Question 14 Marks
A statue $1.46 \ m$ tall, stands on the top of 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. $[ Use \sqrt{3}=1.732]$
AnswerLet $SP$ be the statue and $PB$ be the pedestal. Angles of elevation of $S$ and $P$ are $60^{\circ}$ and $45^{\circ}$ respectively.
Further suppose $A B=x m, PB = h m$
In right $\triangle\text{ABS},$
$\frac{\text{SB}}{\text{AB}}=\tan60^\circ=\sqrt{3}$
$\Rightarrow\frac{\text{h}+1.46}{\text{x}}=\sqrt{3}\dots(1)$
In right $\triangle\text{PAB},$
$\frac{\text{PB}}{\text{AB}}=\tan45^\circ=1$
$\therefore\text{h}=\text{x}\dots(2)$
Putting $x = h$ in $(1)$
$\frac{\text{h}+1.46}{\text{h}}=\sqrt{3}$
$\Rightarrow\text{h}+1.46=\sqrt{3}\text{h}$
Or $\text{h}(\sqrt{3}-1)=1.46$
$\therefore\text{h}=\frac{1.46}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\therefore\ \text{h}=\frac{1.46}{2}\times\big(\sqrt{3}+1)=0.73\times2.732$
$=2\text{m}$ (Nearly)
Thus, height of the pendestal $= 2m.$ View full question & answer→Question 24 Marks
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 m . At a point on the plane, the angle of elevation of the bottom of the flagstaff is $30^{\circ}$ and that of the top of the flagstaff is $60^{\circ}$. Find the height of the tower.
$[ Use \sqrt{3}=1.732]$
Answer
Let AB is the tower of height h meter
In right $\triangle\text{BOA},$
$\frac{\text{OA}}{\text{AB}}=\cot30^\circ$
$\Rightarrow\frac{\text{OA}}{\text{h}}=\sqrt{3}$
$\Rightarrow\text{OA}=\text{h}\sqrt{3}\text{m}\dots(\text{i})$
In right $\triangle\text{COA},$
$\frac{\text{OA}}{\text{AC}}=\cot60^\circ$
$\Rightarrow\frac{\text{OA}}{\text{h}+6}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{OA}=\frac{\big(\text{h}+6\big)}{\sqrt{3}}\text{m}\dots(\text{ii})$
From (i) and (ii),
$\frac{\big(\text{h}+6\big)}{\sqrt{3}}=\text{h}\sqrt{3}$
$\Rightarrow\text{h}+6=3\text{h}$
$\Rightarrow2\text{h}=6$
$\Rightarrow\text{h}=3$
Hence, the height of the tower is 3m. View full question & answer→Question 34 Marks
From a point on the ground $40 \ m$ away from the foot of a tower, the angle ofclevation of the top of the t wer is $30^{\circ}$. The angle of elevation of the top of a water tank (on the top of the tower) is $45^{\circ}$. Find:
- The height of the tower.
- The depth of the tank.
Answer
Let BC is the tower and $CD$ be the water tank.
Let A be the point of observation.
Then, $\angle\text{BAC}=30^\circ,\angle\text{BAD}=45^\circ$ and $\text{AB}=40\text{m}$
In right $\triangle\text{ABD},$
$\frac{\text{BD}}{\text{AB}}=\tan45^\circ$
$\Rightarrow\frac{\text{BD}}{40}=1$
$\Rightarrow\text{BD}=40\text{m}$
In right $\triangle\text{ABC},$
$\frac{\text{BC}}{\text{AB}}=\tan30^\circ$
$\Rightarrow\frac{\text{BC}}{40}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{BC}=\frac{40}{\sqrt{3}}$
Rationalising we get,
$\text{BC}=\frac{40}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{40\sqrt{3}}{3}\text{m}$
- So, height of the tower $=\text{BC}=\frac{40\sqrt{3}}{3}\text{m}=23.1\text{m}$
- Depth of the tank $= CD$
$= (BD - BC)$
$= (40 - 23.1)m$
$= 16.9m$ View full question & answer→Question 44 Marks
A tower stands vertically an the ground.From a point on the ground which is $20\ m$ away from the foot f the tower, th angle of elevation of its top is found tobe $60^\circ $. Find theh ight of the tower. $\big[\text{Take}\sqrt{3}=1.732\big]$
AnswerLet $AB$ be the tower standing on a level ground and $O$ be the position of the observer.
Then $OA = 20m$ and $\angle\text{OAB}=90^\circ$ and $\angle\text{OAB}=60^\circ$
Let AB = h meters
From the right $\triangle\text{OAB},$ we have
$\frac{\text{AB}}{\text{OA}}=\tan60^\circ=\sqrt{3}$
$\Rightarrow\frac{\text{h}}{20}=\sqrt{3}$
$\Rightarrow\text{h}=(20\times\sqrt{3})$
$\Rightarrow\text{h}=20\times1.732$
$\Rightarrow\text{h}=34.64\text{m}$
Hence the height of the tower is $20\sqrt{3}\text{m}=34.64\text{m}$ View full question & answer→Question 54 Marks
An electrician has to repair an electric fault on a pole of height $4$ metres. He needs to reach a point $1$ metre below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use, which when inclined at an angle of $60^\circ $ to the horizontal would enable him to reach the required position? $\big[\text{Use}\sqrt{3}=1.73\big]$
Answer
Let $AB$ be the eletric pole such that $AB = 4m.$
Let $C$ be a point $1m$ below $B.$
$\Rightarrow\text{AC}=4\text{m}-1\text{m}=3\text{m}$
Let OC be the ladder = x metres.
