3 Marks Question

Question 13 Marks

A moving boat is observed from the top of a $150\ m$ high cliff moving away from the cliff. The angle of depression of the boat changes from $60^{\circ}$ to $45^{\circ}$ in $2$ minutes. Find the speed of the boat in $m / h$.

Answer

Let us consider the following diagram.

In $\triangle A B C$
$\tan \theta=\frac{p}{b}=\frac{A B}{B C}$
$\Rightarrow \tan 60^{\circ}=\frac{150}{B C}$
$\Rightarrow \sqrt{3}=\frac{150}{B C}$
$\Rightarrow B C=\frac{150}{\sqrt{3}}$
$=\frac{150}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$=50 \sqrt{3} m \ldots \ldots(i)$
In $\triangle A B D$
$\tan \theta=\frac{p}{b}=\frac{A B}{B D}$
$\Rightarrow \tan 45^{\circ}=\frac{150}{B D}$
$\Rightarrow 1=\frac{150}{B D}$
$\Rightarrow B D=150 m$
Also, $B D=B C+C D$
Putting equation $(i)$ and $(ii),$ in the above equation
$\Rightarrow 150=50 \sqrt{3}+C D$
$\Rightarrow C D=150-50 \sqrt{3} a$
$\Rightarrow C D=150-50 \sqrt{3}$
$=50 \sqrt{3}(\sqrt{3}-1)$
According to the question, time taken to cover the distance from $C$ to $D$ is $2 \min$ .
Converting $2 \min$ in hours
$2 \min =\frac{2}{60}=\frac{1}{30} \text { hours }$
$\therefore $ Speed $=\frac{\text { Distance }}{\text { Time }}$
$\Rightarrow$ Speed $=\frac{C D}{\frac{1}{30}}$
$=\frac{50 \sqrt{3}(\sqrt{3}-1)}{\frac{1}{30}}$
$=30 \times(50 \sqrt{3}(\sqrt{3}-1))$
$=1500 \sqrt{3}(\sqrt{3}-1) m / h$
Hence, the speed of the boat is $1500 \sqrt{3}(\sqrt{3}-1) m / h$

View full question & answer→

Question 23 Marks

On a straight line passing through the foot of a tower, two points $C$ and $D$ are at distances of $4 m$ and $16 m$ from the foot respectively. If the angles of elevation from $C$ and $D$ of the top of the tower are complementary, then find the height of the tower.

Answer

Let $A B$ be the height of tower
Angle of elevation of top of the tower from point $C$ be $\theta$
Angle of elevation of top of the tower from point $D$ be $90-\theta$
In $\triangle \text{ABD}$
$\tan \left(90^{\circ}-\theta\right)=\frac{AB}{BD}=\frac{h}{16}$
$\cot \theta=\frac{h}{16} \ldots \ldots .(i)$
In $\triangle \text{ABC}$,
$\tan \theta=\frac{A B}{B C}=\frac{h}{4} \ldots .$
Multiply $(i)$ by $(ii)$
$\tan \theta \times \cot \theta=\frac{h}{4} \times \frac{h}{16}=\frac{h^2}{64}$
Since $\tan \theta=\frac{1}{\cot \theta}$
$\Rightarrow 1=\frac{h^2}{64}$
$\Rightarrow h^2=64$
$\Rightarrow h=8 m$
Therefore, the height of tower is $h=8 m$

View full question & answer→

Question 33 Marks

The angles of depression of the top and bottom of a $50 m$ high building from the top of a tower are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower and thehorizontal distance between the tower and the building.

Answer

Let the height of the tower $A B$ be $h$ meters and the horizontal distance between the tower and the building $B C$ be $x$ meters.
From the figure, we can see that,
$\text{AE=AB-EB}$
$AE=(h-50) m$
Now, in $\triangle \text{AED}$,
$\tan 45^{\circ}=\frac{AE}{ED}$
Using $\tan 45^{\circ}=1$ we get,
$\Rightarrow 1=\frac{h-50}{x}$
$\Rightarrow x=h-50 \ldots \ldots(1)$
In $\triangle \text{ABC}$,
$\tan 60^{\circ}=\frac{AB}{BC}$
$\tan 60^{\circ}=\sqrt{3}$ we get,
$\Rightarrow \sqrt{3}=\frac{h}{x}$
$\Rightarrow \sqrt{3} x=h \ldots \ldots(2)$
Using $(1)$ and $(2),$
$x=\sqrt{3} x-50$
$\Rightarrow x(\sqrt{3}-1)=50$
Solving and putting $\sqrt{3}=1.73$,
$\Rightarrow x=\frac{50}{(\sqrt{3}-1)}=\frac{50}{0.73}$
$\Rightarrow x=68.49 m$
Now substituting the value of $x$ in $(1),$ we get,
$68.49=(h-50) m$
$\Rightarrow h=68.49+50$
$\Rightarrow h=118.49 m$
Hence, the height of the tower is $118.49 m$ and the horizontal distance between the tower and the building is $68.449 m.$

