MCQ 11 Mark
The angle of depression of a car parked on the road from the top of a $150m$ high tower is $30^\circ$ . The distance of the car from the tower $($in metres$)$ is :
- A
$50\sqrt{3}$
- ✓
$150\sqrt{3}$
- C
$150\sqrt{2}$
- D
$75$
AnswerCorrect option: B. $150\sqrt{3}$
We have a high tower whose height $(AC) = 150m$
Angle of depression $= 30^\circ$
Here given angle of depression $\angle\text{DCB}=30^\circ$
We know,
$\angle\text{DCB}=\angle\text{ACB}\ ($Alternate angles$)$
Let,
Distance the car from tower $AB = x m$
In triangle $\text{ABC}$ we know,
$\Rightarrow\tan30^\circ=\frac{\text{AC}}{\text{AB}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{150}{\text{x}}$
$\Rightarrow\text{x}=150\sqrt3\text{m}$
View full question & answer→MCQ 21 Mark
The angle of depression of a car parked on the road from the top of a $150m$ high tower is $30^\circ$ . The distance of the car from the tower $($in metres$)$ is :
- A
$50\sqrt{3}$
- ✓
$150\sqrt{3}$
- C
$150\sqrt{2}$
- D
$75$
AnswerCorrect option: B. $150\sqrt{3}$
We have a high tower whose height $(AC) = 150m$
Angle of depression $= 30^\circ $
Here given angle of depression $\angle\text{DCB}=30^\circ$
We know,
$\angle\text{DCB}=\angle\text{ACB} \ ($Alternate angles$)$
Let,
Distance the car from tower $AB = x m$
In triangle $\text{ABC}$ we know,
$\Rightarrow\tan30^\circ=\frac{\text{AC}}{\text{AB}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{150}{\text{x}}$
$\Rightarrow\text{x}=150\sqrt3\text{m}$
View full question & answer→MCQ 31 Mark
A kite is flying at a height of $30m$ from the ground. The length of string from the kite to the ground is $60m.$ Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is :
- A
$45^\circ$
- ✓
$30^\circ$
- C
$60^\circ$
- D
$90^\circ$
AnswerCorrect option: B. $30^\circ$
Let $AB$ be the height of the kite above the ground.
Now $, AB = 30m$
Let the length of the string $= AC = 60m$
Let $\angle\text{ABC}=\theta=$ angle of elevation
Now, $\sin\theta=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\sin\theta=\frac{30}{60}$
$\Rightarrow\sin\theta=\frac{1}{2}$
$\Rightarrow\sin\theta=\sin30^\circ$
$\Rightarrow\theta=30^\circ$ View full question & answer→MCQ 41 Mark
A tower stands vertically on the ground. From a point on the ground which is $25 m$ away from the foot of the tower, the angle of elevation of the top of the tower is found to be $45^\circ$ . Then the height $($in meters$)$ of the tower is :
- A
$25\sqrt{2}$
- B
$25\sqrt{3}$
- ✓
$25$
- D
$12.5$
Answer
Let the height of the tower be $h.$
So $, AB = h$
Distance of the point from the foot of the tower $= 25m$
Hence $,CB = 25m$
Now,
$\tan\text{C}=\frac{\text{AB}}{\text{CB}}$
$\tan45^\circ=\frac{\text{AB}}{\text{CB}}$
$1=\frac{\text{h}}{25}$
$25=\text{h}$
Or $\text{h}=25$
Hence, height of the tower $= h = 25m.$ View full question & answer→MCQ 51 Mark
A ladder makes an angle of $60^\circ$ with the ground when placed against a wall. If the foot of the ladder is $2 m$ away from the wall, then the length of the ladder $($in metres$)$ is :
- A
$\frac{4}{\sqrt{3}}$
- B
$4\sqrt{3}$
- C
$2\sqrt{2}$
- ✓
$4$
Answer
Distance between foot of the ladder and wall $= 2m$
Angle of elevation $\theta=60^\circ$
Length of the ladder $= L = AC$
Now figure forms a right angle triangle $\text{ABC}$