In right $\triangle\text{OAC},$
$\text{cosec }60^\circ=\frac{\text{OC}}{\text{AC}}$
$\Rightarrow\frac{2}{\sqrt{3}}=\frac{\text{x}}{3}$
$\Rightarrow\text{x}=\frac{6}{\sqrt{3}}$
On rationalising we get,
$\text{x}=\frac{6}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow\text{x}=\frac{6\sqrt{3}}{3}$
$\Rightarrow\text{x}=2\sqrt{3}$
$\Rightarrow\text{x}=2\times1.73=3.46\text{m}$
Hence, the length of the ladder that he should use is 3.46m. View full question & answer→Question 64 Marks
From the top of a $7-$metre-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. [ $Use \sqrt{3}=1.732$ ]
Answer
Let AB be the building and $CD$ be the cable tawer.
Let the height of the tower be h metres.
Construction: Draw $\text{BE}\perp\text{CD}.$
Then $CE = AB = 7m$ and $DE = (h - 7)m$
In right $\triangle\text{BAC},$
$\cot45^\circ=\frac{\text{AC}}{\text{AB}}$
$\Rightarrow1=\frac{\text{AC}}{7}$
$\Rightarrow\text{AC}=7\text{m}$
So, BE = AC = 7m
In right $\triangle\text{BED},$
$\cot60^\circ=\frac{\text{BE}}{\text{DE}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{7}{(\text{h}-7)}$
$\Rightarrow\text{h}-7=7\sqrt{3}$
$\Rightarrow\text{h}=7\sqrt{3}+7$
$\Rightarrow\text{h}=19.124\text{m}$
Hence, the height of the tower is 19.124m. View full question & answer→Question 74 Marks
On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, $9$ metres away from the foot of the tower, the angle of elevation of the top and bottom the flagpole are $60^{\circ}$ and $30^{\circ}$ respectively. Find the height of the tower and the flagpole mounted on it. $[$ Take $\sqrt{3}=1.732]$
AnswerLet $A B$ be the tower and $B C$ be flagpole, Let $O$ be the point of observetion.
Then, $OA =9 m, \angle AOB =30^{\circ}$ and $\angle AOC =60^{\circ}$
From right angled $\triangle\text{BOA}$
$\frac{\text{AB}}{\text{OA}}=\tan30^\circ$
$\Rightarrow\frac{\text{AB}}{9}=\frac{1}{\sqrt{3}}\Rightarrow\text{AB}=3\sqrt{3}$
From right angled $\triangle\text{OAC}$
$\frac{\text{AC}}{\text{OA}}=\tan60^\circ$
$\frac{\text{AC}}{9}=\sqrt{3}\Rightarrow\text{AC}=9\sqrt{3}\text{m}$
$\therefore\ \text{BC}=(\text{AC}-\text{AB})=6\sqrt{3}\text{m}$
Thus $\text{AB}=3\sqrt{3}\text{m}=5.196\text{m}$ and $\text{BC}=6\sqrt{3}\text{m}=10.392\text{m}$
Hence, hight of the tower = 5.196m and the height of the flagpole = 10392m. View full question & answer→Question 84 Marks
A kite is flying at a height of $75 \ m$ from the level ~round, a ched to a string inclined at $60^{\circ}$ to the hori ontal. Find th length of the string, assuming that there is no slack in it. [Take $\sqrt{3}=1.732]$
AnswerLet $O B$ be the length of the string from the level of ground and $O$ be the point of the observer.
Then, $AB =75 m$ and $\angle OAB =90^{\circ}$ and $\angle OAB =60^{\circ}$, let $OB =1$ meters.
From the right $\triangle OAB$, we have
$\frac{\text{OB}}{\text{AB}}=\text{cosec }60^\circ=\frac{2}{\sqrt{3}}$
$\frac{\text{l}}{75}=\frac{2}{\sqrt{3}}$
$\Rightarrow\text{l}=\Big(75\times\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}\Big)$
$\Rightarrow\text{l}=25\times2\times\sqrt{3}$
$\Rightarrow\text{l}=50\sqrt{3}\text{m}$
$\Rightarrow\text{l}=86.6\text{m}$
Hence, the length of the string 86.6m View full question & answer→Question 94 Marks
The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is $30^{\circ}$. On advancing $150 \ m$ towards the foot of the tower, the angle of elevation becomes $60^{\circ}$ Show that the height of the tower is $129.9$ metres. [Given $\sqrt{3}=1.732$ ]
AnswerLet AB be the tower and let the angle of elevation of its top at C be $30^{\circ}$. Let $D$ be a point at a distance $150 \ m$ from $C$ such that angle of elevation of the tower at $D$ is $60^{\circ}$
Let h m be the height of the tower and $AD = x m$
In $\triangle\text{CAB},$ we have
$\tan30^\circ=\frac{{\text{AB}}}{\text{AC}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}+150}\dots(1)$
In $\triangle\text{DAB},$ we have
$\tan60^\circ=\frac{\text{AB}}{\text{AD}}\Rightarrow\sqrt{3}=\frac{\text{h}}{\text{x}}$
$\Rightarrow1=\frac{\text{h}}{\sqrt{3}}\dots(2)$
Putting the $\text{x}=\frac{\text{h}}{\sqrt{3}}$ in (1), we get
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{h}}{\frac{\text{h}}{\sqrt{3}}+150}\Rightarrow\frac{1}{\sqrt{3}}=\frac{\sqrt{3}\text{h}}{\text{h}+150\sqrt{3}}$
$\Rightarrow\text{h}+150\sqrt{3}=3\text{h}\Rightarrow3\text{h}-\text{h}=150\sqrt{3}$
$2\text{h}=150\sqrt{3}$
$\text{h}=\frac{150}{2}\sqrt{3}=75\sqrt{3}$
$\text{h}=(75\times1.732)\text{m}$
$\text{h}=129.9$
Hence, the height of the tower is $129.9m.$ View full question & answer→Question 104 Marks
The angle of elevation of the top of an unifinished tower at a distance of $75\ m$ from its base is $30^\circ .$ How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60^\circ ? $\big[\text{Take}\sqrt{3}=1.732\big]$
AnswerLet AB be the unfinished tower and let$ AC$ be complete tower.
