\displaystyle \textbf{Question 1: }\text{In the figure given below, }AB\perp BD\text{ and }AB=X\text{ m. Also,}
\displaystyle DC=30\text{ m},\ \angle ADB=30^\circ\text{ and }\angle ACB=45^\circ.\text{ Without using tables, find } X

\displaystyle \textbf{Answer:}
\displaystyle \text{Let }AB=X\text{ m and }BC=y\text{ m.}
\displaystyle \text{In right-angled }\triangle ACB,
\displaystyle \tan45^\circ=\frac{AB}{BC}=\frac{X}{y}
\displaystyle 1=\frac{X}{y}
\displaystyle \therefore y=X
\displaystyle \therefore BC=X\text{ m.}
\displaystyle \text{Now, }DB=DC+CB=30+X\text{ m.}
\displaystyle \text{In right-angled }\triangle ADB,
\displaystyle \tan30^\circ=\frac{AB}{DB}=\frac{X}{30+X}
\displaystyle \frac{1}{\sqrt{3}}=\frac{X}{30+X}
\displaystyle 30+X=\sqrt{3}X
\displaystyle X(\sqrt{3}-1)=30
\displaystyle X=\frac{30}{\sqrt{3}-1}
\displaystyle X=\frac{30(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}
\displaystyle X=\frac{30(\sqrt{3}+1)}{3-1}
\displaystyle X=15(\sqrt{3}+1)\text{ m}
\displaystyle X\approx15(1.732+1)=40.98\text{ m}
\displaystyle \therefore \text{The value of }X\text{ is }15(\sqrt{3}+1)\text{ m, or approximately }40.98\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the height of a tree when it is found that, on walking }20\text{ m away}
\displaystyle \text{from it in a horizontal line through its base, the angle of elevation of its top changes}
\displaystyle \text{from }60^\circ\text{ to }30^\circ.
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the tree }AB=x\text{ m and }BC=y\text{ m.}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}=\frac{x}{y}
\displaystyle \sqrt{3}=\frac{x}{y}
\displaystyle \therefore y=\frac{x}{\sqrt{3}}
\displaystyle \therefore BC=\frac{x}{\sqrt{3}}\text{ m.}
\displaystyle \text{Now }DB=DC+CB=20+\frac{x}{\sqrt{3}}\text{ m.}
\displaystyle \text{In right-angled }\triangle ADB,
\displaystyle \tan30^\circ=\frac{AB}{DB}
\displaystyle \frac{1}{\sqrt{3}}=\frac{x}{20+\frac{x}{\sqrt{3}}}
\displaystyle 20+\frac{x}{\sqrt{3}}=\sqrt{3}x
\displaystyle 20=\sqrt{3}x-\frac{x}{\sqrt{3}}
\displaystyle 20=\frac{2x}{\sqrt{3}}
\displaystyle x=10\sqrt{3}\text{ m}
\displaystyle x\approx17.32\text{ m}
\displaystyle \therefore \text{The height of the tree is }10\sqrt{3}\text{ m, or approximately }17.32\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the height of a building when, on walking }40\text{ m towards it}
\displaystyle \text{in a horizontal line through its base, the angle of elevation of its top changes}
\displaystyle \text{from }30^\circ\text{ to }45^\circ.
