HEIGHTS & DISTANCES

HEIGHTS & DISTANCES

12.1 ANGLE OF ELEVATION :

In order to see an object which is at a higher level compared to the ground level we are to look up. The line joining the object and the eye of the observer is known as the line sight and the angle which this line of sight makes with the horizontal drawn through the eye of the observer is known as the angle of elevation. Therefore, the angle of elevation of an object helps in finding out its height (figure)

HEIGHTS & DISTANCES

12.2 ANGLE OF DEPRESSION :

When the object is at a lower level tan the observer’s eyes, he has to look downwards to have a view of the object. It that case, the angle which the line of sight makes with the horizontal thought the observer’s eye is known as the angle of depression (Figure).

HEIGHTS & DISTANCES

HEIGHTS & DISTANCES

ILLUSTRATIONS :

Ex. 1 A man is standing on the deck of a ship, which is $8m$ above water level. He observes the angle of elevations of the top of a hill as ${60}^{\circ }$ and the angle of depression of the base of the hill as ${30}^{\circ }$. Calculation the distance of the hill from the ship and the height of the hill.

[CBSE - 2005]

HEIGHTS & DISTANCES

Sol. Let $x$ be distance of hill from man and $h+8$ be height of hill which is required. is right triangle ACB.

$\Rightarrow \tan {60}^{\circ }=\frac{AC}{BC}=\frac{h}{x}\Rightarrow \sqrt{3}=\frac{h}{x}$

HEIGHTS & DISTANCES

In right triangle $BCD$.

$\begin{array}{rl}& \Rightarrow \frac{1}{\sqrt{3}}=\frac{8}{x}\Rightarrow x=8\sqrt{3}\\ & \therefore \text{ Height of hill }=h+8=\sqrt{3}\cdot x+8=(\sqrt{3})(8\sqrt{3})+8=32m.\\ & \text{ Distance of ship from hill }=x=8\sqrt{3}m.\end{array}$

HEIGHTS & DISTANCES

Ex. 2 A vertical tower stands on a horizontal plane and is surmounted by vertical flag staff of height 5 meters. At a point on the plane, the angle of elevation of the bottom and the top of the flag staff are respectively ${30}^{\circ }$ and ${60}^{\circ }$ find the height of tower.

[CBSE-2006]

HEIGHTS & DISTANCES

Sol. Let $AB$ be the tower of height $h$ metre and $BC$ be the height of flag staff surmounted on the tower, Let the point of the place be $D$ at a distance $x$ meter from the foot of the tower in $\mathrm{△}ABD$

$\begin{array}{cc}& \tan {30}^{\circ }=\frac{AB}{AD}\\ \Rightarrow & \frac{1}{\sqrt{3}}=\frac{h}{x}\\ \Rightarrow & x=\sqrt{3}h\dots \dots \dots \dots .(i)\end{array}$

HEIGHTS & DISTANCES

In $\mathrm{△}ABD\tan {60}^{\circ }=\frac{AC}{AD}$

$\Rightarrow \sqrt{3}=\frac{5+h}{x}\Rightarrow x=\frac{5+h}{\sqrt{3}}$ …………..(ii)

HEIGHTS & DISTANCES

From (i) and (ii)

$\begin{array}{rl}& \Rightarrow \sqrt{3}h\frac{5+h}{\sqrt{3}}\Rightarrow 3h=5+h\\ & \Rightarrow 2h=5\Rightarrow h=\frac{5}{2}=2.5m\text{ So, the height of tower }=2.5m\end{array}$

HEIGHTS & DISTANCES

Ex. 3 The angles of depressions of the top and bottom of $8m$ tall building from the top of a multistoried building are ${30}^{\circ }$ and ${45}^{\circ }$ respectively. Find the height of multistoried building and the distance between the two buildings.

HEIGHTS & DISTANCES

Sol. Let $AB$ be the multistoried building of height $h$ and let the distance between two buildings be $x$ meters.

