of the lighthouse respectively.
From the given angles of depression,the angles of elevation from the boats to the top of the lighthouse are $25^{\circ}$ and $40^{\circ}$ respectively.
Using the tangent ratio: $\tan(25^{\circ}) = \frac{h}{d_A}$ and $\tan(40^{\circ}) = \frac{h}{d_B}$.
Since $\tan(40^{\circ}) > \tan(25^{\circ})$,it follows that $\frac{h}{d_B} > \frac{h}{d_A}$.
Therefore,$d_A > d_B$. Thus,the distance of boat $A$ from the lighthouse is greater than the distance of boat $B$.

(C) Let $X$ be the height of building $PQ$ and $Y$ be the height of building $SR$. Let $QR$ be the distance between the bases of the two buildings.
In right-angled $\Delta SQR$,$\tan(45^{\circ}) = \frac{SR}{QR} = \frac{Y}{QR}$.
Since $\tan(45^{\circ}) = 1$,we have $Y = QR$.
In right-angled $\Delta PQR$,$\tan(65^{\circ}) = \frac{PQ}{QR} = \frac{X}{QR}$.
Since $\tan(65^{\circ}) > 1$,we have $\frac{X}{QR} > 1$,which implies $X > QR$.
Since $Y = QR$ and $X > QR$,it follows that $X > Y$.
Therefore,the height of building $X$ is greater than the height of building $Y$.

(B) Let the height of the pole be $AB$ and the distance from the base of the pole to the point $C$ be $BC = x \, m$.
In the right-angled triangle $ABC$,the angle of elevation is $\angle C = 60^{\circ}$.
Using the trigonometric ratio $\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}$,we have:
$\tan 60^{\circ} = \frac{AB}{BC}$
Since $\tan 60^{\circ} = \sqrt{3}$,we get:
$\sqrt{3} = \frac{AB}{x}$
Therefore,$AB = \sqrt{3} x \, m$.
Thus,the height of the pole is $\sqrt{3} x \, m$.

(C) Let $AB$ be the height of the tower and $BC = x \, m$ be the distance from the base of the tower to the point of observation $C$.
In the right-angled triangle $ABC$,the angle of elevation is $\angle C = 30^{\circ}$.
Using the trigonometric ratio $\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}$:
$\tan 30^{\circ} = \frac{AB}{BC}$
Substituting the values:
$\frac{1}{\sqrt{3}} = \frac{AB}{x}$
Therefore,the height of the tower $AB = \frac{x}{\sqrt{3}} \, m$.

(D) Let $AB$ be the height of the tower and $BC$ be the distance from the base of the tower to the point $C$ on the ground.
Given: $BC = 100 \, m$ and the angle of elevation $\angle C = 60^\circ$.
In the right-angled triangle $\triangle ABC$,we have:
$\tan C = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{AB}{BC}$
$\tan 60^\circ = \frac{AB}{100}$
Since $\tan 60^\circ = \sqrt{3} \approx 1.732$,
$\sqrt{3} = \frac{AB}{100}$
$AB = 100 \times \sqrt{3}$
$AB = 100 \times 1.732 = 173.2 \, m$.
Rounding to the nearest whole number as per the options,the height is $173 \, m$.

(D) Let $AB$ be the height of the temple,where $AB = 20 \ m$.
Let $BC$ be the breadth of the river.
The angle of elevation from point $C$ to the top $A$ is $\angle ACB = 30^{\circ}$.
In the right-angled triangle $ABC$:
$\tan 30^{\circ} = \frac{AB}{BC}$
$\frac{1}{\sqrt{3}} = \frac{20}{BC}$
$BC = 20 \sqrt{3} \ m$
Using $\sqrt{3} \approx 1.73$:
$BC = 20 \times 1.73 = 34.6 \ m$.
Thus,the breadth of the river is $34.6 \ m$.

