Question 14 Marks
The topmost branch of a tree is tied with a string attached to a pole in the ground. The length of this string Is 200m and it makes an angle of 45° with the ground. Find the distance of the pole to which the string is tied from the base of the tree.
Answer
The topmost branch of the tree is at point $\mathrm{B}$ and $\mathrm{C}$ is the point on the ground to which the topmost branch is tied.
In $\triangle \mathrm{ABC}$,
$
\frac{\mathrm{AC}}{\mathrm{BC}}=\cos 45^{\circ}
$
$
\frac{h}{200}=\frac{1}{\sqrt{2}}
$
$
h=\frac{200}{\sqrt{2}}=\frac{200 \sqrt{2}}{2}=100 \sqrt{2}=100 \times 1.414=141.4
$
Thus, the required distance is $141.4 \mathrm{~m}$. View full question & answer→Question 24 Marks
A kite tied to a string makes an angle of 60° with the ground. Find the perpendicular height of the kite if the length of its string is 250 m.
Answer
Let $\mathrm{K}$ be the kite and the string is tied to point $\mathrm{P}$ on ground.
$\ln \triangle K L P$
$
\begin{aligned}
\frac{\mathrm{KL}}{\mathrm{KP}} & =\sin 60^{\circ} \\
\frac{h}{250} & =\frac{\sqrt{3}}{2}
\end{aligned}
$
$
h=\frac{250 \sqrt{3}}{2}=125 \sqrt{3}
$
Thus, the perpendicular height of the kite is $125 \sqrt{3} \mathrm{~m}$. View full question & answer→Question 34 Marks
An aeroplane takes off at angle of $30^{\circ}$ with the ground. Find the height of the aeroplane above the ground when it has travelled $386 \mathrm{~m}$ without changing direction .
Answer
The plane takes off from point $\mathrm{C}$ on the ground. Let $\mathrm{A}$ be the final position of the plane .
In $\triangle \mathrm{ABC}$
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{AC}}=\sin 30^{\circ} \\
& \frac{\mathrm{AB}}{386}=\frac{1}{2} \\
& h=\frac{386}{2}=193
\end{aligned}
$
Thus, the required height of the aeroplane above the ground is $193 \mathrm{~m}$. View full question & answer→Question 44 Marks
From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be $\alpha$, and $\beta$. If the height of the light house is ' $h$ ' $m$ and the line joining the ships passes through the foot of the light house, show that the distance between the ship is $\frac{\mathrm{h}(\tan \alpha+\tan \beta)}{\tan \alpha \tan \beta} \mathrm{m}$
Answer
Let $A B$ be the lighthouse of height $h \mathrm{~m}$. Let $A C=x$ and $A D=y$.
In $\triangle C A B$,
$
\frac{\mathrm{AB}}{\mathrm{AC}}=\tan \alpha
$
$
\tan \alpha=\frac{\mathrm{h}}{\mathrm{x}}
$
$
\mathrm{x}=\frac{\mathrm{h}}{\tan \alpha} \ldots . \text { (i) }
$
In $\triangle \mathrm{DAB}$,
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{AD}}=\tan \beta \\
& \tan \alpha=\frac{\mathrm{h}}{\mathrm{y}} \\
& y=\frac{\mathrm{h}}{\tan \beta} \ldots(\mathrm{i)}
\end{aligned}
$
In $\triangle \mathrm{DAB}$,
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{AD}}=\tan \beta \\
& \tan \alpha=\frac{\mathrm{h}}{\mathrm{y}} \\
& y=\frac{\mathrm{h}}{\tan \beta} \ldots . \text { (ii) }
\end{aligned}
$
Distance between the ships $=x+y$
$
\begin{aligned}
& =\frac{\mathrm{h}}{\tan \alpha}+\frac{\mathrm{h}}{\tan \beta} \\
& =\mathrm{h}\left(\frac{\tan \beta+\tan \alpha}{\tan \alpha \tan \beta}\right)
\end{aligned}
$ View full question & answer→Question 54 Marks
$
\text { Find the length of the shadow cast by a tree } 60 \mathrm{~m} \text { high when the sun's altitude is } 30^{\circ} \text {. }
$
Answer
Let $A B$ be the tree of height $60 \mathrm{~m}$ and $B C$ be its shadow .
$\ln \triangle \mathrm{ABC}$
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{BC}}=\tan 30^{\circ} \\
& \frac{60}{\mathrm{BC}}=\frac{1}{\sqrt{3}}
\end{aligned}
$
$
\mathrm{BC}=60 \sqrt{3} \mathrm{~m}
$
So, height of tower is $60 \sqrt{3} \mathrm{~m}$. View full question & answer→Question 64 Marks
The length of the shadow of a pillar is $\frac{1}{\sqrt{3}}$ times the height of the pillar . find the angle of elevation of the sun.
Answer
Let $A B$ be the pillar and $B C$ be its shadow.
Let $h$ be the height of the pillar. Then,
$
B C=\frac{1}{\sqrt{3}} h
$
In $\triangle \mathrm{ABC}$,
$
\begin{aligned}
& \tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}} \\
& \Rightarrow \tan \theta=\frac{h}{\frac{h}{\sqrt{3}}}=\sqrt{3}
\end{aligned}
$
But, $\tan 60^{\circ}=\sqrt{3}$
$
\therefore \theta=60^{\circ}
$ View full question & answer→Question 74 Marks
The horizontal distance between towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 60m, find the height of the first tower.
Answer
Let $A B$ and $P Q$ be the two towers.
$
B Q=A R=140 m
$
$
A B=R Q=60 m
$
Let $\mathrm{PR}=\mathrm{h}$
We need to find the height of the first tower, i.e. , PQ.
In $\triangle \mathrm{PRA}$,
$
\begin{aligned}
& \frac{\mathrm{PR}}{\mathrm{AR}}=\tan 30^{\circ} \\
& \frac{\mathrm{h}}{140}=\frac{1}{\sqrt{3}}
\end{aligned}
$
$
\mathrm{h}=\frac{140}{\sqrt{3}}=80.83
$
$
\therefore P Q=P R+R Q=80.83 m+60 m=140.83 m
$
Thus, the height of the first tower is $140.83 \mathrm{~m}$. View full question & answer→Question 84 Marks
From a point $10 \mathrm{~m}$ above the ground, the angle of elevation of the top of a tower is $\alpha$ and the angle of depression is $\beta$. If $\tan \alpha=\frac{5}{2}$ and $\tan \beta=\frac{1}{4}$, calculate the height of the tower to the nearest metre.
AnswerLet AD be the tower and B be the point of observation.
In $\triangle \mathrm{ABC}$,
$
\begin{aligned}
& \tan \alpha=\frac{\mathrm{AC}}{\mathrm{BC}} \\
& \Rightarrow \frac{\mathrm{h}}{\mathrm{x}}=\frac{5}{2} . .\ldots(1)
\end{aligned}
$
In $\triangle \mathrm{BED}$,
$
\tan \beta=\frac{B E}{E D}
$
$
\Rightarrow \frac{1}{4}=\frac{10}{x}
$
$
\Rightarrow x=40
$
From (1),
$
h=100
$
$\therefore$ Height of the tower $=h+10=100+10=110 \mathrm{~m}$ View full question & answer→Question 94 Marks
From the top of a light house $96 \mathrm{~m}$ high, the angles of depression of two ships in the river and at the same level as the base of the light house and on the same side of it, are $\alpha$ and $\beta$. If $\tan$ $\alpha=\frac{1}{4}$ and $\tan \beta=\frac{1}{7}$, find the distance between the ships.
Answer
In the figure, $\mathrm{TO}$ is the light house and $\mathrm{A}$ and $\mathrm{B}$ are the position of the two ships.
$
\begin{aligned}
& \text { In } \triangle \mathrm{AOT} \\
& \frac{\mathrm{OT}}{\mathrm{OA}}=\tan \alpha \\
& \Rightarrow \frac{96}{\mathrm{OA}}=\frac{1}{4} \\
& \Rightarrow \mathrm{OA}=384
\end{aligned}
$
In $\triangle \mathrm{BOT}$,
$
\frac{\mathrm{OT}}{\mathrm{OB}}=\tan \beta
$
$
\Rightarrow \frac{96}{\mathrm{OB}}=\frac{1}{7}
$
$
\Rightarrow \mathrm{OB}=672
$
$\therefore$ Distance between the two ships $=A B=O A-O B=384-672=288 \mathrm{~m}$ View full question & answer→Question 104 Marks
A vertical pole is $90 \mathrm{~m}$ high and the length of its shadow is $90 \sqrt{3}$. what is the angle of elevation of the sun?