Let $O$ be the point of observation.
Then $OA = 75m$
$\angle\text{AOB}=30^\circ$ and $\angle\text{AOC}=60^\circ$
Let AB = h metres
And AC = H metres
$\frac{\text{AB}}{\text{OA}}=\tan30^\circ$
$\Rightarrow\frac{\text{h}}{75}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{h}=\frac{75}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=25\sqrt{3}\text{m}$
And $\frac{\text{AC}}{\text{OA}}=\tan60^\circ$
$\Rightarrow\frac{\text{H}}{75}=\sqrt{3}=\text{H}=75\sqrt{3}\text{m}$
Hence, the required height is $(\text{H}-\text{h})\text{m}=(75\sqrt{3}-25\sqrt{3})\text{m}$
$=50\sqrt{3}\text{m}=86.6$ View full question & answer→Question 114 Marks
The angle of elevation of an aeroplane from a point on the ground is $45^{\circ}$. After flying for $15$ seconds, the elevation changes to $30^{\circ}$. If the aeroplane is flying at a height of $2500$ metres, find the speed of the airoplane.
Answer
Let the first position of the plane be E and after $15$ seconds, $D.$
In right $\triangle\text{ABE},$
$\cot45^\circ-\frac{\text{AB}}{\text{BE}}$
$\Rightarrow1=\frac{\text{AB}}{2500}$
$\Rightarrow\text{AB}=2500\text{m}$
In right $\triangle\text{ACD},$
$\cot30^\circ=\frac{\text{AC}}{\text{CD}}$
$\Rightarrow\sqrt{3}=\frac{\text{AC}}{2500}$
$\Rightarrow\text{AC}=2500\sqrt{3}\text{m}$
Distance travelled
$=\text{AC}-\text{AB}$
$=2500\sqrt{3}-2500$
$=2500(\sqrt{3}-1)\text{m}$
$=2500(1.732-1)$
$=1830\text{m}$
Time taken $= 15$ sec $=\frac{15}{3600}\text{hr}$ (Since $1hr = 3600$ seconds)
Speed $=\frac{\text{Distance}}{\text{Time}}$
$=\frac{1830}{\frac{15}{3600}\times1000}$ (Since $1km = 1000m)$
$=439.2\text{km}/\text{hr}$ View full question & answer→Question 124 Marks
From the top of a hill, the angles of depression of two consecutive kilometre stones due east are fouud to be $45^{\circ}$ and $30^{\circ}$ respectively. Find the height of the hill.
Answer
Let $AB$ is the height of the hill and two stone are $C$ and $D$ resrectively.
Where depression is $45$ degree and $30$ degree. The distance between $C$ and $D$ is 1km.
Here depression and hill has formed right angle triangles with the base.
We have to find the height of the hill with this through trigonomentry.
In $\triangle\text{ABC},\tan45^\circ=\frac{\text{Height}}{\text{Base}}=\frac{\text{AB}}{\text{BC}}$
Again, $\triangle\text{ABD},\tan30=\frac{\text{AB}}{\text{BC}}$
$\frac{1}{\sqrt{3}}=\frac{\text{AB}}{\text{BC}+\text{CD}}$ $\Big[\tan30=\frac{1}{\sqrt{3}}-\frac{1}{1.732}\Big]$
Or $\frac{1}{1.732}=\frac{\text{AB}}{\text{AB}+1}$ [As AB = BC from (i) above]
$1.732 AB = AB + 1$
$1.732 AB - AB = 1$
$AB (1.732 - 1) = 1$
$AB \times 0.732 = 1$
$\text{AB}=\frac{1}{0.732}=1.366$
Hence, height of the hill 1.366km View full question & answer→Question 134 Marks
Aman on the deck of a ship, $16\ m$ above water level, observes that the angles of elevation and depression respectively of the top bottom of a cliff are $60^\circ $and $30^\circ $. Calculate the distance of the cliff from the ship and height of the cliff. $\big[\text{Take}\sqrt{3}=1.732\big]$
AnswerLet $AB$ be the height of the deck and let $CD$ be the cliff.
Let the man be $B$, then, $AB = 16m.$
Let $\text{BE}\perp\text{CD}$ and $\text{AE}\perp\text{CD}$
Then, $\angle\text{EBD}=60$ and $\angle\text{EBC}=30$
$CE = AB = 16m$
Let $CD = h$ metres
Then, $ED = (h = 16)m$
From right $\triangle\text{BED},$ we have
$\frac{\text{BE}}{\text{ED}}=\cot60^\circ$
$\Rightarrow\frac{\text{BE}}{(\text{h}-16)}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{BE}=\frac{(\text{h}-16)}{\sqrt{3}}$
From right $\triangle\text{CAB},$ we have
$\frac{\text{AC}}{\text{AB}}=\cot30^\circ\Rightarrow\frac{\text{AC}}{16}=\sqrt{3}$
$\Rightarrow\text{AC}=16\sqrt{3}\text{m}$
But BE = AC
$\therefore\ \frac{(\text{h}-16)}{\sqrt{3}}=16\sqrt{3}\Rightarrow(\text{h}-16)=48$
$\Rightarrow\text{h}=64\text{m}$
Hence the height of cliff is 64m and the distance between the cliff and ship $=16\sqrt{3}\text{m}=27.71\text{m}$ View full question & answer→Question 144 Marks
A $TV$ tower stands vertically on a bank of 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 anothe point $20\ m$ away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $30^\circ $. Find the height of the ower and the width of the canal.