View full question & answer→

Question 43 Marks

A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as $60^{\circ}$ and the angle of depression of the base of hill as $30^{\circ}$. Find the distance of the hill from the ship and the height of the hill.

Answer

Let $C D$ be the hill and suppose the man is standing on the deck of the ship at point $A$.
The angle of depression of the base $C$ of the hill $C D$ from point $A$ is $30^{\circ}$ and the angle of elevation of the top $D$ of the hill $C D$ is $60^{\circ}$.
So, $\angle E A D=60^{\circ}$ and $\angle B C A=30^{\circ}$
ln $\triangle A E D, \tan 60^{\circ}=\frac{D E}{E A}$
Using $\tan 60^{\circ}=\sqrt{3}$ we get,
$
\sqrt{3}=\frac{h}{x}
$
Hence, $h=\sqrt{3} x \quad$ (equation I)
In $\triangle A B C, \tan 30^{\circ}=\frac{A B}{B C}$
Using $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$ we get,
$
\frac{1}{\sqrt{3}}=\frac{10}{x}
$
Hence, $x=10 \sqrt{3} \quad$ (equation II)
Using equations I and II $
h=\sqrt{3} \times 10 \sqrt{3}=30
$
Hence, $D E=30 m$
So, $C D=C E+E D=10+30=40 m$
Thus, the distance of the hill from the ship is and the height of the hill is 40 m .

View full question & answer→

Question 53 Marks

The angle of elevation of an aero plane from a point $A$ on the ground is $60^{\circ}$. After a flight of $15$ seconds, the angle of elevation changes to $30^{\circ}$. If the aero plane is flying at a constant height of $1500 \sqrt{3} m$, find the speed of the plane in $km / hr$.

Answer

Let, $R$ and $S$ be the two position of the plane and $A$ be the point of observation.
Let, $\text{ATU}$ be the horizontal line through $A$.
The angle of elevation of the plane in two position $P$ and $Q$ from a point $A$ are respectively.
$\angle \text{RAT}=60^{\circ}, \angle \text{SAU}=30^{\circ}$
$R T=1500 \sqrt{3} m($Given$)$
In $\triangle \text{ATR}$,
$\tan 60^{\circ}=\frac{R T}{A T}$
$\Rightarrow \sqrt{3}=\frac{1500 \sqrt{3}}{A T}$
$A T=1500 m$
In $\triangle \text{ASU}$
$\tan 30^{\circ}=\frac{S U}{A U}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1500 \sqrt{3}}{A U}$
$A U=4500 m$
Distance travelled by the plane is $4500 m-1500 m$
$=3000 m$
Speed $=\frac{\text { Distance }}{\text { Time }}$
$=\frac{3000}{15}=200\ m / \sec$
$\Rightarrow 200 \times \frac{18}{5}=720 \ km / hr$
Hence the speed of the plane is $720 \ km / hr$.

View full question & answer→

Question 63 Marks

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 $45^{\circ}$. If the tower is $30 m$ high, find the height of the building.

Answer

Let the height of the building is $PQ=x m$

In $\triangle \text{QRS}$ we have,
$\tan 45^{\circ}=\frac{RS}{QS}$
$\Rightarrow 1=\frac{30}{Q S}$
$\Rightarrow QS=30 m$
Now in $\triangle \text{SPQ}$ we have,
$\tan 30^{\circ}=\frac{PQ}{QS}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{P Q}{30}$
$\Rightarrow PQ=\frac{30}{\sqrt{3}}$
On rationalizing we get,
$P Q=\frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow PQ=\frac{30 \sqrt{3}}{3}=10 \sqrt{3} m$
Hence, the height of the building is $10 \sqrt{3} m$.

View full question & answer→

Question 73 Marks

Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of the two ships as observed from the top of the light house are $60^{\circ}$ and $45^{\circ}$. If the height of the light house is $200 m ,$ find the distance between the two ships. $[$ Use $\sqrt{3}=1.73]$

Answer

Let $d$ be the distance between the two ships. Suppose the distance of one of the ships from the light house is $x$ meters, then the distance of the other ship from the light house is $(d-x)$ meter.