We know,
$\cos\theta=\frac{\text{adjacent side}}{\text{hypotenuse}}$
$\cos60^\circ=\frac{\text{BC}}{\text{AC}}$
$\frac{1}{2}=\frac{2}{\text{AC}}$
$\text{AC}=4\text{m}$
$\therefore$ Length of ladder $(L) = 4m.$ View full question & answer→MCQ 61 Mark
A ladder makes an angle of $60^\circ$ with the ground when placed against a wall. If the foot of the ladder is $2m$ away from the wall, then the length of the ladder $($in metres$)$ is :
- A
$\frac{4}{\sqrt{3}}$
- B
$4\sqrt{3}$
- C
$2\sqrt{2}$
- ✓
$4$
Answer
Distance between foot of the ladder and wall $= 2m$
Angle of elevation $\theta=60^\circ$
Length of the ladder $= L = AC$
Now figure forms a right angle triangle $\text{ABC}$
We know,
$\cos\theta=\frac{\text{adjacent side}}{\text{hypotenuse}}$
$\cos60^\circ=\frac{\text{BC}}{\text{AC}}$
$\text{AC}=4\text{m}$
$\therefore$ Length of ladder $(L) = 4m.$ View full question & answer→MCQ 71 Mark
The angle of depression of a car, standing on the ground, from the top of a $75 m$ high tower, is $30^{\circ}$. The distance of the car from the base of the tower $($in $m.)$ is :
- ✓
$25\sqrt{3}$
- B
$50\sqrt{3}$
- C
$75\sqrt{3}$
- D
$150$
AnswerCorrect option: A. $25\sqrt{3}$
Given that the angle of depression is $30^\circ ,$ then angle of elevation is $(90 - 30 = 60^\circ ).$
Therefore,
We have,
$\tan60^{\circ} =\frac{\text{height of tower}}{\text{distance of the car from the base of the tower}} $
$\Rightarrow\sqrt{3} = \frac{75}{\text{distance of the car from the base of the tower}}$
$\Rightarrow$ distance of the car from the base of the tower $=\frac{75}{\sqrt{3}} $ or $\frac{75\sqrt{3}}{(\sqrt{3}\sqrt{3})}=\frac{25}{\sqrt{3}}$
View full question & answer→MCQ 81 Mark
The length of shadow of a tower on the plane ground is $\sqrt{3}$ times the height of the tower. The angle of elevation of sun is :
- A
$45^\circ$
- ✓
$30^\circ$
- C
$60^\circ$
- D
$90^\circ$
AnswerCorrect option: B. $30^\circ$
Given that the length of shadow of the tower on the plane ground is $\sqrt3$ times the height of the tower.
Let $\theta$ be the angle of elevation.
From the given statement, $\text{BC}=\sqrt3\times\text{AB}$
$\Rightarrow\frac{\text{BC}}{\text{AB}}=\sqrt3$
$\Rightarrow\frac{\text{BC}}{\text{h}}=\cot\theta$
$\Rightarrow\cot\theta=\frac{\text{BC}}{\text{h}}=\sqrt3$
$\Rightarrow\cot\theta=\cot30^\circ$
$\Rightarrow\theta=30^\circ$ View full question & answer→MCQ 91 Mark
The angle of elevation of the top of a tower from a point on the ground. which is $30m$ away from the foot of the tower is $45^{\circ}$. The height of the tower $($in metres$)$ is :
- A
$15$
- B
$30$
- ✓
$30\sqrt{3}$
- D
$10\sqrt{3}$
AnswerCorrect option: C. $30\sqrt{3}$
Given :
Distance of tower from point $C$
Hence $, BC = 30\ cm$
Angle of elevation $\angle\text{ACB}=45^\circ$
$\tan\text{C}=\frac{\text{AB}}{\text{BC}}$
$\tan45^\circ=\frac{\text{AB}}{30}$
$1=\frac{\text{AB}}{30}$
$\text{AB}=30\text{m}$ View full question & answer→MCQ 101 Mark
The angle of elevation and the angle of depression from an object on the ground to an object in the air are related as :
AnswerThe angle of elevation and the angle of depression from an object on the ground to an object in the air are related as equal if the height of objects are the same.
