Height and Distance Questions in English

(B) Let $h$ be the height of the object and $x$ be the distance from the base of the hill to the second observation point.
In $\triangle BCD$,$\tan 60^{\circ} = \frac{h}{x}$ $\Rightarrow \sqrt{3} = \frac{h}{x}$ $\Rightarrow x = \frac{h}{\sqrt{3}}$.
In $\triangle ACD$,$\tan 30^{\circ} = \frac{h}{120 + x}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{120 + x}$ $\Rightarrow 120 + x = h\sqrt{3}$.
Substituting $x = \frac{h}{\sqrt{3}}$ into the equation:
$120 + \frac{h}{\sqrt{3}} = h\sqrt{3}$
$120 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left( \frac{3 - 1}{\sqrt{3}} \right) = \frac{2h}{\sqrt{3}}$
$h = \frac{120 \times \sqrt{3}}{2} = 60\sqrt{3} \ m$.

(B) Let $BC$ be the height of the tower and $CD$ be the height of the flagstaff,where $BC = x$ and $CD = h$.
Let the point be $A$ at a distance $AB = y$ from the foot of the tower $B$.
Given that the tower and the flagstaff subtend equal angles $\theta$ at point $A$.
In $\triangle ABC$,$\tan \theta = \frac{BC}{AB} = \frac{x}{y}$.
In $\triangle ABD$,the total angle is $2\theta$ and the total height is $BD = BC + CD = x + h$.
Thus,$\tan 2\theta = \frac{BD}{AB} = \frac{x+h}{y}$.
Using the formula $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$,we have:
$\frac{2(x/y)}{1-(x/y)^2} = \frac{x+h}{y}$
$\frac{2x/y}{(y^2-x^2)/y^2} = \frac{x+h}{y}$
$\frac{2xy}{y^2-x^2} = \frac{x+h}{y}$
$2xy^2 = (x+h)(y^2-x^2)$
$\frac{2xy^2}{y^2-x^2} = x+h$
$h = \frac{2xy^2}{y^2-x^2} - x$
$h = \frac{2xy^2 - x(y^2-x^2)}{y^2-x^2}$
$h = \frac{2xy^2 - xy^2 + x^3}{y^2-x^2}$
$h = \frac{xy^2 + x^3}{y^2-x^2} = \frac{x(x^2+y^2)}{y^2-x^2}$.

(C) Let the position of the observer be $A$. The aeroplane is at height $h = 1 \ km$. Let the initial position be $D$ and the final position be $E$.
In $\Delta DAP$,$\tan 60^{\circ} = \frac{DP}{AP}$ $\Rightarrow \sqrt{3} = \frac{1}{AP}$ $\Rightarrow AP = \frac{1}{\sqrt{3}} \ km$.
In $\Delta EAQ$,$\tan 30^{\circ} = \frac{EQ}{AQ} = \frac{1}{AP + PQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{1}{\frac{1}{\sqrt{3}} + PQ}$.
Solving for $PQ$: $\frac{1}{\sqrt{3}} + PQ = \sqrt{3}$ $\Rightarrow PQ = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \ km$.
The distance covered by the plane is $PQ = \frac{2}{\sqrt{3}} \ km$ in $10 \ s$.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{2/\sqrt{3} \ km}{10 \ s} = \frac{2}{\sqrt{3}} \times \frac{1}{10} \ km/s$.
Converting to $km/h$: $\text{Speed} = \frac{2}{10\sqrt{3}} \times 3600 \ km/h = \frac{7200}{10\sqrt{3}} = \frac{720}{\sqrt{3}} = 240\sqrt{3} \ km/h$.

(B) In $\triangle ECD$,$\tan 3 \alpha = \frac{h}{CD} \Rightarrow CD = h \cot 3 \alpha \quad \dots(i)$
In $\triangle EBD$,$\tan 2 \alpha = \frac{h}{BD} \Rightarrow BD = h \cot 2 \alpha \quad \dots(ii)$
In $\triangle EAD$,$\tan \alpha = \frac{h}{AD} \Rightarrow AD = h \cot \alpha \quad \dots(iii)$
From Eqs. $(ii)$ and $(iii)$,$AB = AD - BD = h(\cot \alpha - \cot 2 \alpha) \quad \dots(iv)$
From Eqs. $(i)$ and $(ii)$,$BC = BD - CD = h(\cot 2 \alpha - \cot 3 \alpha) \quad \dots(v)$
Dividing $(iv)$ by $(v)$:
$\frac{AB}{BC} = \frac{\cot \alpha - \cot 2 \alpha}{\cot 2 \alpha - \cot 3 \alpha} = \frac{\frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2 \alpha}{\sin 2 \alpha}}{\frac{\cos 2 \alpha}{\sin 2 \alpha} - \frac{\cos 3 \alpha}{\sin 3 \alpha}}$
$= \frac{\frac{\sin(2 \alpha - \alpha)}{\sin \alpha \sin 2 \alpha}}{\frac{\sin(3 \alpha - 2 \alpha)}{\sin 2 \alpha \sin 3 \alpha}} = \frac{\sin \alpha}{\sin \alpha \sin 2 \alpha} \times \frac{\sin 2 \alpha \sin 3 \alpha}{\sin \alpha} = \frac{\sin 3 \alpha}{\sin \alpha}$
$= \frac{3 \sin \alpha - 4 \sin^3 \alpha}{\sin \alpha} = 3 - 4 \sin^2 \alpha = 3 - 2(1 - \cos 2 \alpha) = 1 + 2 \cos 2 \alpha$.

(B) Let the height of the pole be $h$ and the distance from the second point to the base of the pole be $x$.
In $\triangle BDA$,$\tan 45^{\circ} = \frac{h}{x}$ $\Rightarrow 1 = \frac{h}{x}$ $\Rightarrow h = x$.
In $\triangle BCA$,$\tan 30^{\circ} = \frac{h}{20+x}$.
Substituting $x = h$ into the equation:
$\frac{1}{\sqrt{3}} = \frac{h}{20+h}$
$20+h = \sqrt{3}h$
$20 = h(\sqrt{3}-1)$
$h = \frac{20}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$h = \frac{20(\sqrt{3}+1)}{3-1} = \frac{20(\sqrt{3}+1)}{2} = 10(\sqrt{3}+1) \ m$.