Answer
Let $A B$ be the pole and $B C$ be its shadow.
In $\triangle \mathrm{ABC}$,
$
\begin{aligned}
& \tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}} \\
& \Rightarrow \tan \theta=\frac{90}{90 \sqrt{3}}=\frac{1}{\sqrt{3}}
\end{aligned}
$
But, $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
$
\therefore \theta=30^{\circ}
$
Thus, the angle of elevation is $30^{\circ}$. View full question & answer→Question 114 Marks
The horizontal distance between two trees of different heights is 100m. The angle of depression of the top of the first tree when seen from the top of the second tree is 45°. If the height of the second tree is 150m, find the height of the first tree.
Answer
Let $A C$ and $B D$ be two trees.
$
B D=150 \mathrm{~m}, A E=C D=100 \mathrm{~m}
$
In $\triangle \mathrm{BAE}$,
$
\begin{aligned}
& \tan 45^{\circ}=\frac{\mathrm{BE}}{\mathrm{AE}} \\
& \Rightarrow 1=\frac{\mathrm{BE}}{100} \\
& \Rightarrow \mathrm{BE}=100 \\
& \therefore \mathrm{AC}=\mathrm{BD}-\mathrm{BE}=150 \mathrm{~m}-100 \mathrm{~m}=50 \mathrm{~m}
\end{aligned}
$
Thus, the height of the first tree is $50 \mathrm{~m}$. View full question & answer→Question 124 Marks
An observer, 1.5m tall, is 28.5m away from a tower 30m high. Determine the angle of elevation of the top of the tower from his eye.
Answer
Here, $E D$ is the height of the observer and $A C$ is the tower.
$
\begin{aligned}
& B E=C D=28.5 \mathrm{~m} \\
& A B=A C-B C=30 \mathrm{~m}-1.5 \mathrm{~m}=28.5 \mathrm{~m} \\
& \ln \triangle A B E, \\
& \tan <A B E=\frac{A B}{B E} \\
& \Rightarrow \tan <A B E=\frac{28.5 \mathrm{~m}}{28.5 \mathrm{~m}}=1
\end{aligned}
$
But, $\tan 45^{\circ}=1$
$
\therefore<\mathrm{ABE}=45^{\circ}
$
Thus, the required angle of elevation is $45^{\circ}$, View full question & answer→Question 134 Marks
The angle of elevation of the top of a vertical cliff from a point 30 m away from the foot of the cliff is 60°. Find the height of the cliff.
Answer
Let $A B$ be the diff and angle of elevation from point $C$ (on ground) is $30^{\circ}$. In $\triangle \mathrm{ABC}$
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{BC}}=\tan 60^{\circ} \\
& \frac{\mathrm{AB}}{30}=\sqrt{3} \\
& \mathrm{AB}=30 \sqrt{3} \mathrm{~m}
\end{aligned}
$
So, height of tower is $30 \sqrt{3} \mathrm{~m}$ View full question & answer→Question 144 Marks
The height of an observation tower is 180m above sea level. A ship coming towards the tower is observed at an angle of depression of 30°. Calculate the distance of the boat from the foot of the observation tower.
Answer
Let $A B$ be the tower and $C$ be the position of the ship.
Let the distance of the boat from the foot of the observation tower be $\mathrm{x}$.
In triangle $\mathrm{ABC}$,
$
\tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}}
$
$
\Rightarrow \tan 30^{\circ}=\frac{180}{X}
$
$
\Rightarrow \frac{1}{\sqrt{3}}=\frac{180}{X}
$
$
\Rightarrow X=180 \sqrt{3}=180 \times 1.732=311.76
$
Thus, the distance of the boat from the foot of the observation tower is 311.76 or $311.8 \mathrm{~m}$. View full question & answer→Question 154 Marks
Find the angle of depression from the top of a 140m high pillar of a milestone on the ground at a distance of 200m from the foot of the pillar.
Answer$
\text { Let } A B \text { be the pillar. Let the angle of depression be } \theta \text {. }
$
In triangle $\mathrm{ABC}$,
$
\begin{aligned}
& \tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}} \\
& \Rightarrow \tan \theta=\frac{140}{200}=\frac{7}{10}=0.7
\end{aligned}
$
We have : $\tan 35^{\circ}=0.7$
Thus, the angle of depression is $\theta=35^{\circ}$. View full question & answer→Question 164 Marks
A 10 m high pole is kept vertical by a steel wire. The wire is inclined at an angle of 40° with the horizontal ground. If the wire runs from the top of the pole to the point on the ground where Its other end is fixed, find the lenqth of the wire.
Answer
Let $A B$ be the pole and $A C$ be the wire which runs from the top of the pole to the point on the ground where its other end is fixed.
$
\begin{aligned}
& \text { In } \triangle \mathrm{ABC} \\
& \sin 40^{\circ}=\frac{\mathrm{AB}}{\mathrm{AC}} \\
& \Rightarrow 0.6428=\frac{10}{\mathrm{AC}} \\
& \Rightarrow \mathrm{AC}=\frac{10}{0.6428}=15.6
\end{aligned}
$
Thus, the length of the wire is $15.6 \mathrm{~m}$. View full question & answer→Question 174 Marks
A ladder rests against a wall at an angle $a$, to the horizontal. Its foot is pulled away from the wall through a distance ' $a$ ', so that it slides a distance ' $b$ ' down the wall making an angle $\beta$ with the horizontal. Show that $\frac{a}{b}=\frac{\cos \alpha-\cos \beta}{\sin \beta-\sin \alpha}$.
Answer
Let $C P$ and $D W$ be the two position of the ladder such that $C P=D W=x$ (say).
$
C D=a, P W=b, \angle A C P=\alpha \text { and } \angle A D W=\beta
$
In $\triangle \mathrm{APC}$,
$
\frac{\mathrm{AC}}{\mathrm{CP}}=\cos \alpha \Rightarrow \mathrm{AC}=\mathrm{x} \cos \alpha\ldots(i)
$
$
\begin{aligned}
& \operatorname{In} \triangle \mathrm{ADW} \\
& \frac{\mathrm{AD}}{\mathrm{DW}}=\cos \beta \Rightarrow \frac{\mathrm{AC}+\mathrm{CD}}{\mathrm{DW}}=\cos \beta \\
& \Rightarrow \frac{\mathrm{x} \cos \alpha+a}{\mathrm{x}}=\cos \beta \quad[\text { using (i)] } \\
& \Rightarrow \mathrm{x}=\frac{a}{\cos \beta-\cos \alpha}\ldots(ii)
\end{aligned}
$
Again in $\triangle A P C, \frac{A P}{C P}=\sin \alpha$
$\Rightarrow \mathrm{AP}=\mathrm{x} \sin \alpha=\frac{a \sin \alpha}{(\cos \beta-\cos \alpha)}$ ..(iii) [Using (ii)]
Again in $\triangle \mathrm{ADW}, \frac{\mathrm{AW}}{\mathrm{DW}}=\sin \beta$
$
\Rightarrow \mathrm{AW}=\mathrm{x} \sin \beta=\frac{a \sin \beta}{(\cos \beta-\cos \alpha)}\ldots(iv)
$
Now, $P W=A P-A W$
$
\begin{aligned}
& \Rightarrow b=\left(\frac{a \sin \alpha}{\cos \beta-\cos \alpha}\right)-\left(\frac{a \sin \beta}{\cos \beta-\cos \alpha}\right)=\frac{a(\sin \alpha-\sin \beta)}{(\cos \beta-\cos \alpha)} \\
& \Rightarrow \frac{a}{b}=\frac{\cos \beta-\cos \alpha}{\sin \alpha-\sin \beta}
\end{aligned}
$
Hence proved. View full question & answer→Question 184 Marks
From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive milestone on opposite sides of the aeroplane are observed to be $\alpha$, and $\beta$. Show that the height in miles of aeroplane above the road is $\frac{\tan \alpha \tan \beta}{\tan \alpha+\tan \beta}$.
