Heights and Distances — ICSE Class 10 Maths Important Questions

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

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

Answer: (b) $h\sqrt3$h3.

$\tan30^\circ=\dfrac{h}{\text{shadow}}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{\text{shadow}}$30^=h/shadow 1/3=h/shadow, so shadow $=h\sqrt3$=h3.

Answer: (b) $20\sqrt3\text{ m}$203 m.

The angle of depression equals the angle of elevation from the boat, so $\tan60^\circ=\dfrac{60}{d}$60^=60/d, giving $d=\dfrac{60}{\sqrt3}=20\sqrt3\text{ m}\approx34.64\text{ m}$d=60/3=203 m34.64 m.

Answer: (a) $20\sqrt3\text{ m}$203 m.

Let height $=h$=h and nearer distance $=x$=x. Then $\tan60^\circ=\dfrac{h}{x}\Rightarrow x=\dfrac{h}{\sqrt3}$60^=h/x x=h/3 and $\tan30^\circ=\dfrac{h}{x+40}\Rightarrow x+40=h\sqrt3$30^=h/x+40 x+40=h3. Subtracting, $40=h\sqrt3-\dfrac{h}{\sqrt3}=\dfrac{2h}{\sqrt3}$40=h3-h/3=2h/3, so $h=20\sqrt3\text{ m}$h=203 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 $h$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}.$60^=h/6 h=660^=6×3/2=33 m5.20 m.

The string is the hypotenuse of the right triangle, so

$\sin38^\circ=\dfrac{h}{100}\Rightarrow h=100\times0.6157=61.57\text{ m}.$38^=h/100 h=100×0.6157=61.57 m.

The kite is $61.57\text{ m}$61.57 m high.

Let the height of each pole be $h$h, and let $P$P be $x\text{ m}$x m from pole $CA$CA, so it is $(80-x)\text{ m}$(80-x) m from pole $DB$DB.

From $\tan60^\circ$60^: $\sqrt3=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x.\ \ (1)$3=h/x h=3\,x. (1)

From $\tan30^\circ$30^: $\dfrac{1}{\sqrt3}=\dfrac{h}{80-x}\Rightarrow h\sqrt3=80-x.\ \ (2)$1/3=h/80-x h3=80-x. (2)

Substituting (1) in (2): $3x=80-x\Rightarrow x=20$3x=80-x x=20.

$h=20\sqrt3=20\times1.732=34.64\text{ m}.$h=203=20×1.732=34.64 m.

Each pole is $34.64\text{ m}$34.64 m high; $P$P is $20\text{ m}$20 m from $C$C and $60\text{ m}$60 m from $D$D.

Let the height be $h$h.

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

At $30^\circ$30^: $\tan30^\circ=\dfrac{h}{s_2}\Rightarrow s_2=h\sqrt3.$30^=h/s_2 s_2=h3.

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

$s_2-s_1=30\Rightarrow h\sqrt3-h=30\Rightarrow h(\sqrt3-1)=30.$s_2-s_1=30 h3-h=30 h(3-1)=30.

$h=\dfrac{30}{\sqrt3-1}=\dfrac{30(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\dfrac{30(\sqrt3+1)}{2}=15(\sqrt3+1).$h=30/3-1=30(3+1)/(3-1)(3+1)=30(3+1)/2=15(3+1).

$h=15(1.732+1)=15\times2.732=40.98\text{ m}.$h=15(1.732+1)=15×2.732=40.98 m.

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

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

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

$\tan30^\circ=\dfrac{12}{d}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{12}{d}\Rightarrow d=12\sqrt3\text{ m}.$30^=12/d 1/3=12/d d=123 m.

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

$\tan45^\circ=\dfrac{H}{d}\Rightarrow H=d=12\sqrt3\text{ m}.$45^=H/d H=d=123 m.

Total height of cliff $=12+H=12+12\sqrt3=12(1+\sqrt3)=12\times2.732=32.78\text{ m}.$=12+H=12+123=12(1+3)=12×2.732=32.78 m.

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

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

From $C$C: $\tan60^\circ=\dfrac{h}{x}\Rightarrow \sqrt3=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x.\ \ (1)$60^=h/x 3=h/x h=3\,x. (1)

From $D$D: $\tan30^\circ=\dfrac{h}{x+50}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+50}\Rightarrow h\sqrt3=x+50.\ \ (2)$30^=h/x+50 1/3=h/x+50 h3=x+50. (2)

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

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

From (1): $h=\sqrt3\times25=25\sqrt3=25\times1.732=43.30\text{ m}.$h=3×25=253=25×1.732=43.30 m.

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

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

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

$\tan60^\circ=\dfrac{60}{d}\Rightarrow \sqrt3=\dfrac{60}{d}\Rightarrow d=\dfrac{60}{\sqrt3}=20\sqrt3=34.64\text{ m}.$60^=60/d 3=60/d d=60/3=203=34.64 m.

So the horizontal distance is $20\sqrt3\approx34.64\text{ m}$20334.64 m.

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

$\tan30^\circ=\dfrac{60-y}{d}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{60-y}{20\sqrt3}.$30^=60-y/d 1/3=60-y/203.

$60-y=\dfrac{20\sqrt3}{\sqrt3}=20\Rightarrow y=40\text{ m}.$60-y=203/3=20 y=40 m.

(i) Horizontal distance $=20\sqrt3\approx34.64\text{ m}$=20334.64 m.

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

(i) $\tan45^\circ=\dfrac{75}{FX}\Rightarrow FX=75\text{ m}.$45^=75/FX FX=75 m.

(ii) $\tan30^\circ=\dfrac{75}{FY}\Rightarrow FY=75\sqrt3=75\times1.732=129.90\text{ m}.$30^=75/FY FY=753=75×1.732=129.90 m.

(iii) $XY=FY-FX=75\sqrt3-75=75(\sqrt3-1)=75\times0.732=54.90\text{ m}.$XY=FY-FX=753-75=75(3-1)=75×0.732=54.90 m.

(iv) It increases. For a ship at distance $d$d, $\tan\theta=\dfrac{75}{d}$=75/d; as $d$d decreases, $\tan\theta$ increases, so $\theta$ increases.