$\mathrm{\angle }XAC=\mathrm{\angle }ACB={45}^{\circ }$ [Alternate angles $\because AX||DE]$

$XAD=ADE={30}^{\circ }$ [Alternate angles $\because AX||BC]$

In $\mathrm{△}ADE$

$\tan {30}^{\circ }=\frac{AE}{ED}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h-8}{x}$

$(\because CB=DE=x)$

HEIGHTS & DISTANCES

$\Rightarrow x=\sqrt{3}(h-8)$ …………(i)

In $\mathrm{△}ACB$

$\tan {45}^{\circ }=\frac{h}{x}$

$\Rightarrow 1=\frac{h}{x}\Rightarrow x=h$ …………(ii)

HEIGHTS & DISTANCES

Form (i) and (ii)

$\begin{array}{rl}& \sqrt{3}(h-8)=h\Rightarrow \sqrt{3}h-8\sqrt{3}=h\\ & \Rightarrow \sqrt{3}h-h=8\sqrt{3}\Rightarrow h(\sqrt{3}-1)=8\sqrt{3}\\ & \Rightarrow h=\frac{8\sqrt{3}}{\sqrt{3-1}}\times \frac{(\sqrt{3}+1)}{\sqrt{3}+1}\Rightarrow h=\frac{8\sqrt{3}(\sqrt{3}+1)}{2}\Rightarrow h=4\sqrt{3}(\sqrt{3}+1)\\ & \Rightarrow h=4(3+\sqrt{3})\text{ metres }\end{array}$

Form (ii) $x=h$

So, $x=4(3+\sqrt{3})$ metres $$ Hence, height of multistoried building $=4(3+\sqrt{3})$ metres

Distance between two building $=4(3+\sqrt{3})$ metres

HEIGHTS & DISTANCES

Ex. 4 The angle of elevation of an aeroplane from a point on the ground is ${45}^{\circ }$. After a flight of $15sec$, the elevation changes to ${30}^{\circ }$. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.

HEIGHTS & DISTANCES

Sol. Let the point on the ground is $E$ which is y metres from point $B$ and let after $15sec$ flight it covers $x$ metres distance.

In $\mathrm{△}$ AEB.

$\tan {45}^{\circ }=\frac{AB}{EB}\Rightarrow 1=\frac{3000}{y}\Rightarrow y=3000m$ ………..(i)

In $\mathrm{△}$ CED

$\Rightarrow \tan {30}^{\circ }=\frac{CD}{ED}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{3000}{x+y}$

$(\because AB=CD)$

$\Rightarrow x+y=3000\sqrt{3}$ …………(ii)

HEIGHTS & DISTANCES

From equation (i) and (ii)

$\Rightarrow x+3000=3000\sqrt{3}\Rightarrow x=3000\sqrt{3}-3000\Rightarrow x=3000(\sqrt{3}-1)$

$\Rightarrow x=3000\times (1.732-1)\Rightarrow x=2196m$

HEIGHTS & DISTANCES

Hence, the speed of aeroplane is $527.04Km/hr$.

HEIGHTS & DISTANCES

Ex. 5 If the angle of elevation of cloud from a point $h$ metres above a lake is $\alpha$ and the angle of depression of its reflection in the lake is $\beta$, prove that the distance of the cloud from the point of observation is $\frac{2h\sec \alpha }{\tan \beta -\tan \alpha }$.

HEIGHTS & DISTANCES

Sol. Let $AB$ be the surface of the lake and let $C$ be a point of observation such that $AC-h$ metres. Let $D$ be the position of the cloud and $D^{\prime }$ be its reflection in the lake. Then $BD=B{D}^{\prime }$.

In $\mathrm{△}DCE$

$\tan \alpha =\frac{DE}{CE}\Rightarrow CE=\frac{H}{\tan \alpha }\dots \dots \dots .(i)$

In $\mathrm{\Delta }$ CED ${}^{\prime }$

$\begin{array}{cc}& \tan \beta =\frac{E^{\prime }}{EC}\\ \Rightarrow & CE=\frac{h+H+h}{\tan \beta }\\ \Rightarrow & CE=\frac{2h+H}{\tan \beta }\dots \dots \dots .(ii)\end{array}$

HEIGHTS & DISTANCES

From (i) & (ii)

$\begin{array}{rl}& \Rightarrow \frac{H}{\tan \alpha }=\frac{2h+H}{\tan \beta }\Rightarrow H\tan \beta =2h\tan \alpha +H\tan \alpha \\ & \Rightarrow H\tan \beta -H\tan \alpha +2h\tan \alpha \Rightarrow H(\tan \beta -\tan \alpha )=2h\tan \alpha \\ & \Rightarrow H=\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }\dots \dots \dots .(iii)\\ & \text{ In }\mathrm{△}DCE\\ & \text{ Sin}\alpha =\frac{DE}{CD}\\ & \Rightarrow CD=\frac{DE}{\sin \alpha }\\ & \Rightarrow CD=\frac{H}{\sin \alpha }\end{array}$