Answer
Let $P Q$ be $h$
QB be $x$
Given: $A B=1$ mile
$
\begin{aligned}
& Q B=x \\
& A Q=(1-x) \text { mile } \\
& \text { in } \triangle P A Q \\
& \operatorname{Tan} \alpha=\frac{P Q}{A Q}
\end{aligned}
$
in $\triangle \mathrm{PAQ}$
$\operatorname{Tan} \alpha=\frac{\mathrm{PQ}}{\mathrm{AQ}}$
$\operatorname{Tan} \alpha=\frac{\mathrm{h}}{1-\mathrm{x}}$
$1-\mathrm{x}=\frac{\mathrm{h}}{\operatorname{Tan} \alpha}$$\ldots(1)$
$\operatorname{In} \triangle \mathrm{PQB}$
$
\begin{aligned}
& \operatorname{Tan} \beta=\frac{\mathrm{h}}{\mathrm{x}} \\
& \mathrm{x}=\frac{\mathrm{h}}{\operatorname{Tan} \beta}
\end{aligned}
$
Substitute for $\mathrm{x}$ in equation (1)
$
\begin{aligned}
& 1=\frac{\mathrm{h}}{\operatorname{Tan} \beta}+\frac{\mathrm{h}}{\operatorname{Tan} \alpha} \\
& 1=\mathrm{h}\left\{\frac{1}{\operatorname{Tan} \beta}+\frac{1}{\operatorname{Tan} \alpha}\right\} \\
& \frac{1}{\mathrm{~h}}=\frac{\operatorname{Tan} \beta+\operatorname{Tan} \alpha}{\operatorname{Tan} \beta \operatorname{Tan} \alpha}
\end{aligned}
$
Thus, the height in miles of aeroplane above the road is $\frac{\operatorname{Tan} \alpha \operatorname{Tan} \beta}{\operatorname{Tan} \alpha+\operatorname{Tan} \beta}$ View full question & answer→Question 194 Marks
If the angle of elevation of a cloud from a point $\mathrm{h} \mathrm{m}$ above a lake is $\alpha$, and the angle of depression of its reflection in the lake be $\beta$, prove that distance of the cloud from the point of observation is $\frac{2 \mathrm{~h} \sec \alpha}{\tan \alpha-\tan \beta}$.
Answer
Let $A B$ be the surface of the lake and let $P$ be an point of observation such that $A P=h$ meters. Let $c$ be the position of the cloud and $C^{\prime}$ be its reflection in the lake. Then $\angle C P M=\alpha$ and $\angle M P C=\beta$. Let $\mathrm{CM}=\mathrm{x}$
Then, $\mathrm{CB}=\mathrm{CM}+\mathrm{MB}=\mathrm{CM}+\mathrm{PA}=\mathrm{x}+\mathrm{h}$
In $\triangle \mathrm{CPM}$,
$\tan \alpha=\frac{\mathrm{CM}}{\mathrm{PM}}$
$
\Rightarrow \tan \alpha=\frac{\mathrm{x}}{\mathrm{AB}}[\because \mathrm{PM}=\mathrm{AB}]
$
$\Rightarrow \mathrm{AB}=\mathrm{x} \cot \alpha \ldots(1)$
In $\triangle \mathrm{PMC}^{\prime}$,
$
\begin{aligned}
& \tan \beta=\frac{C \prime M}{P M} \\
& \Rightarrow \tan \beta=\frac{\mathrm{x}+2 \mathrm{~h}}{\mathrm{AB}} \\
& \Rightarrow \mathrm{AB}=(\mathrm{x}+2 \mathrm{~h}) \cot \beta\ldots(2)
\end{aligned}
$
From (1) and (2),
$x \cot \alpha=(x+2 h) \cot \beta$
$\Rightarrow \mathrm{x}\left(\frac{1}{\tan \alpha}-\frac{1}{\tan \beta}\right)=\frac{2 \mathrm{~h}}{\tan \beta}$
$\Rightarrow \mathrm{x}\left(\frac{\tan \beta-\tan \alpha}{\tan \alpha \tan \beta}\right)=\frac{2 \mathrm{~h}}{\tan \beta}$
$\Rightarrow \mathrm{x}=\frac{2 \mathrm{~h} \tan \alpha}{\tan \beta-\tan \alpha}$
Again, in $\triangle C P M$,
$\sin \alpha=\frac{\mathrm{CM}}{\mathrm{PC}}=\frac{\mathrm{x}}{\mathrm{PC}}$
$\Rightarrow \mathrm{PC}=\frac{\mathrm{x}}{\sin \alpha}$
$\Rightarrow \mathrm{PC}=\frac{2 \mathrm{~h} \tan \alpha}{\tan \beta-\tan \alpha} \times \frac{1}{\sin \alpha}$
$\Rightarrow \mathrm{PC}=\frac{2 \mathrm{~h} \sec \alpha}{\tan \beta-\tan \alpha}$
$
\text { Hence, the distance of the cloud from the point of observation is } \frac{2 \mathrm{~h} \sec \alpha}{\tan \beta-\tan \alpha} \text {. }
$ View full question & answer→Question 204 Marks
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40m vertically above X, the angle of elevation is 45°. Find the height of the tower PQ and the distance XQ.
Answer
In the figure, $\mathrm{PQ}$ is the tower.
$
\begin{aligned}
& \ln \triangle \mathrm{PQX}, \\
& \therefore \frac{\mathrm{h}}{\mathrm{x}}=\tan 60^{\circ}=\sqrt{3} \\
& \Rightarrow \mathrm{h}=\sqrt{3} \mathrm{x} \quad \ldots(1)
\end{aligned}
$
In $\triangle \mathrm{QRY}$,
$
\begin{aligned}
& \frac{\mathrm{h}-40}{\mathrm{x}}=\tan 45^{\circ}=1 \\
& \Rightarrow \mathrm{h}=40+\mathrm{x} \quad \ldots(2)
\end{aligned}
$
$
\begin{aligned}
& \text { From (1) and (2), } \\
& \sqrt{3} \mathrm{x}=40+\mathrm{x} \\
& \Rightarrow(\sqrt{3}-1) \mathrm{x}=40 \\
& \Rightarrow \mathrm{x}=\frac{40}{\sqrt{3}-1}=\frac{40(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{40}{2}(\sqrt{3}+1)=20(\sqrt{3}+1) \\
& \therefore \mathrm{h}=40+20(\sqrt{3}+1)=20 \sqrt{3}+60=20(\sqrt{3}+3)=20 \times 4.732=94.64
\end{aligned}
$ View full question & answer→Question 214 Marks
The angle of elevation of an aeroplane from a point on the ground is 45°. After 15 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a height of 3000 m, find the speed of the aeroplane.
Answer
Let $A$ be the point of observation on the ground and $B$ and $C$ be the two positions of aeroplane. Let $B L=C M=3000 \mathrm{~m}$.
In $\triangle \mathrm{ALB}$,
$
\tan 45^{\circ}=\frac{\mathrm{BL}}{\mathrm{AL}}
$
$
\Rightarrow 1=\frac{3000}{\mathrm{AL}}
$
$
\Rightarrow \mathrm{AL}=3000
$
In $\triangle \mathrm{AMC}$,
$
\begin{aligned}
& \tan 30^{\circ}=\frac{\mathrm{MC}}{\mathrm{AM}} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{3000}{3000+\mathrm{LM}} \\
& \Rightarrow 3000 \sqrt{3}=(3000+\mathrm{LM}) \\
& \Rightarrow \mathrm{LM}=3000(\sqrt{3}-1) \\
& \therefore \mathrm{BC}=3000(\sqrt{3}-1)
\end{aligned}
$
Now , time taken to travel distance $\mathrm{BC}=15$ seconds
$
\therefore \text { Speed of the aeroplane }=\frac{\text { Distance }}{\text { Time }}=\frac{3000(\sqrt{3}-1)}{15}=200 \times 0.732=146.4
$
Thus, the speed of the aeroplane is $146.4 \mathrm{~m} / \mathrm{sec}$
$
=146.4 \times \frac{\frac{1}{1000}}{\frac{1}{3600}} \mathrm{~km} / \mathrm{hr}=146.4 \times 3.6 \mathrm{~km} / \mathrm{hr}=527.04 \mathrm{~km} / \mathrm{hr}
$ View full question & answer→Question 224 Marks
A man on the top of a tower observes a truck at an angle of depression $\propto$ where $\propto=\frac{1}{\sqrt{5}}$ and sees that it is moving towards the base of the tower. Ten minutes later, the angle of depression of the truck is found to $\beta=\sqrt{5}$. Assuming that the truck moves at a uniform speed, determine how much more ti me it will take to each the base of the tower?
Answer
In the figure, $A B$ is the tower. $A$ is the position of the man. $C$ and $D$ are the two positions of the truck.