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the building }AB=h\text{ m and }BC=x\text{ m.}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan45^\circ=\frac{AB}{BC}=\frac{h}{x}
\displaystyle 1=\frac{h}{x}
\displaystyle \therefore x=h
\displaystyle \therefore BC=h\text{ m.}
\displaystyle \text{Now }DB=DC+CB=40+h\text{ m.}
\displaystyle \text{In right-angled }\triangle ADB,
\displaystyle \tan30^\circ=\frac{AB}{DB}=\frac{h}{40+h}
\displaystyle \frac{1}{\sqrt3}=\frac{h}{40+h}
\displaystyle 40+h=\sqrt3\,h
\displaystyle 40=(\sqrt3-1)h
\displaystyle h=\frac{40}{\sqrt3-1}
\displaystyle h=\frac{40(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}
\displaystyle h=\frac{40(\sqrt3+1)}{2}
\displaystyle h=20(\sqrt3+1)\text{ m}
\displaystyle h\approx20(1.732+1)=54.64\text{ m}
\displaystyle \therefore \text{The height of the building is }20(\sqrt3+1)\text{ m, or approximately }54.64\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{From the top of a lighthouse }100\text{ m high, the angles of depression of}
\displaystyle \text{two ships are }48^\circ\text{ and }36^\circ\text{ respectively. Find the distance between the}
\displaystyle \text{two ships (to the nearest metre) if: }\hfill\text{[ICSE 2010]}
\displaystyle \text{(i) the ships are on the same side of the lighthouse.}
\displaystyle \text{(ii) the ships are on the opposite sides of the lighthouse.}
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) Ships on the same side of the lighthouse}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan48^\circ=\frac{100}{BC}
\displaystyle \therefore BC=\frac{100}{\tan48^\circ}\approx90.04\text{ m}
\displaystyle \text{In right-angled }\triangle ADB,
\displaystyle \tan36^\circ=\frac{100}{DB}
\displaystyle \therefore DB=\frac{100}{\tan36^\circ}\approx137.64\text{ m}
\displaystyle \therefore \text{Distance between the ships}=DB-BC
\displaystyle =137.64-90.04=47.60\text{ m}
\displaystyle \therefore \text{Distance between the ships}\approx48\text{ m.}
\displaystyle \text{(ii) Ships on the opposite sides of the lighthouse}
\displaystyle \therefore \text{Distance between the ships}=DB+BC
\displaystyle =137.64+90.04=227.68\text{ m}
\displaystyle \therefore \text{Distance between the ships}\approx228\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Two pillars of equal heights stand on either side of a roadway which is }150\text{ m}
\displaystyle \text{wide. From a point between the pillars, the angles of elevation of their tops are }60^\circ
\displaystyle \text{and }30^\circ.\text{ Find the height of each pillar and the position of the point.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of each pillar be }h\text{ m.}
\displaystyle \text{Let the distance of the point from the pillar subtending }60^\circ\text{ be }x\text{ m.}
\displaystyle \therefore \text{Its distance from the other pillar}=(150-x)\text{ m.}
\displaystyle \text{From the first right-angled triangle,}
\displaystyle \tan60^\circ=\frac{h}{x}
\displaystyle \therefore h=x\tan60^\circ
\displaystyle \text{From the second right-angled triangle,}
\displaystyle \tan30^\circ=\frac{h}{150-x}
\displaystyle \therefore h=(150-x)\tan30^\circ
\displaystyle \therefore x\tan60^\circ=(150-x)\tan30^\circ
\displaystyle x\sqrt3=(150-x)\frac{1}{\sqrt3}
\displaystyle 3x=150-x
\displaystyle 4x=150
\displaystyle x=37.5\text{ m}
\displaystyle 150-x=150-37.5=112.5\text{ m}
\displaystyle h=x\tan60^\circ
\displaystyle h=37.5\sqrt3\text{ m}
\displaystyle h\approx37.5\times1.732=64.95\text{ m}
\displaystyle \therefore \text{The height of each pillar is }37.5\sqrt3\text{ m, or approximately }64.95\text{ m.}
\displaystyle \therefore \text{The point is }37.5\text{ m from the pillar subtending }60^\circ\text{ and }112.5\text{ m}
\displaystyle \text{from the pillar subtending }30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{From the figure given below, calculate the length of }CD. \displaystyle \textbf{Answer:}
\displaystyle \text{Given, }CB=15\text{ m},\ \angle ACB=47^\circ\text{ and }\angle ADE=22^\circ.
\displaystyle \text{Since }DE\parallel CB\text{ and }AB\perp CB,\text{ we have }DE=CB=15\text{ m.}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan47^\circ=\frac{AB}{BC}
\displaystyle \therefore AB=15\tan47^\circ
\displaystyle \text{In right-angled }\triangle AED,
\displaystyle \tan22^\circ=\frac{AE}{DE}
\displaystyle \therefore AE=15\tan22^\circ
\displaystyle \text{Since }D\text{ and }E\text{ are at the same level, }CD=BE=AB-AE.
\displaystyle \therefore CD=15\tan47^\circ-15\tan22^\circ
\displaystyle =15(1.0724)-15(0.4040)
\displaystyle =16.086-6.060
\displaystyle =10.026\text{ m}
\displaystyle \therefore \text{The length of }CD\text{ is approximately }10.03\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The angle of elevation of the top of a tower from a point is }60^\circ.