Answer
Let the tower height AB = h
In right $\triangle\text{ABC},$
$\cot60^\circ=\frac{\text{x}}{\text{h}}=\frac{1}{\sqrt{3}}\Rightarrow\text{x}=\frac{\text{h}}{\sqrt{3}}\dots(\text{i})$
In right $\triangle\text{DAB},$
$\cot30^\circ=\frac{\text{x}+20}{\text{h}}=\sqrt{3}$
$\Rightarrow\text{x}+20=\text{h}\sqrt{3}$
$\Rightarrow\text{x}=\big(\text{h}\sqrt{3}-20\big)\dots(\text{ii})$
From $(i)$ and $(ii)$,
$\frac{\text{h}}{\sqrt{3}}=\big(\text{h}\sqrt{3}-20\big)$
$\Rightarrow2\text{h}=20\sqrt{3}$
$\Rightarrow\text{h}=10\sqrt{3}\text{m}$
Sunstituting $\text{h}=10\sqrt{3}$ in (i) we get x = 10m.
Hence, the height of the canal is $10\sqrt{3}\text{m},$ and the width of the canal is 10m. View full question & answer→Question 154 Marks
The angles of depression of the top and bottom of a tower as seen from the top of a $60\sqrt{3}-\text{m}-\text{high}$ cliff are $45^\circ $ and $60^\circ $ re!pectively. Find the height of the tower.
Answer
Let $AB$ be the diff and $CD$ be the tower, of height h metres.
So, $AE = CD = h$ metres.
Since AB is given to be $60\sqrt{3}\text{m},\text{BE}=\big(60\sqrt{3}-\text{h}\big)\text{m}.$
In right $\triangle\text{BED},$
$\cot45^\circ=\frac{\text{DE}}{\text{BE}}$
$\Rightarrow1=\frac{\text{DE}}{60\sqrt{3}-\text{h}}$
$\Rightarrow\text{DE}=({60}{\sqrt{3}-\text{h}})\text{m}\dots(\text{i})$
In right $\triangle\text{CAB},$
$\cot60^\circ=\frac{\text{CA}}{\text{AB}}$
$\Rightarrow{\frac{1}{\sqrt{3}}}=\frac{\text{CA}}{60\sqrt{3}}$
$\Rightarrow\text{CA}=60\text{m}\dots(\text{ii})$
Since $CA = DE,$
From $(i)$ and $(ii)$,
${60}{\sqrt{3}}-\text{h}=60$
$\Rightarrow\text{h}=60\sqrt{3}-60$
$\Rightarrow\text{h}=60\times1.732-60$
$\Rightarrow\text{h}=43.92\text{m}$
$\Rightarrow\text{h}=15\text{m}$
Hence, the height of the tower is $43.92\ m.$ View full question & answer→Question 164 Marks
From a point on the ground the angles of elevation of the bottom and top of a communication tower fixed on the top of a $20$-m-high building are $45^\circ $ and $60^\circ $ respectively. Find the eight of the tower. $\big[\text{Use}\sqrt{3}=1.732\big]$
Answer
Height of Bulding $= DB = 20m$
$\angle\text{DCB}=45^\circ$
$\angle\text{ACB}=60^\circ$
In $\triangle\text{DBC}\tan45^\circ=1=\frac{\text{DB}}{\text{BC}}$
Or, $\text{AB}=20\sqrt{3}$
Now, $\text{AD}=\text{AB}-\text{DB}$
$=20\sqrt{3}-20=20(\sqrt{3}-1)$
$=20(1.732-1)=20\times0.732=14.64\text{m}$ View full question & answer→Question 174 Marks
A straight highway leads to the foot of a tower. A man standing on 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 secounds later, the angle of depression of the car is fo d to be $60^\circ $. Find the time taken by the car to reach the foot of the tower form this point.
Answer
Let the tower height $AC = h; BD = y; AB = x$
In right angled triangle $\triangle\text{ABC}$ and $\triangle\text{ADC},$
$\cot60^\circ=\frac{\text{x}}{\text{h}}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{x}=\frac{\text{h}}{\sqrt{3}}$
$\cot30^\circ=\frac{\text{x}+\text{y}}{\text{h}}=\sqrt{3}$
$\Rightarrow\text{x}+\text{y}=\sqrt{3}\text{h}$
$\Rightarrow\text{y}=\sqrt{3}\text{h}-\text{x}$
$\Rightarrow\text{y}=\sqrt{3}\text{h}-\frac{\text{h}}{\sqrt{3}}=\frac{2\text{h}}{\sqrt{3}}$
y distance is covered in 6 seconds
Hence, x distance is covered in $\frac{6\text{x}}{\text{y}}\text{secounds}$
$=\frac{6\times\frac{\text{h}}{\sqrt{3}}}{\frac{2\text{h}}{\sqrt{3}}}=3\text{ secounds}$
Hence, the times taken by the car to reach the foot of the tower from this point is $3$ secounds. View full question & answer→Question 184 Marks
The angles of elevation of the top of a tower from two points at distances of $5$ metres and $20$ metrers. from the base of the tower and in the same straight line with it, are complementry. Find the height of the tower.
Answer
Let One angle of elevation be $\theta.$
Since the angles are complementary, the other angle is $(90^\circ-\theta).$
From right $\triangle\text{CAB},$ we have
$\frac{\text{AB}}{\text{AC}}=\tan\theta$
$\Rightarrow\frac{\text{h}}{5}=\tan\theta$
$\Rightarrow\text{h}=5\tan\theta\dots(\text{i})$
From right $\triangle\text{DAB},$ we have
$\frac{\text{AB}}{\text{AD}}=\tan(90^\circ-\theta)$
$\Rightarrow\frac{\text{h}}{20}=\tan(90^\circ-\theta)$
$\Rightarrow\text{h}=20\tan(90^\circ\theta)\dots(\text{ii})$
Multiplying (i) and (ii), we get
$\text{h}^2=100\tan\theta\times\tan(90^\circ-\theta)$
$\Rightarrow\text{h}^2=100\tan\theta\times\cot\theta$
$\Rightarrow\text{h}^2=100$
$\Rightarrow\text{h}=10\text{m}$
Hence, the height of the tower is 10m. View full question & answer→Question 194 Marks
The angle of elevation of the top of a chimney from the foot of a tower is $60^\circ $and the angle of depression of the foot of the chimney forn the top of the tower is $30^\circ $. If the height of the tower is $40$ metres, find the height of the chimney.