Let the speed of the truck be $x \mathrm{~m} / \mathrm{sec}$
Distance CD $=$ Speed $\times$ time $=600 x$
In right triangle $A B C$,
$
\tan \alpha=\frac{\mathrm{h}}{\mathrm{BC}}
$
It is a given that $\tan \alpha=\frac{1}{\sqrt{5}}$
$
\mathrm{BC}=\mathrm{h} \sqrt{5}
$
In right triangle $A B D$,
$
\tan \beta=\frac{\mathrm{h}}{\mathrm{BD}}
$
It is given that $\tan \beta=\sqrt{5}$
$
\mathrm{h}=\sqrt{5} \mathrm{BD}
$
Now, $C D=B C-B D$
$
\begin{aligned}
& 600 x=5 B D-B D \\
& B D=150 x
\end{aligned}
$
Time taken $=\frac{150 \mathrm{x}}{\mathrm{x}}=150$ seconds
Thus, the time taken by the truck to reach the tower is $150 \mathrm{sec}=2 \mathrm{~min} 30 \mathrm{sec}$. View full question & answer→Question 234 Marks
A man standing on a cliff observes a ship at an angle of depression of the ship is 30°, approaching the shore just beneath him. Three minutes later, the angle of depression of the ship is 60°. How soon will it reach the shore?
Answer
Let $A B$ be the tower.
Initial position of ship is $C$, which changes to $D$ after 3 minutes.
$\ln \triangle \mathrm{ADB}$
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{DB}}=\tan 60^{\circ} \\
& \frac{\mathrm{AB}}{\mathrm{DB}}=\sqrt{3}
\end{aligned}
$
$
\mathrm{DB}=\frac{\mathrm{AB}}{\sqrt{3}}
$
$
\begin{aligned}
& \operatorname{In} \triangle \mathrm{ABC} \\
& \frac{\mathrm{AB}}{\mathrm{BC}}=\tan 30^{\circ} \\
& \frac{\mathrm{AB}}{\mathrm{BD}+\mathrm{DC}}=\frac{1}{\sqrt{3}} \\
& \mathrm{AB} \sqrt{3}=\mathrm{BD}+\mathrm{DC} \\
& \mathrm{AB} \sqrt{3}=\frac{\mathrm{AB}}{\sqrt{3}}+\mathrm{DC} \\
& \mathrm{DC}=\mathrm{AB} \sqrt{3}-\frac{\mathrm{AB}}{\sqrt{3}}=\mathrm{AB}\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) \\
& =\frac{2 \mathrm{AB}}{\sqrt{3}}
\end{aligned}
$
Time taken by car to travel DC distance $\left(\right.$ i.e. $\left.\frac{2 \mathrm{AB}}{\sqrt{3}}\right)=3$ minutes
Time taken by car to travel DB distance (i.e. $\frac{\mathrm{AB}}{\sqrt{3}}$ )
$
=\frac{3}{\frac{2 A B}{\sqrt{3}}} \times \frac{\mathrm{AB}}{\sqrt{3}}=\frac{3}{2}=1 \mathrm{~min} 30 \text { secs }
$
Thus the total time taken is 3 minutes +1 minute 30 seconds $=$ minutes 30 seconds. View full question & answer→Question 244 Marks
The angle of depression of a boat moving towards a diff is 30°. Three minutes later the angle of depression of the boat is 60°. Assuming that the boat is sailing at a uniform speed, determine the time it will take to reach the shore. Also, find the speed of the boat in m/second if the cliff is 450m high.
Answer
Let $A B$ be the diff. Then, $A B=450 \mathrm{~m}$
Initial position of boat is $C$, which changes to $D$ after 3 minutes.
$
\begin{aligned}
& \operatorname{In} \triangle \mathrm{ADB} \\
& \frac{\mathrm{AB}}{\mathrm{DB}}=\tan 60^{\circ} \\
& \frac{450}{\mathrm{DB}}=\sqrt{3} \\
& \mathrm{DB}=\frac{450}{\sqrt{3}}
\end{aligned}
$
$
\begin{aligned}
& \ln \triangle \mathrm{ABC} \\
& \frac{\mathrm{AB}}{\mathrm{BC}}=\tan 30^{\circ} \\
& \frac{450}{\mathrm{BD}+\mathrm{DC}}=\frac{1}{\sqrt{3}} \\
& 450 \sqrt{3}=\mathrm{BD}+\mathrm{DC} \\
& 450 \sqrt{3}=\frac{450}{\sqrt{3}}+\mathrm{DC} \\
& \mathrm{DC}=450 \sqrt{3}-\frac{450}{\sqrt{3}}=450\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) \\
& =\frac{900}{\sqrt{3}}=\frac{900}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=300 \sqrt{3}
\end{aligned}
$
Time taken by car to travel DC distance (i.e., $300 \sqrt{3}$ ) $=3$ minutes
Time taken by car to travel DB distance (i.e. $\frac{450}{\sqrt{3}}$ )
$
=\frac{3}{300 \sqrt{3}} \times \frac{450}{\sqrt{3}}=\frac{450}{300}=1.5
$
Thus, the time it will take to reach the shore is $1 \mathrm{~min} 30$ secs.
$
\begin{aligned}
& \text { Speed of the boat }=\frac{\text { Distance }}{\text { Time }} \\
& =\frac{300 \sqrt{3}}{3}=100 \sqrt{3}=100 \times 1.732=173.2 \mathrm{~m} / \mathrm{min} \\
& =\frac{173.2}{60} \mathrm{~m} / \mathrm{sec}=2.9 \mathrm{~m} / \mathrm{sec}
\end{aligned}
$ View full question & answer→Question 254 Marks
A man in a boat rowing away from a lighthouse 180 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° and 30°. Find the speed of the boat.
Answer
Let $A B$ be the lighthouse.
Initial position of boat is $C$, which changes to $D$ after 2 minutes.
In $\triangle \mathrm{ADB}$
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{DB}}=\tan 60^{\circ} \\
& \frac{180}{\mathrm{x}}=\sqrt{3} \\
& \mathrm{x}=\frac{180}{\sqrt{3}}
\end{aligned}
$
$
\begin{aligned}
& \operatorname{In} \triangle \mathrm{ABC} \\
& \frac{\mathrm{AB}}{\mathrm{BC}}=\tan 30^{\circ} \\
& \frac{180}{\mathrm{x}+\mathrm{y}}=\frac{1}{\sqrt{3}} \\
& 180 \sqrt{3}=\mathrm{x}+\mathrm{y} \\
& 180 \sqrt{3}=\frac{180}{\sqrt{3}}+\mathrm{y} \\
& \mathrm{y}=180\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=180\left(\frac{2}{\sqrt{3}}\right)=\frac{360}{\sqrt{3}}
\end{aligned}
$
Time taken by car to travel DC distance $\left(\right.$ i.e. $\left.\frac{360}{\sqrt{3}}\right)=2$ minutes $=120$ seconds
Speed of the boat $=\frac{\text { Distance }}{\text { Time }}=\frac{\frac{360}{\sqrt{3}}}{120}=\frac{3}{\sqrt{3}}=\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{3 \sqrt{3}}{3}=\sqrt{3}=1.732$
Thus, the speed of the boat is $1.732 \mathrm{~m} / \mathrm{sec}$. View full question & answer→Question 264 Marks
The angle of elevation of a cloud from a point 60m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud.
Answer
Let $C$ be the cloud and $D$ be its reflection. Let the height of the cloud is $h$ metres .
$
\begin{aligned}
& B C=B D=h \\
& B Q=A P=60 m \\
& \therefore C Q=h-60 \text { and } D Q=h+60 \\
& \ln \Delta C Q P \\
& \frac{P Q}{C Q}=\cot 30^{\circ} \\
& \Rightarrow \frac{P Q}{h-60}=\sqrt{3} \\
& \Rightarrow P Q=\sqrt{3}(h-60) \text {....(i) }
\end{aligned}
$
In $\triangle \mathrm{DQP}$,
$
\begin{aligned}
& \frac{\mathrm{PQ}}{\mathrm{DQ}}=\cot 60^{\circ} \\
& \Rightarrow \frac{\mathrm{PQ}}{\mathrm{h}+60}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \mathrm{PQ}=\frac{1}{\sqrt{3}}(\mathrm{~h}+60)\ldots(ii)
\end{aligned}
$
From (i) and (ii),
$
\begin{aligned}
& \Rightarrow \sqrt{3}(h-60)=\frac{1}{\sqrt{3}}(h+60) \\
& \Rightarrow 3 h-180=h+60 \\
& \Rightarrow 2 h=240 \\
& \Rightarrow h=120
\end{aligned}
$
Thus, the height of the cloud is $120 \mathrm{~m}$. View full question & answer→Question 274 Marks
The angle of elevation of a stationary cloud from a point 25m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. What is the height of the cloud above the lake-level?