View full question & answer→MCQ 111 Mark
A pole casts a shadow of length $2\sqrt{3}\text{m}$ on the ground when the sun's elevation is $60^\circ$ . The height of the pole is :
- ✓
$6\text{m}$
- B
$4\sqrt{3}\text{m}$
- C
$12\text{m}$
- D
$3\text{m}$
AnswerCorrect option: A. $6\text{m}$
Let the height of the pole be $h$ metres.
Then, $\frac{\text{h}}{2\sqrt3}=\tan60^\circ=\sqrt{3}$
$\Rightarrow\text{h}=(2\sqrt{3}\times\sqrt{3})=6.$
View full question & answer→MCQ 121 Mark
The angle of elevation of the top of a tower from a point on the ground $30\ m$ away from the foot of the tower is $30^\circ$ . The height of the tower is :
- A
$30\text{m}$
- ✓
$10\sqrt{3}\text{m}$
- C
$20\text{m}$
- D
AnswerCorrect option: B. $10\sqrt{3}\text{m}$
Let $AB$ be the tower and $O$ be the point of observation.
Also $,\angle\text{AOB}=30^\circ,$ and $OB = 30m$
Let : $AB = h \ m$
In $\triangle\text{AOB},$
We have :
$\tan30^\circ=\frac{\text{AB}}{\text{OB}}$
$\Rightarrow\frac{1}{\sqrt{3}}$
$\Rightarrow\frac{\text{h}}{30}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{h}=\frac{30}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$=\frac{30\sqrt{3}}{3}=10\sqrt{3}\text{m}$
Hence, the height of the tower is $10\sqrt{3}\text{m}.$ View full question & answer→MCQ 131 Mark
The angle of elevation of the top of a hill at the foot of a tower is $60^\circ$ and the angle of elevation of the top of the tower from the foot of the hill is $30^\circ$ . If the tower is $50m$ high, then the height of the hill is :
- ✓
$150\text{m}$
- B
$50\sqrt3\text{m}$
- C
$150\sqrt3\text{m}$
- D
$100\sqrt3\text{m}$
AnswerCorrect option: A. $150\text{m}$
Let the height of the hill be $h m$.
$\tan60^\circ=\frac{\text{DC}}{\text{AC}}$
$\Rightarrow\sqrt3=\frac{\text{h}}{\text{AC}}$
$\Rightarrow\sqrt3=\frac{\text{h}}{\sqrt3}\text{m }...(\text{i})$
In right triangle $\text{ABC},$
$\tan30^\circ=\frac{50}{\text{AC}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{50}{\text{AC}}$
$\Rightarrow\text{AC}=50\sqrt3\text{m }...(\text{ii})$ From eq. $(i)$ and $(ii),$
$50\sqrt3=\frac{\text{h}}{\sqrt3}$
$\Rightarrow\text{h}=50\times\sqrt3\times\sqrt3=150\text{m}$ View full question & answer→MCQ 141 Mark
If the height of the tower is $\sqrt3$ times of the length of its shadow, then the angle of elevation of the sun is :
- A
$30^\circ$
- ✓
$60^\circ$
- C
$45^\circ$
- D
$15^\circ$
AnswerCorrect option: B. $60^\circ$
Let the lenght of the shadow be $x$ meters.
Then the height of the tower be $\sqrt{3\text{x}}\text{ meter}.$
$\therefore\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{\sqrt{3\text{x}}}{\text{x}}$
$\Rightarrow\tan\theta=\sqrt3$
$\Rightarrow\tan\theta=\tan60^\circ$
$\Rightarrow\theta=60^\circ$ View full question & answer→MCQ 151 Mark
If the length of the shadow of a tower is $\sqrt3$ times that of its height, then the angle of elevation of the sun is :
- A
$45^\circ$
- B
$60^\circ$
- C
$75^\circ$
- ✓
$30^\circ$
AnswerCorrect option: D. $30^\circ$
Let $AB$ be the tree and $AP$ be the shadow.