According to pollution control norms, the minimum height of a smokeemitting chimney should be $100$ metres. State if the height ,the above mentioned chimney meets the pollution norms. What value is discussed in this question?
Answer
Let $AB$ be the tower and $CD$ be the chimney.
In right $\triangle\text{ACD},$
$\cot60^\circ=\frac{\text{AC}}{\text{CD}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{AC}}{\text{h}}$
$\Rightarrow\text{AC}=\frac{\text{h}}{\sqrt{3}}\dots(\text{i})$
In right $\triangle\text{BAC},$
$\cot30^\circ=\frac{\text{AC}}{\text{AB}}$
$\Rightarrow\sqrt{3}=\frac{\text{AC}}{40}$
$\Rightarrow\text{AC}=40\sqrt{3}$
$\Rightarrow\frac{\text{h}}{\sqrt{3}}=40\sqrt{3}\dots(\text{From}(\text{i}))$
$\Rightarrow\text{h}=120\text{m}$
Hence, the height of the chimney is $120\ m.$
Given that according to pollution control norms, the minimum height of a smoke - emitting chimney should be $100$ metres.
Since $h > 100$, the height meets the pollution norms.
The value discussed is cleanliness and abiduing by rules to avoid polluting the enviroment. View full question & answer→Question 204 Marks
A man observes a car from the to of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from $30^\circ $ to $45^\circ $ in $12$ minutes, find the time taken by the car now to reach the tower.
AnswerConsider the following figure,
Let A be the position of the man and $AB$ be the tower.
Now, consider the triangle, $ABD.$
$\tan30^\circ=\frac{\text{AB}}{\text{BD}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{\text{AB}}{\text{BD}}$
$\Rightarrow\frac{\text{BD}}{\sqrt3}=\text{AB}$
$\Rightarrow\frac{\text{BD + CD}}{\sqrt3}=\text{AB}\ ....(1)$
Consider the triangle $ABC.$
$\tan45^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow1=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\text{AB = BC}\ ....(2)$
Substitute the value of $BC$ from equation $(2)$ in equation $(1)$, we have
$\frac{\text{AB + CD}}{\sqrt3}=\text{AB}$
$\Rightarrow\text{AB + CD}=\sqrt3\text{AB}$
$\Rightarrow\sqrt3\text{AB}-\text{AB}=\text{CD}$
$\Rightarrow\text{AB}\big(\sqrt3-1\big)=\text{CD}\ ....(3)$
Let v m/min be the speed of the car.
Since, the car takes $12$ min to cover the distance $CD$, we have,
$CD = 12v ....(4)$ [Distance = speed × time]
Substitute the value of $CD$ from equation $(4)$ in equation $(3),$
We have,
$\text{AB}\big(\sqrt3-1\big)=12\text{v}$
$\Rightarrow\frac{\text{AB}\big(\sqrt3-1\big)}{12}=\text{v}\ ....(5)$
We need to find the time taken by the car to cover the distance $BC.$
$\text{Time}=\frac{\text{BC}}{\text{v}}$ $\Big[\text{Time}=\frac{\text{Distance}}{\text{Speed}}\Big]$
$\Rightarrow\text{Time}=\frac{12\text{BC}}{\text{AB}\big(\sqrt3-1\big)}$ [From equation $(5)]$
$\Rightarrow\text{Time}=\frac{12\text{AB}}{\text{AB}\big(\sqrt3-1\big)}$ [From equation $(2)]$
$\Rightarrow\text{Time}=\frac{12}{\big(\sqrt3-1\big)}\approx16.4\text{ minutes}$ View full question & answer→Question 214 Marks
The angle of elevation of the top of vertical tower from a point on the ground is $60^\circ $. From another poin $10\ m$ vertically above the first, its angle of elevation is $30^\circ $. Find the height of the tower.
Answer
Let $AB$ be the tower of height, h metres.
In right $\triangle\text{BPQ},$
$\cot30^\circ=\frac{\text{CA}}{\text{PB}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{CA}}{\text{h}}$
$\Rightarrow\text{CA}=\frac{\text{h}}{\sqrt{3}}\dots(\text{i})$
In right $\triangle\text{BED},$
$\cot30^\circ=\frac{\text{DE}}{\text{BE}}$
$\Rightarrow{\sqrt{3}}=\frac{\text{DE}}{(\text{h}-10)}$
$\Rightarrow\text{DE}=\sqrt{3}(\text{h}-10)\dots(\text{ii})$
Since $CA = DE,$
From $(i)$ and $(ii)$,
$\frac{\text{h}}{\sqrt{3}}=\sqrt{3}(\text{h}-10)$
$\Rightarrow\text{h}=3(\text{h}-10)$
$\Rightarrow\text{h}=3\text{h}-30$
$\Rightarrow2\text{h}=30$
$\Rightarrow\text{h}=15\text{m}$
Hence, the height of the tower is $15m.$ View full question & answer→Question 224 Marks
Two poles of equal heights are standing opposite to each other on either side of the road which is $80 \ m$ wide. From a point $P$ between them on the road, the angle of elevation of the top of one pole is $60^{\circ}$ and the angle of depression from the top of another pole at $P$ is $30^{\circ}$. Find the height of each pole and distances of the point $P$ from the poles.
