Solution

It can be observed from the figure that $A B$ is the pole.

In $\triangle ABC$,

$ \begin{aligned} & \dfrac{AB}{AC}=\sin 30^{\circ} \\ & \dfrac{AB}{20}=\dfrac{1}{2} \\ & AB=\dfrac{20}{2}=10 \end{aligned} $

Therefore, the height of the pole is $10 m$.

Solution

Let $A C$ was the original tree. Due to storm, it was broken into two parts. The broken part $A^{\prime} B$ is making $30^{\circ}$ with the ground.

In $\triangle A^{\prime} BC$,

$ \begin{aligned} & \dfrac{BC}{A^{\prime} C}=\tan 30^{\circ} \\ & \dfrac{BC}{8}=\dfrac{1}{\sqrt{3}} \\ & BC=(\dfrac{8}{\sqrt{3}}) m \\ & \dfrac{A^{\prime} C}{A^{\prime} B}=\cos 30^{\circ} \\ & \dfrac{8}{A^{\prime} B}=\dfrac{\sqrt{3}}{2} \\ & A^{\prime} B=(\dfrac{16}{\sqrt{3}}) m \end{aligned} $

Height of tree $=A^{\prime} B+BC$

$ \begin{aligned} & =(\dfrac{16}{\sqrt{3}}+\dfrac{8}{\sqrt{3}}) m=\dfrac{24}{\sqrt{3}} m \\ & =8 \sqrt{3} m \end{aligned} $

Hence, the height of the tree is $8 \sqrt{3} m$.

Solution

It can be observed that AC and PR are the slides for younger and elder children respectively.

In $\triangle ABC$,

$ \dfrac{AB}{AC}=\sin 30^{\circ} $

$\dfrac{1.5}{AC}=\dfrac{1}{2}$

$AC=3 m$

In $\triangle PQR$,

$\dfrac{PQ}{PR}=\sin 60$

$\dfrac{3}{PR}=\dfrac{\sqrt{3}}{2}$

$PR=\dfrac{6}{\sqrt{3}}=2 \sqrt{3} m$

Therefore, the lengths of these slides are $3 m$ and $2 \sqrt{3} m$.

Solution

Let $A B$ be the tower and the angle of elevation from point $C$ (on ground) is

$30^{\circ}$.

In $\triangle ABC$,

$ \begin{aligned} & \dfrac{AB}{BC}=\tan 30^{\circ} \\ & \dfrac{AB}{30}=\dfrac{1}{\sqrt{3}} \\ & AB=\dfrac{30}{\sqrt{3}}=10 \sqrt{3} m \end{aligned} $

Therefore, the height of the tower is $10 \sqrt{3} m$.

Solution

Let $K$ be the kite and the string is tied to point $P$ on the ground.

In $\triangle KLP$, $\dfrac{KL}{KP}=\sin 60^{\circ}$

$\dfrac{60}{K P}=\dfrac{\sqrt{3}}{2}$

$KP=\dfrac{120}{\sqrt{3}}=40 \sqrt{3} m$

Hence, the length of the string is $40 \sqrt{3} m$.

Solution

Let the boy was standing at point $S$ initially. He walked towards the building and reached at point $T$.

It can be observed that

$P R=P Q-R Q$

$=(30-1.5) m=28.5 m=\dfrac{57}{2} m$

In $\triangle PAR$,

$\dfrac{PR}{AR}=\tan 30^{\circ}$

$\dfrac{57}{2 AR}=\dfrac{1}{\sqrt{3}}$

$AR=(\dfrac{57}{2} \sqrt{3}) m$

In $\triangle PRB$,

$ \begin{aligned} & \dfrac{PR}{BR}=\tan 60^{\circ} \\ & \dfrac{57}{2 BR}=\sqrt{3} \\ & BR=\dfrac{57}{2 \sqrt{3}}=(\dfrac{19 \sqrt{3}}{2}) m \end{aligned} $

$S T=A B$

$=AR-BR=(\dfrac{57 \sqrt{3}}{2}-\dfrac{19 \sqrt{3}}{2}) m$

$=(\dfrac{38 \sqrt{3}}{2}) m=19 \sqrt{3} m$

Hence, he walked $19 \sqrt{3} m$ towards the building.

Solution

Let $B C$ be the building, $A B$ be the transmission tower, and $D$ be the point on the ground from where the elevation angles are to be measured.