AnswerLet $C$ be the position of the cloud, I be the surface of the lake and $D$ be the reflection of the cloud. Let $C B=h$, then $O D=25+h$
$
\begin{aligned}
& \operatorname{In} \triangle A B C \\
& \tan 30^{\circ}=\frac{B C}{A B} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} \\
& \Rightarrow \sqrt{3} \mathrm{~h}=\mathrm{x}\ldots(1)
\end{aligned}
$
$
\begin{aligned}
& \ln \triangle \mathrm{ABD} \\
& \tan 60^{\circ}=\frac{\mathrm{BD}}{\mathrm{AB}}=\frac{25+25+\mathrm{h}}{\mathrm{x}} \\
& \sqrt{3} \mathrm{x}=50+\mathrm{h} \ldots(2)
\end{aligned}
$
From (1) and (2),
$
\sqrt{3}(\sqrt{3} \mathrm{~h})=50+\mathrm{h}
$
$
\begin{aligned}
& 2 \mathrm{~h}=50 \\
& \mathrm{~h}=25
\end{aligned}
$
Thus, the height of the cloud above the lake-level is $\mathrm{OC}=50 \mathrm{~m}$. View full question & answer→Question 284 Marks
A parachutist is descending vertically and makes angles of elevation of 45° and 60° from two observing points 100 m apart to his right. Find the height from which he falls and the distance of the point where he falls on the ground from the nearest observation pcint.
Answer
Let $\mathrm{A}$ be the position of the parachutist and $\mathrm{C}$ and $\mathrm{D}$ be the two observation points.
In $\triangle \mathrm{ABC}$,
$
\begin{aligned}
& \tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}} \\
& \Rightarrow \sqrt{3}=\frac{\mathrm{h}}{\mathrm{x}} \\
& \Rightarrow \mathrm{h}=\sqrt{3 \mathrm{x}}
\end{aligned}
$
In $\triangle \mathrm{ABD}$,
$
\begin{aligned}
& \tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{BD}} \\
& \Rightarrow 1=\left(\frac{\mathrm{h}}{\mathrm{x}+100}\right) \\
& \Rightarrow \mathrm{x}+100=\mathrm{h} \\
& \Rightarrow \mathrm{x}+100=\sqrt{3} \mathrm{x} \\
& \Rightarrow \mathrm{x}(\sqrt{3}-1)=100 \\
& \Rightarrow \mathrm{x}=100 \times \frac{1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\
& \Rightarrow \mathrm{x}=100 \times \frac{(\sqrt{3}+1)}{3-1}=50(\sqrt{3}+1)=50 \times 2.732=136.6
\end{aligned}
$
Thus, the distance of the point where he falls on the ground from the nearest observation point (C) is $136.6 \mathrm{~m}$.
Height from which the parachutist fall
$
=\mathrm{h}=\sqrt{3} \mathrm{x}=1.732 \times 136.6=236.59 \approx 236.6 \mathrm{~m} .
$ View full question & answer→Question 294 Marks
An aeroplane when flying at a height of 4km from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the aeroplanes at that instant.
Answer
Let points $A$ and $D$ represent the position of the aeroplanes. Aeroplane A is flying $4 \mathrm{~km}=4000 \mathrm{~m}$ above the ground.
$
\angle \mathrm{ACB}=60^{\circ}, \angle \mathrm{DCB}=45^{\circ}
$
In $\triangle \mathrm{ABC}$,
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{BC}}=\tan 60^{\circ} \\
& \Rightarrow \mathrm{BC}=\frac{4000}{\sqrt{3}}
\end{aligned}
$
$
\begin{aligned}
& \operatorname{In} \triangle \mathrm{DCB} \\
& \frac{\mathrm{DB}}{\mathrm{BC}}=\tan 45^{\circ} \\
& \Rightarrow \mathrm{DB}=\mathrm{BC}=\frac{4000}{\sqrt{3}} \\
& \therefore \mathrm{AD}=\mathrm{AB}-\mathrm{BD} \\
& =4000-\frac{4000}{\sqrt{3}}=4000\left(1-\frac{1}{\sqrt{3}}\right)=4000 \times \frac{\sqrt{3}-1}{\sqrt{3}}=4000 \times \frac{0.732}{1.732}=1690.53 \\
& =\mathrm{h}=\sqrt{3} \mathrm{x}=1.732 \times 136.6=236.59 \approx 236.6 \mathrm{~m}
\end{aligned}
$ View full question & answer→Question 304 Marks
A man on the top of a tower observes that a car is moving directly at a uniform speed towards it. If it takes 720 seconds for the angle of depression to change from 30° to 45°, how soon will the car reach the observation tower?
Answer
Let $A B$ be the tower .
Initial position of car is C, which changes to D after 720 seconds.
$\ln \triangle \mathrm{ADB}$
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{DB}}=\tan 45^{\circ} \\
& \frac{\mathrm{AB}}{\mathrm{DB}}=1 \\
& \mathrm{DB}=\mathrm{AB}
\end{aligned}
$
$\ln \triangle \mathrm{ABC}$
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{BC}}=\tan 30^{\circ} \\
& \frac{\mathrm{AB}}{\mathrm{BD}+\mathrm{DC}}=\frac{1}{\sqrt{3}} \\
& \mathrm{AB} \sqrt{3}=\mathrm{BD}+\mathrm{DC} \\
& \mathrm{AB} \sqrt{3}=\mathrm{AB}+\mathrm{DC} \\
& \mathrm{DC}=\mathrm{AB} \sqrt{3}-\mathrm{AB}=\mathrm{AB}(\sqrt{3}-1)
\end{aligned}
$
Time taken by car to travel DC distance (i.e $\mathrm{AB}(\sqrt{3}-1))=720$ seconds Time taken by car to travel DB distance (i.e. AB)
$
\begin{aligned}
& =\frac{720}{\mathrm{AB}(\sqrt{3}-1)} \times \mathrm{AB}=\frac{720}{(\sqrt{3}-1)} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\
& =\frac{720(\sqrt{3}+1)}{2}=360(\sqrt{3}+1)=360 \times 2.732=983.52
\end{aligned}
$
Thus, the required time taken is 983.52 seconds $=984$ seconds $=16$ mins 24 secs. View full question & answer→Question 314 Marks
The angle of elevation of a tower from a point in line with its base is $45^{\circ}$. On moving $20 \mathrm{~m}$ towards the tower, the angle of elevation changes to $60^{\circ}$. Find the height of the tower.
Answer
Let the height of the tower (AB) be $h$.
$
\begin{aligned}
& \ln \triangle \mathrm{ABC} \\
& \tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}} \\
& \Rightarrow \sqrt{3}=\frac{\mathrm{h}}{\mathrm{BC}} \\
& \Rightarrow \mathrm{BC}=\frac{\mathrm{h}}{\sqrt{3}}
\end{aligned}
$
In $\triangle A B D$,
$
\begin{aligned}
& \tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{BD}} \\
& \Rightarrow 1=\frac{\mathrm{h}}{\mathrm{BC}+20} \\
& \Rightarrow \mathrm{BC}+20=\mathrm{h} \\
& \Rightarrow \frac{\mathrm{h}}{\sqrt{3}}+20=\mathrm{h} \\
& \Rightarrow \mathrm{h}\left(1-\frac{1}{\sqrt{3}}\right)=20 \\
& \Rightarrow \mathrm{h}=20 \times \frac{\sqrt{3}}{\sqrt{3}-1} \\
& \Rightarrow \mathrm{h}=20 \times \frac{\sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\
& =20 \times \frac{\sqrt{3}(\sqrt{3}+1)}{3-1}=10(3+\sqrt{3})=10 \times 4.732=47.32
\end{aligned}
$
Thus, the height of the tower is $47.32 \mathrm{~m}$. View full question & answer→Question 324 Marks
The length of the shadow of a statue increases by 8m, when the latitude of the sun changes from 45° to 30°. Calculate the height of the tower.
Answer
Let the height of the statue $(A B)$ be $h$.
In $\triangle \mathrm{ABC}$,
$\tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}$
$
\Rightarrow \mathrm{BC}=\mathrm{h}
$
In $\triangle \mathrm{ABD}$,
$
\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BD}}
$
$
\frac{1}{\sqrt{3}}=\frac{\mathrm{AB}}{\mathrm{BC}+\mathrm{CD}}
$
$h+8=\sqrt{3} h \quad(\because B C=h)$
$
\Rightarrow \mathrm{h}(\sqrt{3}-1)=8
$
$
\Rightarrow \mathrm{h}=\frac{8}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}
$
$\Rightarrow \mathrm{h}=\frac{8(\sqrt{3}+1)}{2}=4(\sqrt{3}+1)=4 \times 2.732=10.928 \approx 10.93$
Thus, the height of the tower is $10.93 \mathrm{~m}$. View full question & answer→Question 334 Marks
Of the two trees are on either side of a river, one of them is 50m high. From the top of this tree the angles of depression of the top and the foot of the other tree are 30° and 60° respectively. Find the width of the river and the height of the other tree.