Answer
Let $AB$ and $CD$ be the two poles and h be their height.
In right $\triangle\text{PAB},$
$\frac{\text{AB}}{\text{AP}}=\tan60^\circ$
$\Rightarrow\frac{\text{h}}{\text{x}}=\sqrt{3}$
$\Rightarrow\text{h}=\text{x}\sqrt{3}\text{m}\dots(\text{i})$
In right $\triangle\text{PCD},$
$\frac{\text{CD}}{\text{PC}}=\tan30^\circ$
$\Rightarrow\text{h}=\frac{(80-\text{x})}{\sqrt{3}}\text{m}\dots(\text{ii})$
From $(i)$ and $(ii)$,
$\frac{(80-\text{x})}{\sqrt{3}}=\text{x}\sqrt{3}$
$\Rightarrow80-\text{x}=3\text{x}$
$\Rightarrow4\text{x}=80$
$\Rightarrow\text{x}=20\text{m}$
So, $\text{h}=\text{x}\sqrt{3}\text{m}=20\sqrt{3}\text{m}$
Hence, the height of each pole is $20\sqrt{3}\text{m},$ and the distances of pole $AB$ from the point $P$ is $20m$ and that of pole $CD$ from the point $P$ is $60\ m.$ View full question & answer→Question 234 Marks
Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as $30^\circ $ and $45^\circ $ respectively. If the height of the tower is $50$ metres, find the distance between the two men. $\big[\text{Take}\sqrt{3}=1.732\big]$
Answer
Let the men be at $A$ and $B$ respectively.
In right $\triangle\text{APQ},$
$\frac{\text{PQ}}{\text{AP}}=\tan30^\circ$
$\Rightarrow\frac{50}{\text{AP}}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{AP}=50\sqrt{3}$
$\Rightarrow\text{AP}=86.6\text{m}$
In right $\triangle\text{PQB},$
$\frac{\text{PQ}}{\text{PB}}=\tan45^\circ$
$\Rightarrow\frac{50}{\text{PB}}=1$
$\Rightarrow\text{PB}=50\text{m}$
Now, $\text{AB}=\text{AP}+\text{PB}$
$=86.6+50$
$=136.6\text{m}$
Hence, the distance between the two men is $136.6m.$ View full question & answer→Question 244 Marks
An observer $1.5\ m$ talI is $30\ m$ away from a chimney. the angle of elevation of the top of the chimney from his eye is $60^\circ $. Find the height of the chimney.
Answer
Let $AB$ be the observe and $CD = h$ metres be the tower.
$BE = AC = 30m$
From right $\triangle\text{BED},$ we have
$\frac{\text{DE}}{\text{BE}}=\tan60^\circ$
$\Rightarrow\frac{\text{DE}}{\text{BE}}=\sqrt{3}$
$\Rightarrow\text{DE}=\sqrt{3}\text{ BE}$
$\Rightarrow\text{h}-1.5=\sqrt{3}(30)$
$\Rightarrow\text{h}-1.5=30\sqrt{3}$
$\Rightarrow\text{h}=30\sqrt{3}+1.5\approx53.4\text{m}$
Hence, the height of the chimney is $53.4m.$ View full question & answer→Question 254 Marks
If at some time of the day the ratio of the height of a vertically standing pole to the length of its shadow on the ground is $\sqrt{3}:1$ then find the angle of elevation of the sun at that times.
Answer
Let the height of the tower be $x$ and $y$ the length of the shadow on the ground be $x : y.$
The angle of elevation of the sun from the ground is $\theta.$
We have, $\text{x}:\text{y} =\sqrt{3}:1$
Now, in $\triangle\text{ABC},$
$\tan\theta=\frac{\text{Height}}{\text{Base}}$
$\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\tan\theta=\frac{\sqrt{3}}{1}$
$\therefore\tan\theta=60^\circ$ View full question & answer→Question 264 Marks
A ladder of length $6$ metres makes angle of $45^\circ $ with the floor while leaning against one wall of a room. lf the foot of the ladder is kept fixed on the floor and it is made to lean against the opposite wall of the room, it makes an angle of $60^\circ $ with the floor. Find the distance between two walls of the room.
Answer
Let the floors be $AB$ and $CD$ and $E$ and $E$ and $F$ be the positions of the ladder leaning against each wall.
From right $\triangle\text{MAE},$ we have
$\frac{\text{AM}}{\text{EM}}=\cos60^\circ$
$\Rightarrow\frac{\text{AM}}{6}=\frac12$
$\Rightarrow\text{AM}=3\text{m}$
From right $\triangle\text{MCF},$ we have
$\frac{\text{CM}}{\text{MF}}=\cos45^\circ$
$\Rightarrow\frac{\text{CM}}{6}=\frac{1}{\sqrt{2}}$
$\Rightarrow\text{CM}=\frac{6}{\sqrt{2}}$
On rationalising, we get
$\text{CM}=\frac{6}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{6\sqrt{2}}{2}=3\sqrt{2}\text{m}$
So, $\text{AC}=\text{AM}+\text{CM}$
$=3+3\sqrt{2}$
$=3+(3\times1.414)$
$=7.242\text{m}$
Thus, the distance between the two walls of the room is $7.242m.$ View full question & answer→Question 274 Marks
The angle of elevation of the top of tower at a distance of $120 \ m$ from a point $A$ on the ground is $45^{\circ}$. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is $60^{\circ}$, then find the height of the flagstaff.
$[Use \sqrt{3}=1.732]$
Answer
Let AB is the tower of height h meter and $AC$ is flagstaff of height x meter.