Let $AB = x$ meters.
Then $\text{AP}=\text{x}\sqrt3\text{ meters}$
Also, $\angle\text{APB}=\theta$
In right angled triangle $\text{ABP}$
$\tan\theta=\frac{\text{AB}}{\text{AP}}$
$\Rightarrow\tan\theta=\frac{\text{x}}{\text{x}\sqrt3}=\frac{1}{\sqrt3}$
$\Rightarrow\tan\theta=\tan30^\circ$
$\Rightarrow\theta=30^\circ$
Therefore, the angle of elevation of the Sun is $30^\circ $. View full question & answer→MCQ 161 Mark
If the angle of elevation of a cloud from a point $200m$ above a lake is $30^\circ$ and the angle of depression of its reflection in the lake is $60^\circ,$ then the height of the cloud above the lake is :
- A
$200m$
- B
$500m$
- C
$30m$
- ✓
$400m$
AnswerCorrect option: D. $400m$
Let $AB$ be the surface of the lake and $P$ be the point of observation.
So $AP = 60m$.
The given situation can be represented as,
Here $, C$ is the position of the cloud and $C\ '$ is the reflection in the lake.
Then $CB = C'B.$
Let PM be the perpendicular from $P$ on $CB.$
Then $\angle\text{CPM}=30^\circ$ and $\angle\text{C}'\text{PM}=60^\circ.$
Let $CM = h, PM = x,$ then $CB = h + 200$ and $C'B = h + 200$
Here, we have to find the height of cloud.
So we use trigonometric ratios.
In $\triangle\text{CMP,}$
$\Rightarrow\ \tan30^\circ=\frac{\text{CM}}{\text{PM}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\sqrt{3}\text{h}$
Again in $\triangle\text{PMC}\ ',$
$\Rightarrow\ \tan60^\circ=\frac{\text{C}'\text{M}}{\text{PM}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{C}'\text{B}+\text{BM}}{\text{PM}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}+200+200}{\text{x}}$
$\Rightarrow\ \sqrt{3}\text{x}=\text{h}+400$
Put, $\text{x}=\sqrt{3}\text{h}$
$\Rightarrow 3h = h + 400$
$\Rightarrow 2h = 400$
$\Rightarrow h = 200$
Now,
$\Rightarrow CB = h + 200$
$\Rightarrow CB = 200 + 200$
$\Rightarrow CB = 400m$ View full question & answer→MCQ 171 Mark
A pole $6m$ high casts a shadow $2\sqrt3\text{m}$ long on the ground, then the sun’s elevation is :
- ✓
$60^\circ$
- B
$30^\circ$
- C
$45^\circ$
- D
$75^\circ$
AnswerCorrect option: A. $60^\circ$
Let height $= 6m$ lenght of shadow $=2\sqrt{3\text{m}}$
$\theta$ is angle of elevation $\tan\theta=\frac{\text{(height)}}{\text{(Shadow length)}}$
$=\frac{6}{2}\sqrt3=\sqrt3$
$\theta=\frac{\pi}{3}$
Angle of inclination is $= 60^\circ .$
View full question & answer→MCQ 181 Mark
A pole of height $60m$ has a shadow of length $20\sqrt3\text{m}$ at a particular instant of time. The angle of elevation of the sun at this point of time.