Answer
Let $A B$ and $C D$ be the two trees.
$
\begin{aligned}
& \ln \triangle \mathrm{AEC}, \\
& \tan 30^{\circ}=\frac{\mathrm{EA}}{\mathrm{EC}} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{EA}}{\mathrm{EC}} \\
& \Rightarrow \mathrm{EC}=\sqrt{3} \mathrm{EA} \ldots .(1) \\
& \ln \triangle \mathrm{ABD}, \\
& \tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{BD}}=\sqrt{3} \\
& \Rightarrow \frac{50}{\mathrm{BD}}=\sqrt{3} \\
& \Rightarrow \mathrm{BD}=\frac{50}{\sqrt{3}}
\end{aligned}
$
In $\triangle \mathrm{ABD}$,
$\tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{BD}}=\sqrt{3}$
$\Rightarrow \frac{50}{\mathrm{BD}}=\sqrt{3}$
$\Rightarrow \mathrm{BD}=\frac{50}{\sqrt{3}}$
Thus, the width of the river is $\mathrm{BD}=\frac{50}{\sqrt{3}}=28.8 \mathrm{~m}$.
From (1),
$
\mathrm{EA}=\frac{\mathrm{EC}}{\sqrt{3}}=\frac{\mathrm{BD}}{\sqrt{3}}=\frac{50}{3}=16.67
$
Height of the other tree $=C D=50-E A=50-16.67=33.33 \mathrm{~m}$ View full question & answer→Question 344 Marks
A man on the deck of a ship is 10 m above the water level. He observes that the angle of elevation of the top of a diff is 45° and the angle of depression of the base is 30°. Find the distance of the diff from the ship and the height of the cliff.
AnswerLet $\mathrm{B}$ be the position of the man, $\mathrm{D}$ the base of the cliff, $\mathrm{x}$ be the distance of cliff from the ship and $\mathrm{h}$ +10 be the height of the hill. $\angle \mathrm{ABC}=45^{\circ}$ and $\angle \mathrm{DBC}=30^{\circ}$
Therefore, $\angle \mathrm{BDE}=30^{\circ}$
$
\begin{aligned}
& \operatorname{In} \triangle \mathrm{ABC}, \\
& \tan 45^{\circ}=\frac{\mathrm{AC}}{\mathrm{BC}} \\
& \Rightarrow \frac{\mathrm{h}}{\mathrm{x}}=1 \\
& \Rightarrow \mathrm{h}=\mathrm{x} \text { (1) } \\
& \operatorname{In} \Delta \mathrm{BED}, \\
& \tan 30^{\circ}=\frac{\mathrm{BE}}{\mathrm{ED}} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{10}{\mathrm{x}} \\
& \Rightarrow \mathrm{x}=10 \sqrt{3}=10 \times 1.732=17.32
\end{aligned}
$
Thus, the distance of the diff from the ship is $17.32 \mathrm{~m}$.
From (1),
$
h=x=17.32
$
$\therefore$ Height of the diff $=17.32+10=27.32$
Thus, the height of the diff is $27.32 \mathrm{~m}$. View full question & answer→Question 354 Marks
The angle of elevation of a tower from a point $200 \mathrm{~m}$ from its base is $\theta$, when $\tan \theta=\frac{2}{5}$. The angle of elevation of this tower from a point $120 \mathrm{~m}$ from its base is $\Phi$. Calculate the height of tower and the value of $\Phi$.
Answer
Let OT be the tower .
A and B be the two points from where the angle of elevation to the top of the tower is measured.
In $\triangle A O T$,
$
\begin{aligned}
& \frac{\mathrm{OT}}{\mathrm{OA}}=\tan \theta \\
& \Rightarrow \frac{\mathrm{h}}{200}=\frac{2}{5} \\
& \Rightarrow \mathrm{h}=80 \ldots(1)
\end{aligned}
$
Thus, the height of the tower is $80 \mathrm{~m}$.
In $\triangle \mathrm{BOT}$,
$
\begin{aligned}
& \frac{\mathrm{OT}}{\mathrm{OB}}=\tan \Phi \\
& \Rightarrow \frac{\mathrm{h}}{120}=\tan \Phi \\
& \Rightarrow \frac{80}{120}=\tan \Phi \quad \text { [Using (1)] } \\
& \Rightarrow \frac{2}{3}=\tan \Phi
\end{aligned}
$
From the table, we get $\Phi=34^{\circ}$. View full question & answer→Question 364 Marks
From the top of a 60m high building the angles of depression of the top and bottom of a lamp post are 30° and 60° respectively. Find the distance on the ground between the building and the lamp post and the difference in their heights.
Answer
Let $A B$ be the building. Then, $A B=60 \mathrm{~m}$.
Let the height of the lamp post (CD) be h.
Let the distance between the building and the lamp post be $x$. In $\triangle \mathrm{ACB}$,
$
\begin{aligned}
& \tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}} \\
& \therefore \sqrt{3}=\frac{60}{X} \\
& \therefore X=\frac{60}{\sqrt{3}}=20 \sqrt{3}=20 \times 1.732=34.64\ldots(1)
\end{aligned}
$
Thus, the distance between the building and the lamp post is $34.64 \mathrm{~m}$ In $\triangle \mathrm{ADE}$,
$
\tan 30^{\circ}=\frac{\mathrm{AE}}{\mathrm{DE}}
$
$
\therefore \frac{1}{\sqrt{3}}=\frac{60-h}{X}
$
$
\therefore X=\sqrt{3}(60-h)\ldots(2)
$
From (1) and (2):
$
\begin{aligned}
& \sqrt{3}(60-h)=20 \sqrt{3} \\
& 60-h=20 \\
& h=40
\end{aligned}
$ View full question & answer→Question 374 Marks
Two boats approaching a light house in mid sea from opposite directions observe the angle of elevation of the top of the light house as 30° and 45° respectively. If the distance between the two boats is 150m, find the height of the light house.
Answer
Let the position of the two boats be at points $A$ and $C$. Let BD be the lighthouse of height $h$. Let $A D=x$. Then, $C D=150-x$ In $\triangle B A D$,
$
\begin{aligned}
& \tan 45^{\circ}=\frac{\mathrm{BD}}{\mathrm{AD}} \\
& \Rightarrow 1=\frac{\mathrm{h}}{\mathrm{X}} \\
& \Rightarrow \mathrm{h}=\mathrm{X} \ldots(1) \\
& \ln \triangle \mathrm{BDC}, \\
& \tan 30^{\circ}=\frac{\mathrm{BD}}{\mathrm{DC}} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{150-X}
\end{aligned}
$
$
\Rightarrow 150-\mathrm{X}=\sqrt{3} \mathrm{~h}\ldots(2)
$
From (1) and (2),
$
\begin{aligned}
& 150-\mathrm{h}=\sqrt{3} \mathrm{~h} \\
& 150=(\sqrt{3}+1) \mathrm{h}
\end{aligned}
$
$\mathrm{h}=\frac{150}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{150(\sqrt{3}-1)}{3-1}$
$=75(\sqrt{3}-1)$
$=75 \times 0.732=54.9$
Thus, the height of the light house is $54.9 \mathrm{~m}$. View full question & answer→Question 384 Marks
An observer point for ships moving in the sea 500m above the sea level. The person manning this point observes the angle of depression of twc boats as 45° and 30°. Find the distance between the boats when they are on the same side of the observation point and when they are on opposite sides of the observation point.
AnswerCase 1: When the boats are on same side of the observation point.
Let the position of the two ships be $\mathrm{C}$ and $\mathrm{D}$. Let $\mathrm{A}$ be the point of observation.
$
\begin{aligned}
& A B=500 m \\
& \text { In } \triangle B A C
\end{aligned}
$
$
\begin{aligned}
& \tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}} \\
& \Rightarrow 1=\frac{500}{\mathrm{BC}} \\
& \Rightarrow \mathrm{BC}=500 \ldots .(1)
\end{aligned}
$
In $\triangle A B D$,
$
\begin{aligned}
& \tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BD}} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{500}{\mathrm{BD}} \\
& \Rightarrow \mathrm{BD}=500 \sqrt{3}\ldots(2)
\end{aligned}
$
From (1) and (2),
Thus, in this case, the distance between the boats is 366 m.
Case 2: When the boats are on different side of the observation point.
Let the position of the two ships be A and C. Let B be the point of observation.