$\angle\text{APB}=45^\circ$ and $\angle\text{BPC}=60^\circ$
$\tan60^\circ=\frac{\text{x}+\text{h}}{120}$
$\Rightarrow\sqrt{3}=\frac{\text{x}+\text{h}}{120}$
$\Rightarrow\text{x}=120\sqrt{3}-\text{h}$
$\tan45^\circ=\frac{\text{h}}{120}$
$\Rightarrow1=\frac{\text{h}}{120}$
$\Rightarrow\text{h}=120$
$\therefore$ height of the flagstaff, x
$=120\sqrt{3}-120$
$=120(\sqrt{3}-1)$
$=120\times0.732$
$=87.8\text{m}$
Hence, the height of the flastaff is$ 87.8\ m.$ View full question & answer→Question 284 Marks
From the top of a vertical tower, the angle of depression of two cars in the same straight line with the base of the tower, at an instant are found to be $45^{\circ}$ and $60^{\circ}$.. If the cars are $100\ m$ apart and are on the same side of the tower, find the height of the tower.
Answer
In the above figure, let $AB$ be the tower and $P$ and $D$ be the positions of the two cars at an instant, observed from A.
Join A and $E.$
The angles of depression are $\angle\text{DAE}$ and $\angle\text{PAE}.$
Given that, $PD = 100m.$
Since AE is parallel to$ BD,$
So,$\angle\text{ADB}=\angle\text{DAE}=45^\circ$ and $\angle\text{APB}=\angle\text{PAE}=60^\circ.$
Join P,D and A,B. We get two right-angled triangles $\triangle\text{ABD}$ and $\triangle\text{ABP}.$
We are to find AB. We use trigonometric ratio tan for both the triangles, using AB as height and BP as base for $\triangle\text{ABP}.$ and AB as height and BD as base for $\triangle\text{ABD}.$
From $\triangle\text{ABD},$
$\tan\angle\text{ADB}=\tan45^\circ=\frac{\text{AB}}{\text{BD}}$
Or, $\text{AB}=\text{BD}$
From $\triangle\text{APB},$
$\tan\angle\text{APB}=\tan60^\circ=\frac{\text{AB}}{\text{BP}}$
Or, $\text{BD}=\text{BP}\sqrt{3}$
$\text{BP}+100=\text{BP}\sqrt{3}$
$\text{BP}(\sqrt{3}-1)=100$
$\text{BP}=\frac{100}{(\sqrt{3}-1)}=136.61\text{m}.$ View full question & answer→Question 294 Marks
The angle of elevation of the top of building from the foot of a tower is $30^{\circ}$. The angle of elevation of the top of the tower from the foot of the building is $60^{\circ}$. If the tower is $60^{\circ}$. If the tower is $60 \ m$ high, Find the height of the building.
Answer
Let AB be the building and $CD$ be the tower.
In right $\triangle\text{ACD},$
$\cot60^\circ=\frac{\text{AC}}{\text{CD}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{x}}{60}$
$\Rightarrow\text{x}=\frac{60}{\sqrt{3}}$
On rationalising, we get
$\text{x}=\frac{60}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=20\sqrt{3}\text{m}$
In right $\triangle\text{BAC},$
$\tan30^\circ=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\text{h}=\frac{\text{x}}{\sqrt{3}}$
$\Rightarrow\text{h}=\frac{20\sqrt{3}}{\sqrt{3}}=20\text{m}$
Hence, the height of the building is $20\ m.$ View full question & answer→Question 304 Marks
As observed from the top of a lighthouse, $100\ m$ above sea level, the angle of depression of a ship, sailing directly towards it, changes from $30^\circ $ to $60^\circ $. Determine the distance travelled by the ship during the period of observation. $\big[\text{Use}\sqrt{3}=1.732\big]$
AnswerLet $AB$ be the light house and let $C$ and $D$ be the positions of the ship.
Let $AD = x, CD = y$
In $\triangle\text{BDA},$
$\frac{\text{x}}{100}=\cot60^\circ$
$\text{x}=\frac{100}{\sqrt{3}}\text{m}$
Similarlly in $\triangle\text{BCA},\frac{\text{x}+\text{y}}{100}=\cot30^\circ$
$\Rightarrow(\text{x}+\text{y})=100\sqrt{3}\text{m}$
$\text{y}=(\text{x}+\text{y})-\text{x}$
$=\Big(100\sqrt{3}-\frac{100}{\sqrt{3}}\Big)\text{m}=\Big(\frac{200}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}\Big)\text{m}$
$=115.46\text{m}$
The distance travelled by the ship during the period of observation $= 115.46\ m.$ View full question & answer→Question 314 Marks
The angle of elevation of the top $Q$ of a vertical tower $P Q$ from a point $X$ on the ground is $60^{\circ}$. At a point $Y, 40 m$ vertically above $X$, the angle of elevation is $45^{\circ}$. Find the height of tower. [Take $\left.\sqrt{3}=1.732\right]$
Answer
Given that $PQ$ is a vertical tower.
Let h be the height of the tower.