HEIGHTS & DISTANCES

Substituting the value of $H$ from (iii)

$\begin{array}{rl}& CD=\frac{2h\tan \alpha }{(\tan \beta -\tan \alpha )\sin \alpha }\Rightarrow CD=\frac{2h\frac{\sin \alpha }{\cos \alpha }}{(\tan \beta -\tan \alpha )\sin \alpha }\\ & CD=\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }\end{array}$

Hence, the distance of the cloud from the point of observation is $\frac{2h\sec \alpha }{\tan \beta -\tan \alpha }$

Hence Proved.

HEIGHTS & DISTANCES

Ex. 6 A boy is standing on the ground and flying a kite with $100m$ of string at an elevation of ${30}^{\circ }$. Another boy is standing on the roof of a $10m$ high building and is flying his kite at an elevation of ${45}^{\circ }$. 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.

HEIGHTS & DISTANCES

Sol. Let the length of second string $b\times m$.

In $\mathrm{△}ABC$

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

$\frac{1}{2}=\frac{AC}{100}\Rightarrow AC=50m$

HEIGHTS & DISTANCES

In $\mathrm{△}AEF$

$Sin{30}^{\circ }=\frac{AF}{AE}$

$\frac{1}{\sqrt{2}}=\frac{AC-FC}{x}$

$\frac{1}{\sqrt{2}}=\frac{50-10}{x}[\therefore AC=50m,FC=ED=10m]\frac{1}{\sqrt{2}}=\frac{40}{x}$

$x=40\sqrt{2}m$ (So the length of string that the second boy must have so that the two kites meet $=40\sqrt{2}m$.)

DAILY PRACTICE PROBLEMS 12

OBJECTIVE DPP - 12.1

1. Upper part of a vertical tree which is broken over by the winds just touches the ground and makes an angle of ${30}^{\circ }$ with the ground. If the length of the broken part is 20 metres, then the remaining part of the trees is of length

(A) 20 metres

(B) $10\sqrt{3}$ metres

(C) 10 metres

(D) $10\sqrt{2}$ metres

HEIGHTS & DISTANCES

HEIGHTS & DISTANCES

2. The angle of elevation of the top of a tower as observed from a point on the horizontal ground is ’ $x$ ‘. If we move a distance ’ $d$ ’ towards the foot of the tower, the angle of elevation increases to ’ $y$ ‘, then the height of the tower is

(A) $\frac{d\tan x\tan y}{\tan y-\tan x}$

(B) $d(\tan y+\tan x)$

(C) $d(\tan y-\tan x)$

(D) $\frac{d\tan x\tan y}{\tan y+\tan x}$

HEIGHTS & DISTANCES

HEIGHTS & DISTANCES

3. The angle of elevation of the top of a tower, as seen from two points A & B situated in he same line and at distances ’ $p$ ’ and ’ $q$ ’ respectively from the foot of the tower, are complementary, then the height of the tower is

$(A)$ pq

(B) $\frac{p}{q}$

(C) $\sqrt{pq}$

(D) noen of these

HEIGHTS & DISTANCES

HEIGHTS & DISTANCES

4. The angle of elevation of the top of a tower at a distance of $\frac{50\sqrt{3}}{3}$ metres from the foot is ${60}^{\circ }$. Find the height of the tower

(A) $50\sqrt{3}$ metres

(B) $\frac{20}{\sqrt{3}}$ metres

(C) -50 metres

(D) 50 metres

HEIGHTS & DISTANCES

HEIGHTS & DISTANCES

5. The Shadow of a tower, when the angle of elevation of the sun is ${30}^{\circ }$, is found to be $5m$ longer than when its was ${45}^{\circ }$, then the height of tower in metre is

(A) $\frac{5}{\sqrt{3}+1}$

(B) $\frac{5}{2}(\sqrt{3}-1)$

(C) $\frac{5}{2}(\sqrt{3}+1)$

(D) None of these.

HEIGHTS & DISTANCES

HEIGHTS & DISTANCES

SUBJECTIVE DPP - 12.2

1. From the top a light house, the angles of depression of two ships of the opposite sides of it are observed to be $\alpha$ and $\beta$. If the height of the light house be $h$ meters and the line joining the ships passes thought the foot of the light house. Show that the distance between the ships is $\frac{h(\tan \alpha +\tan \beta )}{\tan \alpha \tan \beta }$ meters.