\displaystyle \text{From another point }30\text{ m vertically above the first point, the angle of elevation}
\displaystyle \text{is }45^\circ.\text{ Find: (i) the height of the tower, and (ii) its horizontal distance}
\displaystyle \text{from the points of observation.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the tower }AB=h\text{ m and its horizontal distance }BC=x\text{ m.}
\displaystyle \text{Since the upper observation point is }30\text{ m above the lower point,}
\displaystyle BE=30\text{ m and }AE=(h-30)\text{ m.}
\displaystyle \text{In right-angled }\triangle ADE,
\displaystyle \tan45^\circ=\frac{AE}{DE}=\frac{h-30}{x}
\displaystyle 1=\frac{h-30}{x}
\displaystyle \therefore x=h-30
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}=\frac{h}{x}
\displaystyle \therefore h=x\tan60^\circ
\displaystyle \text{Substituting }x=h-30,
\displaystyle h=(h-30)\tan60^\circ
\displaystyle h=(h-30)\sqrt3
\displaystyle h=\sqrt3h-30\sqrt3
\displaystyle h(\sqrt3-1)=30\sqrt3
\displaystyle h=\frac{30\sqrt3}{\sqrt3-1}
\displaystyle h=\frac{30\sqrt3(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}
\displaystyle h=15\sqrt3(\sqrt3+1)
\displaystyle h=45+15\sqrt3\text{ m}
\displaystyle h\approx45+15(1.732)=70.98\text{ m}
\displaystyle x=h-30
\displaystyle x=15+15\sqrt3=15(\sqrt3+1)\text{ m}
\displaystyle x\approx40.98\text{ m}
\displaystyle \therefore \text{The height of the tower is }45+15\sqrt3\text{ m, or approximately }70.98\text{ m.}
\displaystyle \therefore \text{Its horizontal distance from the observation points is }15(\sqrt3+1)\text{ m,}
\displaystyle \text{or approximately }40.98\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{From the top of a cliff }60\text{ m high, the angles of depression of the}
\displaystyle \text{top and bottom of a tower are }30^\circ\text{ and }60^\circ\text{ respectively. Find the height}
\displaystyle \text{of the tower.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the tower }CD=h\text{ m.}
\displaystyle \text{Given, }AB=60\text{ m.}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}=\frac{60}{BC}
\displaystyle \therefore BC=\frac{60}{\tan60^\circ}
\displaystyle \text{Since }DE\parallel CB\text{ and }AB\parallel CD,\ DE=BC.
\displaystyle AE=AB-BE=60-h
\displaystyle \text{In right-angled }\triangle AED,
\displaystyle \tan30^\circ=\frac{AE}{DE}=\frac{60-h}{BC}
\displaystyle 60-h=BC\tan30^\circ
\displaystyle 60-h=\frac{60}{\tan60^\circ}\tan30^\circ
\displaystyle 60-h=60\left(\frac{\tan30^\circ}{\tan60^\circ}\right)
\displaystyle 60-h=60\left(\frac{\frac{1}{\sqrt3}}{\sqrt3}\right)
\displaystyle 60-h=60\times\frac{1}{3}=20
\displaystyle h=60-20=40\text{ m}
\displaystyle \therefore \text{The height of the tower is }40\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A man on a cliff observes a boat at an angle of depression of }30^\circ.
\displaystyle \text{The boat is sailing towards the shore to the point immediately beneath him. Three minutes}
\displaystyle \text{later, the angle of depression is }60^\circ.\text{ Assuming that the boat sails at a uniform}
\displaystyle \text{speed, determine: (i) the additional time required to reach the shore, and (ii) the speed}
\displaystyle \text{of the boat in metres per second, if the height of the cliff is }500\text{ m.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }AB=500\text{ m and let }D\text{ and }C\text{ be the initial and later positions of the boat.}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \tan30^\circ=\frac{AB}{BD}=\frac{500}{BD}
\displaystyle \therefore BD=\frac{500}{\tan30^\circ}=500\sqrt3\text{ m}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}=\frac{500}{BC}
\displaystyle \therefore BC=\frac{500}{\tan60^\circ}=\frac{500}{\sqrt3}\text{ m}
\displaystyle \text{Distance travelled by the boat in }3\text{ minutes}=DC=BD-BC
\displaystyle =500\sqrt3-\frac{500}{\sqrt3}
\displaystyle =\frac{1500-500}{\sqrt3}=\frac{1000}{\sqrt3}\text{ m}
\displaystyle \text{Since the boat travels at a uniform speed,}
\displaystyle \frac{\text{Additional time}}{3}=\frac{BC}{DC}
\displaystyle \text{Additional time}=3\times\frac{\frac{500}{\sqrt3}}{\frac{1000}{\sqrt3}}
\displaystyle =3\times\frac{1}{2}=1.5\text{ minutes}
\displaystyle \text{Speed of the boat}=\frac{DC}{3\times60}