- A
$30^\circ$
- B
$60^\circ$
- ✓
$45^\circ$
- D
AnswerCorrect option: C. $45^\circ$
Let the height of the pole $AB = 60m,$
The lenght of the shadow $\text{B}=20\sqrt3\text{m}$ and angle of elevation be $\theta$
$\therefore\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{60}{20\sqrt3}$
$\Rightarrow\tan\theta=\frac{3}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$\Rightarrow\tan\theta=\sqrt3$
$\Rightarrow\tan\theta=\tan60^\circ$
$\Rightarrow\theta=60^\circ$ View full question & answer→MCQ 191 Mark
A tree $12m$ high is broken by the wind in such a way that its top touches the ground and makes an angle $30^\circ$ with the ground. The height at which from the bottom the tree is broken by the wind is :
Answer
Let tree broke from the height of $x$ from point $A,$ then length of the broken tree be $(12 - x)$ meters and angle of elevation $= \theta=30^\circ$
In triangle $\text{ABC} , \sin30^\circ=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\frac{1}{2}=\frac{\text{x}}{12-\text{x}}$
$\Rightarrow2\text{x}=12-\text{x}$
$\Rightarrow3\text{x}=12$
$\Rightarrow\text{x}=4\text{m}$ View full question & answer→MCQ 201 Mark
A ladder $15m$ long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the wall, then the height of the wall is :
AnswerCorrect option: C. $\frac{15}{2}\text{m}$
Suppose $AB$ is the wall and $AC$ is the ladder.
It is given that, $AC = 15m$ and $\angle\text{CAB}=60^\circ.$
In right $\triangle\text{ABC,}$
$\cos60^\circ=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\frac{1}{2}=\frac{\text{AB}}{15}$
$\Rightarrow\text{AB}=\frac{15}{2}\text{m}$
Thus, the height of the wall is $\frac{15}{2}\text{m.}$ View full question & answer→MCQ 211 Mark
The tops of two towers of heights $x$ and $y,$ standing on a level ground subtend angles of $30^\circ$ and $60^\circ,$ respectively at the centre of the line joining their feet. Then, $x : y$ is :
- A
$1 : 2$
- B
$2 : 1$
- ✓
$1 : 3$
- D
$3 : 1.$
AnswerCorrect option: C. $1 : 3$
Let $AB$ and $CD$ be the two towers such that $AB = x$ and $CD = y.$
We have,
$\angle\text{AEB}=30^\circ,\angle\text{CED}=60^\circ$ and $BE = DE$
In $\triangle\text{ABE},$
$\tan30^\circ=\frac{\text{AB}}{\text{BE}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{x}}{\text{BE}}$
$\Rightarrow\text{BE}=\text{x}{\sqrt{3}}$
Also, in $\triangle\text{CDE},$
$\Rightarrow\tan60^\circ=\frac{\text{CD}}{\text{DE}}$
$\Rightarrow\sqrt{3}=\frac{\text{y}}{\text{DE}}$
$\Rightarrow\text{DE}=\frac{\text{y}}{\sqrt{3}}$
As $, BE = DE$
$\Rightarrow\text{x}\sqrt{3}=\frac{\text{y}}{\sqrt{3}}$
$\Rightarrow\frac{\text{x}}{\text{y}}=\frac{1}{\sqrt{3}\times\sqrt{3}}$
$\Rightarrow\frac{\text{x}}{\text{y}}=\frac13$
$\therefore\ \text{x}:\text{y}=1:3$ View full question & answer→MCQ 221 Mark
In a rectangle, the angle between a diagonal and a side is $30^\circ$ and the length of this diagonal is $8\ cm.$ the area of the rectangle is :
AnswerCorrect option: C. $16\sqrt{3}\text{ cm}^2$
Let $\text{ABCD}$ be the rectangle in which $\angle\text{BAD}=30^\circ$ and $AC = 8\ cm.$
In $\triangle\text{BAC},$
We have :
$\frac{\text{AB}}{\text{AC}}=\cos30^\circ=\frac{\sqrt{3}}{2}$
$\Rightarrow\frac{\text{BC}}{8}=\frac{1}{2}$
$\Rightarrow\text{BC}=\frac82=4\text{m}$
$\therefore$ Area of the rectangle $=(\text{AB}\times\text{BC})$
$=(4\sqrt{3}\times4)=16\sqrt{3} \ \text{cm}^2$ View full question & answer→MCQ 231 Mark
From the top of a cliff $20m$ high, the angle of elevation of the top of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is :
Answer
Let $AB$ be the cliff and $CD$ be the tower.