In $\triangle BCD$,

$\dfrac{BC}{CD}=\tan 45^{\circ}$

$\dfrac{20}{CD}=1$

$CD=20 m$

In $\triangle ACD$,

$ \begin{aligned} & \dfrac{AC}{CD}=\tan 60^{\circ} \\ & \dfrac{AB+BC}{CD}=\sqrt{3} \\ & \dfrac{AB+20}{20}=\sqrt{3} \\ & AB=(20 \sqrt{3}-20) m \\ & \quad=20(\sqrt{3}-1) m \end{aligned} $

Therefore, the height of the transmission tower is $20(\sqrt{3}-1)$ m.

Solution

Let $A B$ be the statue, $B C$ be the pedestal, and $D$ be the point on the ground from where the elevation angles are to be measured.

In $\triangle BCD$,

$ \begin{aligned} & \dfrac{B C}{C D}=\tan 45^{\circ} \\ & \dfrac{B C}{C D}=1 \\ & B C=C D \end{aligned} $

In $\triangle ACD$,

$ \begin{aligned} & \dfrac{AB+BC}{CD}=\tan 60^{\circ} \\ & \dfrac{AB+BC}{BC}=\sqrt{3} \\ & 1.6+BC=BC \sqrt{3} \\ & BC(\sqrt{3}-1)=1.6 \\ & BC=\dfrac{(1.6)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\ & \quad=\dfrac{1.6(\sqrt{3}+1)}{(\sqrt{3})^{2}-(1)^{2}} \\ & \quad=\dfrac{1.6(\sqrt{3}+1)}{2}=0.8(\sqrt{3}+1) \end{aligned} $

Therefore, the height of the pedestal is $0.8(\sqrt{3}+1) m$.

Solution

Let $A B$ be the building and $C D$ be the tower.

In $\triangle CDB$, $\dfrac{CD}{BD}=\tan 60^{\circ}$

$\dfrac{50}{BD}=\sqrt{3}$

$BD=\dfrac{50}{\sqrt{3}}$

In $\triangle A B D$,

$\dfrac{AB}{BD}=\tan 30^{\circ}$

$AB=\dfrac{50}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}}=\dfrac{50}{3}=16 \dfrac{2}{3}$

Therefore, the height of the building is

$ 16 \dfrac{2}{3} m $

Solution

Let $A B$ and $C D$ be the poles and $O$ is the point from where the elevation angles are measured.

In $\triangle ABO$,

$ \begin{aligned} & \dfrac{AB}{BO}=\tan 60^{\circ} \\ & \dfrac{AB}{BO}=\sqrt{3} \\ & BO=\dfrac{AB}{\sqrt{3}} \end{aligned} $

In $\triangle CDO$, $\dfrac{CD}{DO}=\tan 30^{\circ}$

$\dfrac{CD}{80-BO}=\dfrac{1}{\sqrt{3}}$

$CD \sqrt{3}=80-BO$

$CD \sqrt{3}=80-\dfrac{AB}{\sqrt{3}}$

$CD \sqrt{3}+\dfrac{AB}{\sqrt{3}}=80$

Since the poles are of equal heights,

$C D=A B$

$CD[\sqrt{3}+\dfrac{1}{\sqrt{3}}]=80$

$CD(\dfrac{3+1}{\sqrt{3}})=80$

$CD=20 \sqrt{3} m$

$BO=\dfrac{AB}{\sqrt{3}}=\dfrac{CD}{\sqrt{3}}=(\dfrac{20 \sqrt{3}}{\sqrt{3}}) m=20 m$

$DO=BD-BO=(80-20) m=60 m$

Therefore, the height of poles is $20 \sqrt{3} m$ and the point is $20 m$ and $60 m$ far from these poles.