In $\triangle B A D$,
$\tan 45^{\circ}=\frac{\mathrm{BD}}{\mathrm{AD}}$
$
\Rightarrow 1=\frac{500}{\mathrm{AD}}
$
$
\begin{aligned}
& \Rightarrow \mathrm{AD}=500 \ldots(1) \\
& \operatorname{In} \triangle \mathrm{BDC} \\
& \tan 30^{\circ}=\frac{\mathrm{BD}}{\mathrm{DC}} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{500}{\mathrm{DC}} \\
& \Rightarrow \mathrm{DC}=500 \sqrt{3} \ldots .(2)
\end{aligned}
$
From (1) and (2),
$
\mathrm{AC}=\mathrm{AD}+\mathrm{DC}=500(1+\sqrt{3})=500 \times 2.732=1366
$
Thus, in this case, the distance between the boats is $1366 \mathrm{~m}$. View full question & answer→Question 394 Marks
An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. Find the width of the river.
Answer
Let $A D$ be the height of the aeroplane and $B C=x \mathrm{~m}$ be the width of the nver.
Given: $A D=200 \mathrm{~m}$
In $\triangle \mathrm{ABD}$
$
\begin{aligned}
& \frac{A D}{B D}=\tan 45^{\circ} \\
& \Rightarrow \frac{A D}{B D}=1 \\
& \Rightarrow A D=B D \\
& \Rightarrow B D=200 m(\because A D=200 m)
\end{aligned}
$
Now,
In $\triangle \mathrm{ACD}$
$
\begin{aligned}
& \frac{\mathrm{AC}}{\mathrm{CD}}=\tan 60^{\circ} \\
& \Rightarrow \frac{\mathrm{AC}}{\mathrm{CD}}=\sqrt{3} \\
& \Rightarrow \mathrm{CD}=\frac{\mathrm{AC}}{\sqrt{3}}=\frac{200}{\sqrt{3}} \\
& \Rightarrow \mathrm{BC}=\mathrm{BD}+\mathrm{CD}=200+\frac{200}{\sqrt{3}}=200+115.47 \\
& \Rightarrow \mathrm{BC}=315.4 \mathrm{~m}
\end{aligned}
$
Thus, the width of the river is $315.4 \mathrm{~m}$. View full question & answer→Question 404 Marks
The angles of depression of two cars on a straight road as observed from the top of a 42m high building are 60° and 75° respectively. Find the distance between the cars if they are on opposite sides of the building.
Answer
$
\text { Let the position of the two cars be } \mathrm{A} \text { and } \mathrm{C} \text {. Let } \mathrm{BO} \text { be the building of height } 42 \mathrm{~m} \text {. }
$
$
\begin{aligned}
& \tan 75^{\circ}=\frac{\mathrm{BD}}{\mathrm{AD}} \\
& \Rightarrow 3.7321=\frac{42}{\mathrm{AD}} \\
& \Rightarrow \mathrm{AD}=\frac{42}{3.7321} \\
& \Rightarrow \mathrm{AD}=11.25\ldots(1)
\end{aligned}
$
$
\begin{aligned}
& \operatorname{In} \triangle B D C \\
& \tan 60^{\circ}=\frac{B D}{D C} \\
& \Rightarrow \sqrt{3}=\frac{42}{D C} \\
& \Rightarrow D C=\frac{42}{1.732}=24.25\ldots(2) \\
& \therefore A C=A D+D C=11.25 \mathrm{~m}+24.25 \mathrm{~m}=35.5 \mathrm{~m}
\end{aligned}
$
Thus, the distance between the cars is $67.63 \mathrm{~m}$. View full question & answer→Question 414 Marks
Two persons standing on opposite sides of a tower observe the angles of elevation of the top of the tower to be 60° and 50° respectively. Find the distance between them, if the height of the tower is 80m.
Answer
Let the position of the two persons be A and C. Let BD be the tower of height $80 \mathrm{~m}$. In $\triangle \mathrm{BAD}$,
$
\begin{aligned}
& \tan 60^{\circ}=\frac{\mathrm{BD}}{\mathrm{AD}} \\
& \Rightarrow \sqrt{3}=\frac{80}{\mathrm{AD}} \\
& \Rightarrow \mathrm{AD}=\frac{80}{\sqrt{3}} \\
& \Rightarrow \mathrm{AD}=\frac{80 \sqrt{3}}{3}=\frac{80 \times 1.732}{3}=46.19\ldots(1)
\end{aligned}
$
$\ln \triangle B D C$
$\tan 50^{\circ}=\frac{\mathrm{BD}}{\mathrm{DC}}$
$
\begin{aligned}
& \Rightarrow 1.1918=\frac{80}{\mathrm{DC}} \\
& \Rightarrow \mathrm{DC}=\frac{80}{1.1918}=67.13\ldots(2)
\end{aligned}
$
$
\therefore A C=A D+D C=46.19 m+67.13 m=113.32 m
$
Thus, the horizontal distance between the two persons is $113.32 \mathrm{~m}$. View full question & answer→Question 424 Marks
A boy is standing on the ground and flying a kite with 100m of sting at an elevation of 30°. Another boy is standing on the roof of a 10m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.
Answer
Let $C$ be the position of the first boy and $D$ be the position of the second boy who is standing on the roof of a $10 \mathrm{~m}$ high building.
Let $B$ be the position of the kites of both the boys.
Let $A B=h$ and $C A=x$.
$\ln \triangle \mathrm{ABC}$
$
\begin{aligned}
& \sin 30^{\circ}=\frac{\mathrm{h}}{100} \\
& \Rightarrow \frac{1}{2}=\frac{\mathrm{h}}{100} \\
& \Rightarrow \mathrm{h}=50 \ldots \text { (1) } \\
& \ln \triangle B D E_{,} \\
& \tan 45^{\circ}=\frac{B E}{B D} \\
& \Rightarrow 1=\frac{\mathrm{h}-10}{\mathrm{X}} \\
& \Rightarrow \mathrm{X}=(\mathrm{h}-10) \ldots(2)
\end{aligned}
$
$
\begin{aligned}
& \text { From (1) and (2), } \\
& x=50-10=40
\end{aligned}
$
In $\triangle \mathrm{BDE}^{\prime}$
$
\begin{aligned}
& \sin 45^{\circ}=\frac{\mathrm{BE}}{\mathrm{BD}} \\
& \Rightarrow \frac{1}{\sqrt{2}}=\frac{\mathrm{h}-10}{\mathrm{BC}}
\end{aligned}
$
$
\Rightarrow \mathrm{BC}=\sqrt{2}(50-10)=40 \sqrt{2}
$
Thus, the required length of the string that the second boy must have $40 \sqrt{2} \mathrm{~m}$ View full question & answer→Question 434 Marks
A boy is 1.54 m tall. Standing at a distance of 3m in front of a 4.54 m high wall he can just manage to see the sun. Find the angle of elevation of the sun.
Answer
Let the position of the boy be at point $\mathrm{T}$ and $\mathrm{P}$ be the position of the sun.
$
\begin{aligned}
& \mathrm{BR}=\mathrm{TQ}=3 \mathrm{~m} \\
& \mathrm{PQ}=4.54 \mathrm{~m} \\
& \mathrm{BT}=1.54 \mathrm{~m} \\
& \therefore \mathrm{PR}=4.54 \mathrm{~m}-1.54 \mathrm{~m}=3 \mathrm{~m}
\end{aligned}
$
In $\triangle \mathrm{PRB}$
$
\begin{aligned}
& \frac{\mathrm{PR}}{\mathrm{BR}}=\tan \theta \\
& \frac{3}{3}=\tan \theta \\
& \tan \theta=1
\end{aligned}
$
We know that $\tan 45^{\circ}=1$.
Thus, the angle of elevation is $\theta=45^{\circ}$. View full question & answer→Question 444 Marks
A 1.4m tall boy stands at a point 50m away from a tower and observes the angle of elevation of the top of the tower to be 60°. Find the height of the tower.
Answer
Let the position of the boy be at point $T$.
$
\begin{aligned}
& \mathrm{BR}=\mathrm{TQ}=50 \mathrm{~m} \\
& \mathrm{RQ}=\mathrm{BT}=1.5 \mathrm{~m}
\end{aligned}
$
In $\triangle \mathrm{PRB}$
$
\begin{aligned}
& \frac{\mathrm{PR}}{\mathrm{BR}}=\tan 60^{\circ} \\
& \frac{\mathrm{PR}}{50}=\sqrt{3} \\
& \mathrm{PR}=50 \sqrt{3}
\end{aligned}
$
Height of the tower
$
=\mathrm{PQ}=\mathrm{PR}+\mathrm{RQ}=50 \sqrt{3}+1.5=50 \times 1.732+1.5=86.6+1.5=88.1 \mathrm{~m}
$
Thus, the height of the tower is approximately $88 \mathrm{~m}$. View full question & answer→Question 454 Marks
The angle of elevation of the top of an unfinished tower at a point 150 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60°?