We have,
$AB = 20m$
Also, $CE = AB = 20m$
Let $\angle\text{ACB}=\angle\text{CAE}=\angle\text{DAE}=\theta$
In $\triangle\text{ABC},$
$\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{20}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{20}{\text{AE}}\ ($As, $BC = AE)$
$\Rightarrow\text{AE}=\frac{20}{\tan\theta}\dots(\text{i})$
Also, in $\triangle\text{ADE},$
$\tan\theta=\frac{\text{DE}}{\text{AE}}$
$\Rightarrow\tan\theta=\frac{\text{DE}}{\Big(\frac{20}{\tan\theta}\Big)}\ [$Using $(i)]$
$\Rightarrow\tan\theta=\frac{\text{DE}\times\tan\theta}{20}$
$\Rightarrow\text{DE}=\frac{20\times\tan\theta}{\tan\theta}$
$\Rightarrow\text{DE}=20\text{m}$
Now, $\text{CD} = \text{DE} + \text{CE}$
$=20+20$
$\therefore\ \text{CD}=40\text{m}$
Disclaimer : The answer given in the textbook is incorrect,
The same has been rectified above. View full question & answer→MCQ 241 Mark
The angle of elevation of the top of a tower at a point on the ground $50m$ away from the foot of the tower is $45^\circ$ . Then the height of the tower $($in metres$)$ is :
- ✓
$50$
- B
$50\sqrt{3}$
- C
$\frac{50}{\sqrt{2}}$
- D
$\frac{50}{\sqrt{3}}$
AnswerLet $AB$ be tower and $C$ is as a point on the ground $52m$ away
From foot of tower $B$
Angle of elevation is $45^\circ$
let $h$ be height of tower $= x m$
$\therefore\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan45^\circ=\frac{\text{AB}}{5}$
$\Rightarrow1=\frac{\text{AB}}{50}$
$\Rightarrow=50\text{m}$ View full question & answer→MCQ 251 Mark
If altitude of the sun is $60^\circ,$ the height of a tower which casts a shadow of length $30m$ is :
- A
$10\sqrt3\text{m}$
- B
$15\sqrt3\text{m}$
- C
$20\sqrt3\text{m}$
- ✓
$30\sqrt3\text{m}$
AnswerCorrect option: D. $30\sqrt3\text{m}$
Given : Angle of elevation $\theta=60^\circ$
Let the height of the tower $AB$ be $h$ meters.
And the lenght of the shodow $BC$ is $30$ meters.
$\therefore\tan60^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\sqrt3=\frac{\text{h}}{30}$
$\Rightarrow\text{h}=30\sqrt3\text{ meters}.$
Therefore the height of the tower is $30\sqrt3\text{ meters}.$ View full question & answer→MCQ 261 Mark
The shadow of a tower on a level plane is found to be $60$ metres longer when the sun's altitude is $30^\circ$ than that when it is $45^\circ$ . The height of the tower in metres is :
- ✓
$30(\sqrt{3}+1)$
- B
$30(\sqrt{3}-1)$
- C
$30(3+\sqrt{3})$
- D
$30(3-\sqrt{3})$
AnswerCorrect option: A. $30(\sqrt{3}+1)$
$30(\sqrt{3}+1)$
View full question & answer→MCQ 271 Mark
A circus artist is climbing a long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. The ratio of the height of the pole to the length of the string is $ 1 : \sqrt2.$ The angle made by the rope with the ground level is :
- ✓
$45^\circ$
- B
$60^\circ$
- C
$30^\circ$
- D
AnswerCorrect option: A. $45^\circ$
Let the height of the pole $AB$ be $x$ meters and length of string $AC$ be $\sqrt2\text{x}\text{ meters}.$
Angle of elevation $=\theta$
$\therefore\sin\theta=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\sin\theta=\frac{\text{x}}{\sqrt2\text{x}}$
$\Rightarrow\sin\theta=\frac{1}{\sqrt2}$
$\Rightarrow\sin\theta=\sin45^\circ$
$\Rightarrow\theta=45^\circ$ View full question & answer→MCQ 281 Mark
A tower subtends an angle of $30^\circ$ at a point on the same level as its foot. At a second point hmetres above the first, the depression of the foot of the tower is $60^\circ$ . The height of the tower is :
- A
$\frac{\text{h}}{2}\text{m}$
- B
$\sqrt{3}\text{h m}$
- ✓
$\frac{\text{h}}{3}\text{m}$
- D
$\frac{\text{h}}{\sqrt{3}}\text{m}$
AnswerCorrect option: C. $\frac{\text{h}}{3}\text{m}$
Let $AB$ be the tower and $C$ is a point on the same level as its foot such that $\angle\text{ACB}=30^\circ$
The given situation can be represented as,
Here $D$ is a point $h m$ above the point $C$.