\displaystyle =\frac{\frac{1000}{\sqrt3}}{180}=\frac{50}{9\sqrt3}\text{ m/s}
\displaystyle =\frac{50\sqrt3}{27}\text{ m/s}\approx3.21\text{ m/s}
\displaystyle \therefore \text{The boat will take an additional }1.5\text{ minutes to reach the shore.}
\displaystyle \therefore \text{The speed of the boat is approximately }3.21\text{ m/s.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A man in a boat rowing away from a lighthouse }150\text{ m high takes}
\displaystyle 2\text{ minutes for the angle of elevation of the top of the lighthouse to change from }60^\circ
\displaystyle \text{to }45^\circ.\text{ Find the speed of the boat.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }C\text{ and }D\text{ be the initial and final positions of the boat respectively.}
\displaystyle \text{Given, }AB=150\text{ m and }CD\text{ is the distance travelled in }2\text{ minutes.}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}=\frac{150}{BC}
\displaystyle \therefore BC=\frac{150}{\tan60^\circ}=\frac{150}{\sqrt3}=50\sqrt3\text{ m}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \tan45^\circ=\frac{AB}{BD}=\frac{150}{BD}
\displaystyle \therefore BD=\frac{150}{\tan45^\circ}=150\text{ m}
\displaystyle \therefore CD=BD-BC
\displaystyle =150-50\sqrt3\text{ m}
\displaystyle \approx150-86.60=63.40\text{ m}
\displaystyle \text{Time taken}=2\text{ minutes}=120\text{ seconds}
\displaystyle \text{Speed of the boat}=\frac{\text{Distance travelled}}{\text{Time taken}}
\displaystyle =\frac{150-50\sqrt3}{120}\text{ m/s}
\displaystyle \approx\frac{63.40}{120}=0.53\text{ m/s}
\displaystyle \therefore \text{The speed of the boat is approximately }0.53\text{ m/s.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A person standing on the bank of a river observes that the angle of}
\displaystyle \text{elevation of the top of a tree on the opposite bank is }60^\circ.\text{ When he moves }40\text{ m}
\displaystyle \text{away from the bank, the angle of elevation becomes }30^\circ.\text{ Find: (i) the height of}
\displaystyle \text{the tree, correct to two decimal places, and (ii) the width of the river.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the tree }AB=h\text{ m and the width of the river }BC=x\text{ m.}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}=\frac{h}{x}
\displaystyle \therefore h=x\sqrt3
\displaystyle \text{After moving }40\text{ m away from the bank, }DB=x+40\text{ m.}
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \tan30^\circ=\frac{AB}{DB}=\frac{h}{x+40}
\displaystyle \therefore h=\frac{x+40}{\sqrt3}
\displaystyle \therefore x\sqrt3=\frac{x+40}{\sqrt3}
\displaystyle 3x=x+40
\displaystyle 2x=40
\displaystyle x=20\text{ m}
\displaystyle \therefore h=x\sqrt3=20\sqrt3\text{ m}
\displaystyle h\approx20\times1.732=34.64\text{ m}
\displaystyle \therefore \text{The height of the tree is approximately }34.64\text{ m.}
\displaystyle \therefore \text{The width of the river is }20\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The horizontal distance between two towers is }75\text{ m. The angle}
\displaystyle \text{of depression of the top of the first tower, as seen from the top of the second tower}
\displaystyle \text{which is }160\text{ m high, is }45^\circ.\text{ Find the height of the first tower.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the first tower be }h\text{ m.}
\displaystyle \text{The difference between the heights of the two towers}=(160-h)\text{ m.}
\displaystyle \text{The horizontal distance between the towers}=75\text{ m.}
\displaystyle \tan45^\circ=\frac{160-h}{75}
\displaystyle 1=\frac{160-h}{75}
\displaystyle 160-h=75
\displaystyle h=160-75=85\text{ m}
\displaystyle \therefore \text{The height of the first tower is }85\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The length of the shadow of a tower standing on level ground is found to be}
\displaystyle 2y\text{ m longer when the sun's altitude is }30^\circ\text{ than when it is }45^\circ.\text{ Prove that}
\displaystyle \text{the height of the tower is }y(\sqrt3+1)\text{ m.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the tower }AB=h\text{ m.}
\displaystyle \text{Let }BC\text{ be the length of the shadow when the sun's altitude is }45^\circ.