(B) Let the height of the tower be $h$. The observer is $1.5\, m$ tall,so the height of the triangle formed is $AE = h - 1.5$.
Given the distance from the tower is $DE = 28.5\, m$ and the angle of elevation is $45^\circ$.
In $\triangle ADE$,$\tan 45^\circ = \frac{AE}{DE}$.
Since $\tan 45^\circ = 1$,we have $1 = \frac{AE}{28.5}$.
Therefore,$AE = 28.5\, m$.
The total height of the tower $h = AE + EB = 28.5 + 1.5 = 30\, m$.

(C) The angle of elevation is defined as the angle formed by the line of sight (ray of vision) with the horizontal line (horizontal ray) when the object is located above the horizontal level of the observer's eye.
Therefore,if the angle of elevation is formed,the object must be situated above the horizontal ray.

(C) Let the height of the pole be $h$ and the length of the shadow be $s$.
According to the problem,$s = \frac{1}{\sqrt{3}} h$.
Let $\theta$ be the angle of elevation of the sun.
In the right-angled triangle formed by the pole and its shadow,we have $\tan(\theta) = \frac{\text{height}}{\text{shadow}} = \frac{h}{s}$.
Substituting the value of $s$,we get $\tan(\theta) = \frac{h}{\frac{1}{\sqrt{3}} h} = \sqrt{3}$.
Since $\tan(60^\circ) = \sqrt{3}$,the angle of elevation $\theta = 60^\circ$.

(D) Let the height of the shorter pole be $h$ and the height of the taller pole be $2h$.
Let the distance between the two poles be $x$.
The midpoint of the line joining their feet divides the distance $x$ into two equal parts,each of length $\frac{x}{2}$.
Let the angles of elevation from the midpoint be $\theta$ and $90^\circ - \theta$.
For the shorter pole: $\tan(90^\circ - \theta) = \frac{h}{x/2} = \frac{2h}{x}$.
Since $\tan(90^\circ - \theta) = \cot \theta$,we have $\cot \theta = \frac{2h}{x}$.
For the taller pole: $\tan \theta = \frac{2h}{x/2} = \frac{4h}{x}$.
Multiplying the two equations: $\tan \theta \cdot \cot \theta = \left(\frac{4h}{x}\right) \cdot \left(\frac{2h}{x}\right)$.
$1 = \frac{8h^2}{x^2}$.
$h^2 = \frac{x^2}{8}$.
$h = \sqrt{\frac{x^2}{8}} = \frac{x}{2\sqrt{2}}$.

(B) Let the height of the tower be $h$ and the foot of the tower be $O$.
Let $x_A$ be the distance of house $A$ from the tower and $x_B$ be the distance of house $B$ from the tower.
From the geometry of the problem,the angle of depression equals the angle of elevation.
For house $A$: $\tan(30^{\circ}) = \frac{h}{x_A} \implies \frac{1}{\sqrt{3}} = \frac{h}{x_A} \implies x_A = h\sqrt{3}$.
For house $B$: $\tan(60^{\circ}) = \frac{h}{x_B} \implies \sqrt{3} = \frac{h}{x_B} \implies x_B = \frac{h}{\sqrt{3}}$.
Comparing the distances,since $\sqrt{3} > \frac{1}{\sqrt{3}}$,it follows that $x_A > x_B$.
Therefore,house $B$ is closer to the tower than house $A$.

(B) Let the hill be represented by a right-angled triangle where the hypotenuse is the distance walked on the hill,$x$ metres.
The angle of elevation with the ground is $\theta = 30^{\circ}$.
The height reached from the ground is the side opposite to the angle,$y$ metres.
Using the trigonometric ratio $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$,we have:
$\sin(30^{\circ}) = \frac{y}{x}$
Since $\sin(30^{\circ}) = \frac{1}{2}$,we get:
$\frac{1}{2} = \frac{y}{x}$
Cross-multiplying gives $x = 2y$.
Therefore,the correct option is $B$.