HEIGHTS & DISTANCES

2. A ladder rests against a wall at angle $\alpha$ to the horizontal. Its foot is pulled away from the previous point through a distance ’ $a$ ‘, so that is slides down a distance ’ $b$ ’ on the wall making an angle $\beta$. With the horizontal show that $\frac{a}{b}=\frac{\cos \alpha -\cos \beta }{\sin \beta -\sin \alpha }$

HEIGHTS & DISTANCES

3. From an aeroplanne vertically above a straight horizontal road, the angle of depression of two consecutive kilometer stone on opposite side of aeroplane are observed to be $\alpha$ and $\beta$. Show that the height of aeroplane above the road is $\frac{\tan \alpha \tan \beta }{\tan \alpha +\tan \beta }$ kilometer.

HEIGHTS & DISTANCES

4. A round balloon of radius ’ $r$ ’ subtends an angle $\theta$ at the eye of an observer while the angle of elevation of its centre is $\varphi$. Prove that the height of the centre of the balloon is $r\sin \varphi$ cosec $\frac{\theta }{2}$.

HEIGHTS & DISTANCES

5. A window in a building is at a height of $10m$ from the ground. The angle of depression of a point $P$ on the ground from the window is ${30}^{\circ }$. The angle of elevation of the top of the building from the point $P$ is ${60}^{\circ }$. Find the height of the building.

HEIGHTS & DISTANCES

Sol. 5. $30$ m

HEIGHTS & DISTANCES

6. A man on a cliff observers a boat at an angle of depression of ${30}^{\circ }$ which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression

HEIGHTS & DISTANCES

Sol. 6. $9$ min

HEIGHTS & DISTANCES

7. The angles of elevation of the top of a tower two points ’ $P$ ’ and ’ $Q$ ’ at distances of ’ $a$ ’ and ’ $b$ ’ respectively of the boat is found to be ${60}^{\circ }$. Find the total time taken by the boat from the initial point to reach the shore.

from the base and in the same straight line with it, are complementary. Prove that the height of the tower is $\sqrt{ab}$.

[CBSE - 2004]

HEIGHTS & DISTANCES

8. Two pillars of equal height are on either side of a road, which is $100m$ wide. The angles of elevation of the top the pillars are ${60}^{\circ }$ and ${30}^{\circ }$ at a point on the road between the pillar. Find the position of the pint between the pillars. Also find the height of each pillar,

[CBSE - 2005]

HEIGHTS & DISTANCES

Sol. 8. Height $=43.3$ $m$, Position - point is $25$ m $\text{ from}$ $1^{\text{st }}$ end and $75$ $m$ from $2^{\text{nd }}$ end.

HEIGHTS & DISTANCES

9. At a point, the angle of elevation of a tower is such that its tangent is $\frac{5}{12}$, On walking 240 mnearer the tower, the tangent to the angle of elevation becomes $\frac{3}{4}$, Find the height of the tower.

[CBSE - 2006]

HEIGHTS & DISTANCES

Sol. 9. $225$ m

HEIGHTS & DISTANCES

10. From a window ’ $x$ ‘mtres high above the ground in a street, the angles of elevation and depression of the top and foot of the other hose on the opposite side of the street are $\alpha$ and $\beta$ respectively, Show that the opposite house is $x(1+\tan \alpha \cot \beta )$ metres.

[CBSE - 2006]

HEIGHTS & DISTANCES

11. A pole $5m$ high is fixed on the top of a towel, the angle of elevation of the top of the pole observed from a point ’ $A$ ’ on the ground is ${60}^{\circ }$ an the angle of depression the point ; $A$; from the top of the tower is ${45}^{\circ }$ Find the height of the tower.

[CBSE - 2007]

HEIGHTS & DISTANCES

Sol. 11. $6.82$ m

HEIGHTS & DISTANCES

12. The angle of elevation of a jet fighter from a point $A$ on the ground is ${60}^{\circ }$ After a flight of 15 seconds, the angle $O$ elevation changes to ${30}^{\circ }$ If the jet is flying at a spies of $720km/fr$, find the constant height at which the jet is flying. [use $\sqrt{3}=1.732$ ]

[CBSE - 2008]

HEIGHTS & DISTANCES

Sol. 12. $2598$ m