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan45^\circ=\frac{AB}{BC}=\frac{h}{BC}
\displaystyle 1=\frac{h}{BC}
\displaystyle \therefore BC=h
\displaystyle \text{The shadow at an altitude of }30^\circ\text{ is }2y\text{ m longer.}
\displaystyle \therefore BD=BC+CD=h+2y
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \tan30^\circ=\frac{AB}{BD}=\frac{h}{h+2y}
\displaystyle \frac{1}{\sqrt3}=\frac{h}{h+2y}
\displaystyle h+2y=\sqrt3h
\displaystyle 2y=h(\sqrt3-1)
\displaystyle h=\frac{2y}{\sqrt3-1}
\displaystyle h=\frac{2y(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}
\displaystyle h=\frac{2y(\sqrt3+1)}{3-1}
\displaystyle h=y(\sqrt3+1)\text{ m}
\displaystyle \therefore \text{The height of the tower is }y(\sqrt3+1)\text{ m. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{An aeroplane is flying horizontally at a height of }1\text{ km. Its angle}
\displaystyle \text{of elevation from a point on the ground is initially }60^\circ.\text{ After }10\text{ seconds, its}
\displaystyle \text{angle of elevation is observed to be }30^\circ.\text{ Find its uniform speed in kilometres}
\displaystyle \text{per hour.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }E\text{ and }A\text{ be the initial and final positions of the aeroplane respectively.}
\displaystyle \text{Given, }EC=AB=1\text{ km and }\angle EDC=60^\circ.
\displaystyle \text{In right-angled }\triangle EDC,
\displaystyle \tan60^\circ=\frac{EC}{DC}=\frac{1}{DC}
\displaystyle \therefore DC=\frac{1}{\tan60^\circ}=\frac{1}{\sqrt3}\text{ km}
\displaystyle \text{After }10\text{ seconds, }\angle ADB=30^\circ.
\displaystyle \text{In right-angled }\triangle ADB,
\displaystyle \tan30^\circ=\frac{AB}{DB}=\frac{1}{DB}
\displaystyle \therefore DB=\frac{1}{\tan30^\circ}=\sqrt3\text{ km}
\displaystyle \therefore \text{Distance travelled}=CB=DB-DC
\displaystyle =\sqrt3-\frac{1}{\sqrt3}
\displaystyle =\frac{3-1}{\sqrt3}=\frac{2}{\sqrt3}\text{ km}
\displaystyle \text{Time taken}=10\text{ seconds}=\frac{10}{3600}\text{ hour}
\displaystyle \text{Speed of the aeroplane}=\frac{\text{Distance travelled}}{\text{Time taken}}
\displaystyle =\frac{\frac{2}{\sqrt3}}{\frac{10}{3600}}
\displaystyle =\frac{720}{\sqrt3}=240\sqrt3\text{ km/h}
\displaystyle \approx240\times1.732=415.69\text{ km/h}
\displaystyle \therefore \text{The uniform speed of the aeroplane is approximately }415.69\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{From the top of a hill, the angles of depression of two consecutive}
\displaystyle \text{kilometre stones due east are }30^\circ\text{ and }45^\circ\text{ respectively. Find the distances}
\displaystyle \text{of the two stones from the foot of the hill.}\hfill\text{[ICSE 2007]}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the hill }AB=h\text{ km.}
\displaystyle \text{Let }BC=x\text{ km be the distance of the nearer stone from the foot of the hill.}
\displaystyle \text{Since the stones are consecutive, }CD=1\text{ km.}
\displaystyle \therefore BD=(x+1)\text{ km.}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan45^\circ=\frac{AB}{BC}=\frac{h}{x}
\displaystyle 1=\frac{h}{x}
\displaystyle \therefore h=x
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \tan30^\circ=\frac{AB}{BD}=\frac{h}{x+1}
\displaystyle \frac{1}{\sqrt3}=\frac{x}{x+1}
\displaystyle \sqrt3x=x+1
\displaystyle x(\sqrt3-1)=1
\displaystyle x=\frac{1}{\sqrt3-1}
\displaystyle x=\frac{\sqrt3+1}{2}\text{ km}
\displaystyle x\approx1.366\text{ km}
\displaystyle \therefore BC\approx1.366\text{ km}
\displaystyle BD=x+1=\frac{\sqrt3+1}{2}+1
\displaystyle BD=\frac{\sqrt3+3}{2}\text{ km}
\displaystyle BD\approx2.366\text{ km}
\displaystyle \therefore \text{The two stones are approximately }1.366\text{ km and }2.366\text{ km}
\displaystyle \text{from the foot of the hill.}
\displaystyle \\

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