In right$-$angled $\triangle \text{ADO}$, we have,
$\tan 45^{\circ}=\frac{OD}{AD}=\frac{200}{x}$
$1=\frac{200}{x}$
$x=200 m$
In right-angled $\triangle \text{BDO}$, we have
$\tan 60^{\circ}=\frac{OD}{BD}=\frac{200}{d-x}$
Using $\tan 60^{\circ}=\sqrt{3}$
$\sqrt{3}=\frac{200}{d-x}$
$d-x=\frac{200}{\sqrt{3}}$
Putting $x=200$
$\Rightarrow d=\frac{200}{\sqrt{3}}+200$
$d=\frac{200+200 \sqrt{3}}{\sqrt{3}}$
$d=\frac{200(1+\sqrt{3})}{\sqrt{3}}$
$d=200 \times 1.58$
$d=316$
Thus, the distance between two ships is approximately $316 m.$

View full question & answer→

Question 83 Marks

The angle of elevation of an aeroplane from a point on the ground is $60^{\circ}$. After a flight of $30$ seconds the angle of elevation becomes $30^{\circ}$. If the aeroplane is flying at a constant height of $3000 \sqrt{3} m$, find the speed of the aeroplane.

Answer

Here $AD=3000 \sqrt{3} m$
Consider the $\triangle \text{BCE}$,
$\tan 30^{\circ}=\frac{B C}{E C}$
$\tan 30^{\circ}=\frac{3000 \sqrt{3}}{EC}$
$\frac{1}{\sqrt{3}}=\frac{3000 \sqrt{3}}{EC}$
$EC=9000 m$
Now consider $\triangle \text{ADE}$,
$\tan 60^{\circ}=\frac{AD}{ED}$
$\sqrt{3}=\frac{3000 \sqrt{3}}{ED}$
$ED=3000 m$
Distance covered by the aeroplane in $30$ seconds $\text{=AB=CD=EC-ED}$
Distance covered by the aeroplane in $30$ seconds $=9000-3000=6000 m$
Speed of aeroplane $=\frac{\text { Distance }}{\text { Time }}$
Speed of aeroplane $=\frac{6000}{30}=200\ m / s$

View full question & answer→

Question 93 Marks

The horizontal distance between two poles is $15 m .$ The angle of depression of the top of first pole as seen from the top of second pole is $30^{\circ}$. If the height of the second pole is $24 ,$ find the height of the first pole. $[\sqrt{3}=1.732]$

Answer

Let two poles $\text{AB}$ and $\text{CD}$ are apart and $\text{CD}=24 m$ According to the question,
Angle of depression of the top of first pole as seen from the top of second pole is $30^{\circ}$.
Using the figure according to the question,

$\angle \text{CAL}=30^{\circ}$ and $\text{BD}=15 m$
$\because \text{BD=AL}$
$\therefore \text{AL}=15 m$
Now,
Let the height of the pole $\text{AB}$ be $h$
$\Rightarrow \text{LD}=h$
$\therefore \text{CL}=24-h$
In the right angled $\triangle \text{ACL}$,
$\tan A=\frac{CL}{AL}$
$\tan 30^{\circ}=\frac{24-h}{15}$
Using $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{24-h}{15}$
$\frac{15}{\sqrt{3}}=24-h$
$24-\frac{15 \sqrt{3}}{3}=h$
$24-5 \sqrt{3}=h$
$24-(5 \times 1.732)=h$
$h=15.34$
Thus the height of the pole $\text{(AB)}$ is $15.34 m$

View full question & answer→

Question 103 Marks

The angles of depression of the top and bottom of a tower as seen from the top of a $60 \sqrt{3} \ m$ high cliff are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower.

Answer

Let the height of the tower be $A B$ be $x$
$\Rightarrow A B=D E=x$
It is given that the height of the cliff $C D=60 \sqrt{3} m$
$\Rightarrow C E=60 \sqrt{3}-x$
Now, In right triangle $\text{AEC}$,
$\tan 45^{\circ}=\frac{C E}{E A}$
$\Rightarrow 1=\frac{60 \sqrt{3}-x}{E A}$
$\Rightarrow E A=60 \sqrt{3}-x$
In right triangle $\text{CDB}$
$\tan 60^{\circ}=\frac{C D}{D B}$
$\Rightarrow \sqrt{3}=\frac{60 \sqrt{3}}{D B}$
$\Rightarrow D B=60 m$
$\Rightarrow D B=E A=60 m$
Now, from Equation $(1),$ we have
$\Rightarrow E A=60 \sqrt{3}-x$
$\Rightarrow 60=60 \sqrt{3}-x$
$x=60(\sqrt{3}-1) m$
Thus, the height of the tower is $60(\sqrt{3}-1) m$