Solution

In $\triangle ABC$,

$\dfrac{AB}{BC}=\tan 60^{\circ}$

$\dfrac{AB}{BC}=\sqrt{3}$

$BC=\dfrac{AB}{\sqrt{3}}$

In $\triangle ABD$,

$\dfrac{AB}{BD}=\tan 30^{\circ}$

$\dfrac{AB}{BC+CD}=\dfrac{1}{\sqrt{3}}$

$\dfrac{AB}{\dfrac{AB}{\sqrt{3}}+20}=\dfrac{1}{\sqrt{3}}$

$\dfrac{AB \sqrt{3}}{AB+20 \sqrt{3}}=\dfrac{1}{\sqrt{3}}$

$3 AB=AB+20 \sqrt{3}$

$2 AB=20 \sqrt{3}$

$AB=10 \sqrt{3} m$

$BC=\dfrac{AB}{\sqrt{3}}=(\dfrac{10 \sqrt{3}}{\sqrt{3}}) m=10 m$

Therefore, the height of the tower is $10 \sqrt{3} m$ and the width of the canal is $10 m$.

Solution

Let $A B$ be a building and $C D$ be a cable tower.

In $\triangle ABD$,

$\dfrac{AB}{BD}=\tan 45^{\circ}$

$\dfrac{7}{BD}=1$

$BD=7 m$

In $\triangle ACE$,

$AE=BD=7 m$

$\dfrac{CE}{AE}=\tan 60^{\circ}$

$\dfrac{CE}{7}=\sqrt{3}$

$CE=7 \sqrt{3} m$

$CD=CE+ED=(7 \sqrt{3}+7) m$

$ =7(\sqrt{3}+1) m $

Therefore, the height of the cable tower is $7(\sqrt{3}+1) m$.

Solution

Let $A B$ be the lighthouse and the two ships be at point $C$ and $D$ respectively.

In $\triangle ABC$,

$ \begin{aligned} & \dfrac{AB}{BC}=\tan 45^{\circ} \\ & \dfrac{75}{BC}=1 \\ & BC=75 m \end{aligned} $

In $\triangle ABD$,

$ \dfrac{AB}{BD}=\tan 30^{\circ} $

$\dfrac{75}{BC+CD}=\dfrac{1}{\sqrt{3}}$

$\dfrac{75}{75+CD}=\dfrac{1}{\sqrt{3}}$

$75 \sqrt{3}=75+CD$

$75(\sqrt{3}-1) m=CD$

Therefore, the distance between the two ships is $75(\sqrt{3}-1) m$.

Solution

Let the initial position $A$ of balloon change to $B$ after some time and $C D$ be the girl.

In $\triangle ACE$,

$\dfrac{AE}{CE}=\tan 60^{\circ}$ $\dfrac{AF-EF}{CE}=\tan 60^{\circ}$

$\dfrac{88.2-1.2}{C E}=\sqrt{3}$

$\dfrac{87}{CE}=\sqrt{3}$

$CE=\dfrac{87}{\sqrt{3}}=29 \sqrt{3} m$

In $\triangle BCG$,

$\dfrac{BG}{CG}=\tan 30^{\circ}$

$\dfrac{88.2-1.2}{CG}=\dfrac{1}{\sqrt{3}}$

$87 \sqrt{3} m=CG$

Distance travelled by balloon $=EG=CG-CE$

$=(87 \sqrt{3}-29 \sqrt{3}) m$

$=58 \sqrt{3} m$

Solution

Let $A B$ be the tower.

Initial position of the car is $C$, which changes to $D$ after six seconds.

In $\triangle ADB$,

$\dfrac{AB}{DB}=\tan 60^{\circ}$

$\dfrac{AB}{DB}=\sqrt{3}$

$DB=\dfrac{AB}{\sqrt{3}}$

In $\triangle ABC$,

$\dfrac{AB}{BC}=\tan 30^{\circ}$

$\dfrac{AB}{BD+DC}=\dfrac{1}{\sqrt{3}}$

$ \begin{aligned} & AB \sqrt{3}=BD+DC \\ & AB \sqrt{3}=\dfrac{AB}{\sqrt{3}}+DC \\ & DC=AB \sqrt{3}-\dfrac{AB}{\sqrt{3}}=AB(\sqrt{3}-\dfrac{1}{\sqrt{3}}) \\ & =\dfrac{2 AB}{\sqrt{3}} \end{aligned} $

Time taken by the car to travel distance DC $(.$ i.e., $.\dfrac{2 AB}{\sqrt{3}})=6$ seconds

Time taken by the car to travel distance $D B(.$ i.e., $.\dfrac{AB}{\sqrt{3}})=\dfrac{6}{\dfrac{2 AB}{\sqrt{3}}} \times \dfrac{AB}{\sqrt{3}}$

$=\dfrac{6}{2}=3$ seconds