Mix Examples - Some Applications of Trigonometry Questions in English

(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) In the right-angled triangle formed,the height of the building is $y$ (opposite side) and the distance from the base is $x$ (adjacent side).
Using the trigonometric ratio,$\tan(70^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{x}$.
We know that $\tan(45^\circ) = 1$.
Since the tangent function is increasing in the interval $(0^\circ, 90^\circ)$,and $70^\circ > 45^\circ$,it follows that $\tan(70^\circ) > \tan(45^\circ)$.
Therefore,$\tan(70^\circ) > 1$.
Substituting the ratio,$\frac{y}{x} > 1$.
Since $x$ and $y$ are lengths (positive values),we can multiply both sides by $x$ to get $y > x$,which is equivalent to $x < y$.

(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) In a right-angled triangle,the side opposite to the $30^{\circ}$ angle is given by the formula: $\text{Side opposite to } 30^{\circ} = \text{Hypotenuse} \times \sin(30^{\circ})$.
Given that the hypotenuse is $12$ and $\sin(30^{\circ}) = \frac{1}{2}$.
Therefore,the measure of the side opposite to the angle is $12 \times \frac{1}{2} = 6$.

(B) Consider a right-angled triangle $ABC$ where $\angle B = 90^{\circ}$ and $\angle C = 60^{\circ}$.
In this triangle,the side opposite to $\angle C$ is $AB$ and the hypotenuse is $AC$.
Using the trigonometric ratio $\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}$,we have:
$\sin 60^{\circ} = \frac{AB}{AC}$
Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get:
$\frac{\sqrt{3}}{2} = \frac{AB}{AC}$
Therefore,$AB = \frac{\sqrt{3}}{2} AC$.
Thus,the measure of the side opposite to the angle of $60^{\circ}$ is $\frac{\sqrt{3}}{2}$ times the measure of the hypotenuse.

(C) In $\Delta ABC$,$\angle B = 90^{\circ}$.
Given $AC = 16$ (hypotenuse) and $BC = 8$ (side adjacent to $\angle C$).
Using the trigonometric ratio for cosine:
$\cos C = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{8}{16} = \frac{1}{2}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have $m \angle C = 60^{\circ}$.

(D) Let the right triangle be $\triangle ABC$,where $\angle B = 90^{\circ}$ and $\angle C = 45^{\circ}$.
In $\triangle ABC$,the side opposite to $\angle C$ is $AB$ and the hypotenuse is $AC$.
Using the trigonometric ratio $\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}$:
$\sin 45^{\circ} = \frac{AB}{AC}$
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,we have:
$\frac{1}{\sqrt{2}} = \frac{AB}{AC}$
Therefore,$AB = \frac{1}{\sqrt{2}} AC$.
Thus,the measure of the side opposite to the angle of $45^{\circ}$ is $\frac{1}{\sqrt{2}}$ times the measure of the hypotenuse.

(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) In the right-angled $\Delta ABC$,we are given the hypotenuse $AC = 20$ and the angle $m\angle C = 30^\circ$.
We need to find the adjacent side $BC$.
Using the trigonometric ratio $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$,we have:
$\cos 30^\circ = \frac{BC}{AC}$
Substituting the values:
$\frac{\sqrt{3}}{2} = \frac{BC}{20}$
$BC = 20 \times \frac{\sqrt{3}}{2}$
$BC = 10\sqrt{3}$
Using $\sqrt{3} \approx 1.73$:
$BC = 10 \times 1.73 = 17.3$

(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 $\theta = 60^\circ$.

(A) In right-angled $\Delta ABC$,we are given the hypotenuse $AC = 30 \text{ cm}$ and $\angle C = 30^\circ$.
We need to find the side $AB$,which is opposite to $\angle C$.
Using the trigonometric ratio $\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}$:
$\sin 30^\circ = \frac{AB}{AC}$
Since $\sin 30^\circ = \frac{1}{2}$,we have:
$\frac{1}{2} = \frac{AB}{30}$
$AB = \frac{30}{2} = 15 \text{ cm}$.

(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 $B$.