Mix Examples - Some Applications of Trigonometry Questions in English

(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 $\overline{AB}$ 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 $86.5 \, m$).

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

(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 $x < y$.