Answer
Let $B C$ be the length of unfinished tower. Let the tower be raised upto point $A$ so that the angle of elevation at point $\mathrm{A}$ is $60^{\circ}$. $\mathrm{D}$ is the point on ground from where elevation angles are measured.
In $\triangle B C D$
$
\begin{aligned}
& \frac{\mathrm{BC}}{\mathrm{CD}}=\tan 30^{\circ} \\
& \frac{\mathrm{BC}}{\mathrm{CD}}=\frac{1}{\sqrt{3}} \\
& \mathrm{BC}=\frac{\mathrm{CD}}{\sqrt{3}} \\
& \mathrm{BC}=\frac{150}{\sqrt{3}}
\end{aligned}
$
In $\triangle A C D$
$
\begin{aligned}
& \frac{\mathrm{AB}+\mathrm{BC}}{\mathrm{CD}}=\tan 60^{\circ} \\
& \Rightarrow \frac{\mathrm{AB}+\mathrm{BC}}{\mathrm{CD}}=\sqrt{3} \\
& \Rightarrow \frac{\mathrm{AB}+\frac{150}{\sqrt{3}}}{150}=\sqrt{3} \ldots(\text { Using (i) )) }
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \mathrm{AB}=150 \sqrt{3}-\frac{150}{\sqrt{3}}=\frac{150 \times 3-150}{\sqrt{3}} \\
& \Rightarrow \mathrm{AB}=\frac{300}{\sqrt{3}}=\frac{300}{1.732}=173.2 \mathrm{~m}
\end{aligned}
$
Thus, the required height is $300 \mathrm{~m}$. View full question & answer→Question 464 Marks
A vertical tow er stands on a horizontal plane and is surmounted by a flagstaff of height 7m. From a point on the plane the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the ft agstaff is 45°. Find the height of the tower.
Answer
Let $A B$ be the flagstaff, $B C$ be the tower and $D$ be the point on ground from where elevation angles are measured.
$
\begin{aligned}
& \ln \triangle \mathrm{BCD} \\
& \frac{\mathrm{BC}}{\mathrm{CD}}=\tan 30^{\circ} \\
& \frac{\mathrm{BC}}{\mathrm{CD}}=\frac{1}{\sqrt{3}} \\
& \sqrt{3} \mathrm{BC}=\mathrm{CD} \\
& \ln \triangle \mathrm{ACD} \\
& \frac{\mathrm{AB}+\mathrm{BC}}{\mathrm{CD}}=\tan 45^{\circ} \\
& \frac{\mathrm{AB}+\mathrm{BC}}{\sqrt{3} \mathrm{BC}}=1 \\
& 7+\mathrm{BC}=\sqrt{3} \mathrm{BC}
\end{aligned}
$
$
\begin{aligned}
& \mathrm{BC}(\sqrt{3}-1)=7 \\
& \mathrm{BC}=\frac{(7)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\
& =\frac{7(\sqrt{3}+1)}{(\sqrt{3})^2-(1)^2} \\
& =\frac{7(\sqrt{3}+1)}{2}=3.5(\sqrt{3}+1)=3.5 \times 2.732=9.562
\end{aligned}
$
Thus, the height of the tower is $9.562 \mathrm{~m}=9.56 \mathrm{~m}$. View full question & answer→Question 474 Marks
A flagstaff stands on a vertical pole. The angles of elevation of the top and the bottom of the flagstaff from a point on the ground are found to be 60° and 30° respectively. If the height of the pole is 2.5m. Find the height of the flagstaff.
Answer
Let $A B$ be the building and $C D$ be the tower.
$\ln \triangle \mathrm{ABD}$
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{BD}}=\tan 30^{\circ} \\
& \frac{12}{\mathrm{BD}}=\frac{1}{\sqrt{3}} \\
& \mathrm{BD}=12 \sqrt{3}
\end{aligned}
$
$\ln \triangle \mathrm{ACE}$
$
\begin{aligned}
& \mathrm{AE}=\mathrm{BD}=12 \sqrt{3} \\
& \frac{\mathrm{CE}}{\mathrm{AE}}=\tan 60^{\circ} \\
& \frac{\mathrm{CE}}{12 \sqrt{3}}=\sqrt{3} \\
& \mathrm{CE}=12 \sqrt{3} \times \sqrt{3}=12 \times 3=36 \\
& \mathrm{CD}=\mathrm{CE}+\mathrm{ED}=36+12=48
\end{aligned}
$
So, height of the tower is $48 \mathrm{~m}$ and its distance from the building is $12 \sqrt{3} \mathrm{~m}=12 \times 1.732 \mathrm{~m}=20.78$ m(approximately) View full question & answer→Question 484 Marks
The angles of depression and elevation of the top of a 12m high building from the top and the bottom of a tower are 60° and 30° respectively. Find the height of the tower, and its distance from the building.
Answer
Let $A B$ be the building and $C D$ be the tower.
$\ln \triangle \mathrm{ABD}$
$
\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{BD}}=\tan 30^{\circ} \\
& \frac{12}{\mathrm{BD}}=\frac{1}{\sqrt{3}} \\
& \mathrm{BD}=12 \sqrt{3}
\end{aligned}
$
$\ln \triangle \mathrm{ACE}$
$
\begin{aligned}
& \mathrm{AE}=\mathrm{BD}=12 \sqrt{3} \\
& \frac{\mathrm{CE}}{\mathrm{AE}}=\tan 60^{\circ} \\
& \frac{\mathrm{CE}}{12 \sqrt{3}}=\sqrt{3} \\
& \mathrm{CE}=12 \sqrt{3} \times \sqrt{3}=12 \times 3=36 \\
& \mathrm{CD}=\mathrm{CE}+\mathrm{ED}=36+12=48
\end{aligned}
$
So, height of the tower is $48 \mathrm{~m}$ and its distance from the building is $12 \sqrt{3} \mathrm{~m}=12 \times 1.732 \mathrm{~m}=20.78$ m(approximately) View full question & answer→Question 494 Marks
The top of a palm tree having been broken by the wind struck the ground at an angle of 60° at a distance of 9m from the foot of the tree. Find the original height of the palm tree.
Answer
Let $\mathrm{AC}$ was original tree. It was broken into two parts. The broken part $\mathrm{A}^{\prime} \mathrm{B}$ is making $60^{\circ}$ with ground. $\ln \triangle A^{\prime} B C$
$
\begin{aligned}
& \frac{\mathrm{BC}}{\mathrm{A}^{\prime} \mathrm{C}}=\tan 60^{\circ} \\
& \frac{\mathrm{BC}}{9}=\sqrt{3} \\
& B C=9 \sqrt{3} \\
& \frac{\mathrm{A}^{\prime} \mathrm{C}}{\mathrm{A}^{\prime} \mathrm{B}}=\cos 60^{\circ} \\
& \frac{9}{\mathrm{~A}^{\prime} \mathrm{B}}=\frac{1}{2} \\
& A^{\prime} B=18
\end{aligned}
$
Height of tree $=A^{\prime} B+B C$
$
=9 \sqrt{3}+18=9 \times 1.732+18=15.588+18=33.588
$
Hence, the height of tree was $33.588 \mathrm{~m}=33.6 \mathrm{~m}$ (approximately). View full question & answer→Question 504 Marks
Due to a heavy storm, a part of a banyan tree broke without separating from the main. The top of the tree touched the ground 15 m from the base making an angle of 45° with the ground. Calculate the height of the tree before it was broken.
Answer
Let $\mathrm{AC}$ was original tree. Due to storm it was broken into two parts. The broken part $\mathrm{A}^{\prime} \mathrm{B}$ is making $45^{\circ}$ with ground.
In $\triangle A^{\prime} B C$
$
\begin{aligned}
& \frac{\mathrm{BC}}{\mathrm{A}^{\prime} \mathrm{C}}=\tan 45^{\circ} \\
& \frac{\mathrm{BC}}{15}=1 \\
& \mathrm{BC}=15 \\
& \frac{\mathrm{A}^{\prime} \mathrm{C}}{\mathrm{A}^{\prime} \mathrm{B}}=\cos 45^{\circ} \\
& \frac{15}{\mathrm{~A}^{\prime} \mathrm{B}}=\frac{1}{\sqrt{2}} \\
& A^{\prime} B=15 \sqrt{2}
\end{aligned}
$
Height of tree $=A \prime B+B C=15+15 \sqrt{2}=15(1+\sqrt{2})=15 \times 2.414=36.21$
Hence, the height of tree was $36.21 \mathrm{~m}$. View full question & answer→