View full question & answer→

Question 113 Marks

From the top of a tower $100 m$ high, a man observes two cars on the opposite sides of the tower with angles of depression $30^{\circ}$ and $45^{\circ}$ respectively. Find the distance between the cars.$[$Use $\sqrt{3}=1.73 ]$

Answer

Consider the diagram,

Let $\text{MN}$ be the tower of height $100 m . C$ and $D$ are the position of two cars whose angle of depression from the top of the tower is $30^{\circ}$ and $40^{\circ}$ respectively.
Clearly, from the diagram $\angle \text{YAC}=\angle \text{ACB}=30^{\circ}$ and $\angle \text{XAD}=\angle \text{ADB}=45^{\circ}$
$($Alternate interior angles$)$
In $\triangle \text{ABC}$,
$\tan 30^{\circ}=\frac{AB}{CB}$
$\frac{1}{\sqrt{3}}=\frac{100}{CB}$
$\Rightarrow \text{CB}=100 \sqrt{3}$
$\operatorname{In} \triangle \text{ABD,}$
$\tan 45^{\circ}=\frac{AB}{BD}$
$1=\frac{100}{BD}$
$\Rightarrow \text{BD}=100$
The distance between two cars $=\text{CD}$
$=\text{CB+BD} ($ from the diagram $)$
$=100 \sqrt{3}+100$
$=100(\sqrt{3}+1)$
$=100(1.73+1)$
$=100(2.73)$
$=273 m$
Therefore, the distance between the two cars is $273 m .$

View full question & answer→

Question 123 Marks

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^{\circ}$ and $45^{\circ}$. If the bridge is at a height of $8 m$ from the banks, then find the width of the river.

Answer

In $\triangle \text{ABC,} \tan 45^{\circ}=\frac{A C}{B C}$
$\Rightarrow 1=\frac{8}{B C}$
$\Rightarrow B C=8 m$

In $\triangle \text{ACD,} \tan 30^{\circ}=\frac{A C}{C D}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{8}{C D}$
$\Rightarrow CD=8 \sqrt{3}$
Now, $\text{BD=BC=CD}$
$=8+8 \sqrt{3}=8(1+\sqrt{3})$
$=8(1+1.732)$
$=8 \times 2.732$
$=13.856 m$

View full question & answer→

Question 133 Marks

The area of a circular play ground is $22176\ cm^2$. Find the cost of fencing this ground at the rate of $Rs. 50$ per metre.

Answer

$\pi r^2=22176 \ cm^2$
$\Rightarrow r^2=\frac{22176 \times 7}{22}$
$\Rightarrow r=\sqrt{7056}$
$\Rightarrow r=84 \ cm$
Length of fencing $=$ Circumference of ground
$=2 \pi r$
$=2 \times \frac{22}{7} \times 84$
$=528 \ cm$
$=5.28 m$
Rate $=\text { Rs. } 50 / m$
Cost $=$ Length $\times$ rate
$ =5.28 \times 50$
$=\text { Rs. } 264$

View full question & answer→

Question 143 Marks

Two poles of heights $25 \ m$ and $35 \ m$ stand vertically on the ground. The tops of two poles are connected by a wire, which is inclined to the horizontal at an angle of $30^{\circ}$. Find the length of the wire and the distance between the poles.

Answer

In $\triangle A B C, \tan 45^{\circ}$
$=\frac{A C}{B C}$
$\Rightarrow 1=\frac{8}{B C}$
$ \Rightarrow B C=8\ m$

$\therefore DE=35-25=10\ m$
In $\triangle CDE$,
$\tan 30^{\circ}=\frac{DE}{CE}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{10}{CE}$
$\Rightarrow CE=10 \sqrt{3} \ m$
Thus, distance between the poles
$=BA=CE=10 \sqrt{3} \ m$
Again, in $\triangle CDE$,
$\sin 30^{\circ}=\frac{DE}{CD}$
$\Rightarrow \frac{1}{2}=\frac{10}{CD}$
$\Rightarrow CD=20 \ m$
Hence, length of the wire is $20 \ m$ .

View full question & answer→

Maths 3 Marks Question

Test yourself on this topic