In right $\triangle\text{QPX},$
$\tan60^\circ-\frac{\text{PQ}}{\text{XP}}$
$\Rightarrow\sqrt{3}=\frac{\text{h}}{\text{XP}}$
$\Rightarrow\text{XP}=\frac{\text{h}}{\sqrt{3}}\text{m}\dots(\text{i})$
In right $\triangle\text{QRY},$
$\tan45^\circ=\frac{\text{QR}}{\text{YR}}$
$\Rightarrow1=\frac{(\text{h}-40)}{\text{YR}}$
$\Rightarrow\text{YR}=(\text{h}-40)\text{m}\dots(\text{ii})$
Since $XP = YR$, From $(i)$ and $(ii)$, we have
$\frac{\text{h}}{\sqrt{3}}=\text{h}-40$
$\Rightarrow\text{h}=\sqrt{3}\text{h}-40\sqrt{3}$
$\Rightarrow\sqrt{3}\text{h}-\text{h}=40\sqrt{3}$
$\Rightarrow\text{h}=\frac{40\sqrt{3}}{(\sqrt{3}-1)}$
On rationalising we get,
$\text{h}=\frac{40\sqrt{3}}{(\sqrt{3}-1)}\times\frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$\Rightarrow\text{h}=\frac{120+40\sqrt{3}}{2}$
$=40\bigg(\frac{3+\sqrt{3}}{2}\bigg)$
$=20(3+\sqrt{3})=94.6$
Hence, the height of the tower is $94.6\ m.$ View full question & answer→Question 324 Marks
An aeroplane is flying at a height of $300 \ m$ above the ground. Flying at this height the angles of depression on from the two aeoplane of two points on both banks of a river in opposite directions are $45^{\circ}$ and $60^{\circ}$ respectively. Find the width of the river. [ $Use \sqrt{3}=1.732$ ]
Answer
Correct Figure,
$\tan 45^{\circ} = \frac{300}{\text{y}}$
$\Rightarrow 1 = \frac{300}{\text{y}} \text{ or } \text{y} = 300$
$\tan 60^{\circ} = \frac{300}{\text{x}}$
$\Rightarrow \sqrt{3} = \frac{300}{\text{x}} \text{ or } \text{x} = \frac{300}{\sqrt{3}} = 100\sqrt{3}$
Width of river $= 300 + 100 \sqrt{3} = 300 + 173.2$
$= 473.2 \text{ m}$ View full question & answer→Question 334 Marks
From the top of a building $AB , 60 m$ high, the an es of depression of the top and bottom of a vertical lamp-post $CD$ are obseved to be $30^{\circ}$ and $60^{\circ}$ respetively. Find:
i. The horizontal distance between $A B$ and $C D$,
ii. The height of the lamp post,
iii. The difference between the heights of the building and the lamp-post.
Answer
Given that $AB$ is a bulding that is $60\ m$, high.
Let $BD = CE = x$ and $CD = BE = y$
$\Rightarrow\text{AE}=\text{AB}-\text{BE}=60-\text{y}$
- In right $\triangle\text{ACE},$
$\tan30^\circ=\frac{\text{AE}}{\text{CE}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{60-\text{y}}{\text{x}}$
$\Rightarrow\text{x}=60\sqrt{3}-\text{y}\sqrt{3}\dots(\text{i})$
In right $\triangle\text{ACE},$
$\tan60^\circ=\frac{\text{AB}}{\text{BD}}$
$\Rightarrow\sqrt{3}=\frac{60}{\text{x}}$
$\Rightarrow\text{x}=\frac{60}{\sqrt{3}}$
On rationalising we get,
$\text{x}=\frac{60}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow\text{x}=\frac{60\sqrt{3}}{{3}}$
$\Rightarrow\text{x}={20\sqrt{3}}$
$\Rightarrow\text{x}=20\times1.732=34.64\text{m}$
Thus, the horizontal distance between $AB$ and $CD$ is $34.64m.$
- From $(i)$, we get the height of the lamp-post
$= CD = y$
$\text{x}=60\sqrt{3}-\text{y}\sqrt{3}$
$\Rightarrow20\sqrt{3}=60\sqrt{3}=\text{y}\sqrt{3}$
$\Rightarrow20=60-\text{y}$
$\Rightarrow\text{y}=40\text{m}$
Thus, the height of the lamp-post is 40m.
- The difference between of the building and the lamp-post
$= AB - CD$
$= 60 - 40$
$= 20m$ View full question & answer→Question 344 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 $ resbectively. If the bridge is at a height of $2.5\ m$ from the banks, find the width of the river.$\big[\text{Take}\sqrt{3}=1.732\big]$
Answer
Let the width of the river be $AC.$
$D$ is the point on the bridge and $BD = 5m$
In right $\triangle\text{ABD},$
$\Rightarrow\tan45^\circ=\frac{\text{BD}}{\text{AB}}$
$\Rightarrow1=\frac{2.5}{\text{AB}}$
$\Rightarrow\text{AB}=2.5\text{m}$
In right $\triangle\text{CBD},$
$\tan30^\circ=\frac{\text{BD}}{\text{BC}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{2.5}{\text{BC}}$
$\Rightarrow\text{BC}=2.5\sqrt{3}\text{m}$
$\text{AC}=\text{AB}+\text{BC}$
$=2.5+2.5\sqrt{3}$
$=2.5(1+\sqrt{3})$
$=2.5(1+1.732)$
$=6.83\text{m}$
$\therefore$ Width of the river is$ 6.83m.$ View full question & answer→Question 354 Marks
The horizontal distance between two towers is $60$ metres. The angle of depression of the top of the first tower when seen from the top of the second tower is $30^\circ $. If the height of the second tower is $90$ metres, find the height of the first tower. $\big[\text{Use}\sqrt{3}=1.732\big]$
AnswerLet AB and $CD$ be the first and secound towers respectively.
Then, $CD = 90m$ and $AC = 60m.$
Let $DE$ be the horizontal line through $D.$
Draw $\text{BF}\perp\text{CD},$
Then, $\text{BF}=\text{AC}=60\text{m}$
$\angle\text{FBD}=\angle\text{EDB}=30^\circ$
Now, $\frac{\text{FD}}{\text{BF}}=\tan30^\circ-\frac{\text{FD}}{60}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{FD}=\Big(60\times\frac{1}{\sqrt{3}}\Big)\text{m}=20\sqrt{3}\text{m}$
$\therefore\text{AB}=\text{FC}=(\text{CD}-\text{FD})$
$=(90-20\sqrt{3})\text{m}=55.36\text{m}$ View full question & answer→