Chapter 09 Some Applications of Trigonometry Exercise-01
EXERCISE 9.1
1. A circus artist is climbing a $20 \mathrm{~m}$ long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30^{\circ}$ (see Fig. 9.11).
Fig. 9.11
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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 $10 m$.
2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^{\circ}$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8 \mathrm{~m}$. Find the height of the tree.
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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 $8 \sqrt{3} m$.
3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of $1.5 \mathrm{~m}$, and is inclined at an angle of $30^{\circ}$ to the ground, whereas for elder children, she wants to have a steep slide at a height of $3 \mathrm{~m}$, and inclined at an angle of $60^{\circ}$ to the ground. What should be the length of the slide in each case?
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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 $2 \sqrt{3} m$.
4. The angle of elevation of the top of a tower from a point on the ground, which is $30 \mathrm{~m}$ away from the foot of the tower, is $30^{\circ}$. Find the height of the tower.
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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 $10 \sqrt{3} m$.
5. A kite is flying at a height of $60 \mathrm{~m}$ above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^{\circ}$. Find the length of the string, assuming that there is no slack in the string.
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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 $40 \sqrt{3} m$.
6. A $1.5 \mathrm{~m}$ tall boy is standing at some distance from a $30 \mathrm{~m}$ tall building. The angle of elevation from his eyes to the top of the building increases from $30^{\circ}$ to $60^{\circ}$ as he walks towards the building. Find the distance he walked towards the building.
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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.
7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $20 \mathrm{~m}$ high building are $45^{\circ}$ and $60^{\circ}$ respectively Find the height of the tower.
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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.
8. A statue, $1.6 \mathrm{~m}$ tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^{\circ}$ and from the same point the angle of elevation of the top of the pedestal is $45^{\circ}$. Find the height of the pedestal.
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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 $0.8(\sqrt{3}+1) m$.
9. The angle of elevation of the top of a building from the foot of the tower is $30^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $60^{\circ}$. If the tower is $50 \mathrm{~m}$ high, find the height of the building.
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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 $
10. Two poles of equal heights are standing opposite each other on either side of the road, which is $80 \mathrm{~m}$ wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^{\circ}$ and $30^{\circ}$, respectively. Find the height of the poles and the distances of the point from the poles.
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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.
11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^{\circ}$. From another point $20 \mathrm{~m}$ away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is $30^{\circ}$ (see Fig. 9.12). Find the height of the tower and the width of the canal.
Fig. 9.12
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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 $10 m$.
12. From the top of a $7 \mathrm{~m}$ high building, the angle of elevation of the top of a cable tower is $60^{\circ}$ and the angle of depression of its foot is $45^{\circ}$. Determine the height of the tower.
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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 $7(\sqrt{3}+1) m$.
13. As observed from the top of a $75 \mathrm{~m}$ high lighthouse from the sea-level, the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
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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 $75(\sqrt{3}-1) m$.
14. A $1.2 \mathrm{~m}$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2 \mathrm{~m}$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $60^{\circ}$. After some time, the angle of elevation reduces to $30^{\circ}$ (see Fig. 9.13). Find the distance travelled by the balloon during the interval.
Fig. 9.13
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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 $=EG=CG-CE$
$=(87 \sqrt{3}-29 \sqrt{3}) m$
$=58 \sqrt{3} m$
15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^{\circ}$, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be $60^{\circ}$. Find the time taken by the car to reach the foot of the tower from this point.
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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