In $\angle\text{BCD,}$
$\Rightarrow\ \tan\text{B}=\frac{\text{CD}}{\text{CB}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{h}}{\text{CB}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}}{\text{CB}}$
$\Rightarrow\ \text{CB}=\frac{\text{h}}{\sqrt{3}}\ .....(1)$
Again in triangle $\text{ABC},$
$\tan\text{C}=\frac{\text{AB}}{\text{CB}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{AB}}{\big(\frac{\text{h}}{\sqrt{3}}\big)} \ [$Using $(1)]$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{AB}}{\big(\frac{\text{h}}{\sqrt{3}}\big)}$
$\Rightarrow\ \text{AB}=\frac{\text{h}}{3}$ View full question & answer→MCQ 291 Mark
The string of a kite is $100m$ long and it makes an angle of $60^\circ$ with the horizontal. If these is no slack in the string, the height of the kite from the ground is :
- ✓
$50\sqrt{3}\text{m}$
- B
$100\sqrt{3}\text{m}$
- C
$50\sqrt{2}\text{m}$
- D
AnswerCorrect option: A. $50\sqrt{3}\text{m}$
Let $AB$ be the string of the kite and $AX$ br the horizontal line.
If $\text{BC}\perp\text{AX},$ then $AB = 100m$ and $\angle\text{BAC}=60^\circ.$
Le t:
$BC = h m$
In the right $\triangle\text{ACB},$
We have :
$\frac{\text{BC}}{\text{AB}}=\sin60^\circ=\frac{\sqrt{3}}{2}$
$\Rightarrow\frac{\text{h}}{100}=\frac{\sqrt{3}}{2}$
$\Rightarrow\text{h}=\frac{100\sqrt{3}}{2}$
$=50\sqrt{3}\text{m}$
Hence, the height of the kite is $=50\sqrt{3}\text{m}.$ View full question & answer→MCQ 301 Mark
An electric pole is tied from the top to a point $($some distance away from the base$)$ on the ground using a string. The ratio of the height of pole to the string is $\sqrt3 : 22, $ then the angle of elevation of the top from the point on the ground is :
- A
$45^\circ$
- ✓
$60^\circ$
- C
$30^\circ$
- D
AnswerCorrect option: B. $60^\circ$
Let the height of the pole $\text{AB}=\sqrt3\text{x}\text{ meters}$ and length of the string $AC = 2x$ meters.
