Heights and Distances — Important Questions

Answer: (b) $45^\circ$.

$\tan\theta=\dfrac{\text{height}}{\text{distance}}=\dfrac{10}{10}=1$, so $\theta=45^\circ$.

Answer: (b) $h\sqrt3$.

$\tan30^\circ=\dfrac{h}{\text{shadow}}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{\text{shadow}}$, so shadow $=h\sqrt3$.

Answer: (b) $20\sqrt3\text{ m}$.

The angle of depression equals the angle of elevation from the boat, so $\tan60^\circ=\dfrac{60}{d}$, giving $d=\dfrac{60}{\sqrt3}=20\sqrt3\text{ m}\approx34.64\text{ m}$.

Answer: (a) $20\sqrt3\text{ m}$.

Let height $=h$ and nearer distance $=x$. Then $\tan60^\circ=\dfrac{h}{x}\Rightarrow x=\dfrac{h}{\sqrt3}$ and $\tan30^\circ=\dfrac{h}{x+40}\Rightarrow x+40=h\sqrt3$. Subtracting, $40=h\sqrt3-\dfrac{h}{\sqrt3}=\dfrac{2h}{\sqrt3}$, so $h=20\sqrt3\text{ m}$.

Answer: (a) The two horizontals are parallel and the line of sight is a transversal, so the angle of depression and angle of elevation are equal alternate angles. R correctly explains A.

Let the height reached be $h$. The ladder is the hypotenuse, so

$\sin60^\circ=\dfrac{h}{6}\Rightarrow h=6\sin60^\circ=6\times\dfrac{\sqrt3}{2}=3\sqrt3\text{ m}\approx5.20\text{ m}.$

Let the height be $h$.

$\tan45^\circ=\dfrac{h}{30}\Rightarrow 1=\dfrac{h}{30}\Rightarrow h=30\text{ m}.$

The height of the tower is $30\text{ m}$.

Let height $=h$ and the nearer distance from the foot $=x$.

From the nearer point: $\tan60^\circ=\dfrac{h}{x}\Rightarrow \sqrt3=\dfrac{h}{x}\Rightarrow x=\dfrac{h}{\sqrt3}.$

From the farther point: $\tan30^\circ=\dfrac{h}{x+20}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+20}\Rightarrow x+20=h\sqrt3.$

Subtracting the first from the second:

$20=h\sqrt3-\dfrac{h}{\sqrt3}=\dfrac{3h-h}{\sqrt3}=\dfrac{2h}{\sqrt3}.$

$h=\dfrac{20\sqrt3}{2}=10\sqrt3=10\times1.732=17.32\text{ m}.$

The height of the tower is $17.32\text{ m}$.

Let the height be $h$.

At $45^\circ$: $\tan45^\circ=\dfrac{h}{s_1}\Rightarrow s_1=h.$

At $30^\circ$: $\tan30^\circ=\dfrac{h}{s_2}\Rightarrow s_2=h\sqrt3.$

The shadow increases by $30\text{ m}$:

$s_2-s_1=30\Rightarrow h\sqrt3-h=30\Rightarrow h(\sqrt3-1)=30.$

$h=\dfrac{30}{\sqrt3-1}=\dfrac{30(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\dfrac{30(\sqrt3+1)}{2}=15(\sqrt3+1).$

$h=15(1.732+1)=15\times2.732=40.98\text{ m}.$

The height of the tower is about $40.98\text{ m}$.

Let the horizontal distance from the ship to the cliff be $d$. Let the deck be at height $12\text{ m}$.

Depression of base ($30^\circ$): the base is $12\text{ m}$ below the deck, so

$\tan30^\circ=\dfrac{12}{d}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{12}{d}\Rightarrow d=12\sqrt3\text{ m}.$

Elevation of top ($45^\circ$): let the top be $H$ above the deck level, so

$\tan45^\circ=\dfrac{H}{d}\Rightarrow H=d=12\sqrt3\text{ m}.$

Total height of cliff $=12+H=12+12\sqrt3=12(1+\sqrt3)=12\times2.732=32.78\text{ m}.$

The cliff is about $32.78\text{ m}$ high.

Let the height $AB=h$ and $BC=x$. Then $BD=x+50$.

From $C$: $\tan60^\circ=\dfrac{h}{x}\Rightarrow \sqrt3=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x.\ \ (1)$

From $D$: $\tan30^\circ=\dfrac{h}{x+50}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+50}\Rightarrow h\sqrt3=x+50.\ \ (2)$

Substitute (1) into (2): $(\sqrt3\,x)\sqrt3=x+50\Rightarrow 3x=x+50\Rightarrow 2x=50\Rightarrow x=25.$

So $BC=25\text{ m}$.

From (1): $h=\sqrt3\times25=25\sqrt3=25\times1.732=43.30\text{ m}.$

The tower is $43.30\text{ m}$ high and $BC=25\text{ m}$.

Let the building be $PQ=60\text{ m}$ with $P$ at the top, and let the lamp-post be $RS$ of height $y$, standing at horizontal distance $d$ from the building.

Depression of the foot $S$ ($60^\circ$): considering the whole $60\text{ m}$ drop,

$\tan60^\circ=\dfrac{60}{d}\Rightarrow \sqrt3=\dfrac{60}{d}\Rightarrow d=\dfrac{60}{\sqrt3}=20\sqrt3=34.64\text{ m}.$

So the horizontal distance is $20\sqrt3\approx34.64\text{ m}$.

Depression of the top $R$ ($30^\circ$): the top $R$ is $y$ above the ground, so the vertical drop from $P$ to $R$'s level is $(60-y)$, over the same horizontal distance $d$:

$\tan30^\circ=\dfrac{60-y}{d}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{60-y}{20\sqrt3}.$

$60-y=\dfrac{20\sqrt3}{\sqrt3}=20\Rightarrow y=40\text{ m}.$

(i) Horizontal distance $=20\sqrt3\approx34.64\text{ m}$.

(ii) Height of the lamp-post $=40\text{ m}$.

(i) At $B$: $\tan60^\circ=\dfrac{h}{x}\Rightarrow \sqrt3=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x.$

(ii) At $A$ (distance $x+60$ from the foot): $\tan30^\circ=\dfrac{h}{x+60}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+60}\Rightarrow h\sqrt3=x+60.$

(iii) Substitute $h=\sqrt3\,x$ into $h\sqrt3=x+60$:

$(\sqrt3\,x)\sqrt3=x+60\Rightarrow 3x=x+60\Rightarrow 2x=60\Rightarrow x=30\text{ m}.$

(iv) $h=\sqrt3\,x=30\sqrt3=30\times1.732=51.96\text{ m}.$

The drone is about $51.96\text{ m}$ above the road.