An angle of elevation be $\theta$
$\therefore\sin\theta=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\sin\theta=\frac{\sqrt3\text{x}}{2\text{x}}$
$\Rightarrow\sin\theta=\frac{\sqrt3}{2}$
$\Rightarrow\sin\theta=\sin60^\circ$
$\Rightarrow\theta=60^\circ$ View full question & answer→MCQ 311 Mark
From a light house the angles of depression of two ships on opposite sides of the light house are observed to be $30^\circ$ and $45^\circ$ . If the height of the light house is $h$ metres, the distance between the ships is :
- ✓
$(\sqrt{3}+1)\text{h meters}$
- B
$(\sqrt{3}-1)\text{h meters}$
- C
$\sqrt{3}\text{h meters}$
- D
$1+\Big(1+\frac{1}{\sqrt{3}}\Big)\text{h meters}$
AnswerCorrect option: A. $(\sqrt{3}+1)\text{h meters}$
Let $AB$ be light house and $P$ and $Q$ are two ships on its opposite sides which form angle of elevation of $A$ as $45^\circ$ and $30^\circ$ respectively $AB = h.$
Let $PB = x$ and $QB = y$
Now in right $\triangle\text{APB,}$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{AB}}{\text{PB}}$
$\Rightarrow\ \tan45^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ 1=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\text{h}\ .....(\text{i})$
Similarly in right $\triangle\text{AQB,}$
$\tan30^\circ=\frac{\text{AB}}{\text{QB}}=\frac{\text{h}}{\text{y}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{y}}$
$\Rightarrow\ \text{y}=\sqrt{3}\text{h}\ ......(\text{ii})$
Adding $(i)$ and $(ii)$
$\text{PQ}=\text{x}+\text{y}=\text{h}+\sqrt{3}\text{h}$
$(\sqrt{3}+1)\text{h meters}$ View full question & answer→MCQ 321 Mark
The angle of elevation of the top of a tower from a point on the ground and at a distance of $30m$ from its foot is $30^\circ$ . The height of the tower is :
- A
$30\sqrt{3}\text{m}$
- B
$10\text{m}$
- ✓
$10\sqrt{3}\text{m}$
- D
$30\text{m}$
AnswerCorrect option: C. $10\sqrt{3}\text{m}$
In right triangle $\text{ABC},$
$\tan30^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{AB}}{30}$
$\text{AB}=\frac{30}{\sqrt{3}}\text{m}$
$\Rightarrow\frac{30}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$=10\sqrt{3}\text{m}$
Hence the height of the tower is $10\sqrt{3}\text{ meters}.$ View full question & answer→MCQ 331 Mark
The measure of the angle of elevation of the top of a tower $75\sqrt3\text{m}$ high from a point at a distance of $75m$ from the foot of the tower in a horizontal plane is :
- ✓
$60^\circ$
- B
$45^\circ$
- C
$15^\circ$
- D
$30^\circ$
AnswerCorrect option: A. $60^\circ$
Given : distance from a point to the foot of the tower $= 75m$ and the height of the tower $=75\sqrt3\text{m}$
$\therefore\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{75\sqrt3}{75}$
$\Rightarrow\tan\theta=\sqrt3$
$\Rightarrow\tan\theta=\tan60^\circ$
$\Rightarrow\theta=60^\circ$
View full question & answer→MCQ 341 Mark
A vertical stick $20\ cm$ long casts a shadow $15\ cm$ long. At the same time, a tower casts a shadow $30m$ long. The height of the tower is :
Answer
Let the height of the vertical stick be $AB = 20\ cm,$ the length of the shadow of the stick be $BC = 15\ cm$ and angle of elevation $\theta$
At the same time the height of the tower be $h$ meters and the shadow of the tower $= QR = 30\ cm$ and the angle of elevation.
$\angle\text{PRQ}=\theta$
Now, in triangle $\text{ABC},$
$\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{20}{15}=\frac{4}{3}.....(\text{i})$
Again in triangle $\text{PQR},$
$\tan\theta=\frac{\text{PR}}{\text{QR}}$
$\Rightarrow\frac{4}{3}=\frac{\text{h}}{30}\ [$From eq. $(i)]$
$\Rightarrow\text{h}=\frac{30\times4}{3}=40\text{ meters}$ View full question & answer→MCQ 351 Mark
If the height of a tower is half the height of the flagstaff on it and the angle of elevation of the top of the tower as seen from a point on the ground is $30^\circ ,$ then the angle of elevation of the top of the flagstaff as seen from the same point is :
- A
$30^\circ$
- ✓
$60^\circ$
- C
$45^\circ$
- D
AnswerCorrect option: B. $60^\circ$
Here height of the tower $=\text{CD}=\frac{\text{h}}{2}\text{ meters},$ height of the flagstaff $= AD = h$ meters, angle of elevation of top of the tower $=\angle\text{DBC}=30^\circ$ and angle of elevation of the top of the flagstaff from ground $\angle\text{ABC}=\theta$
Now, in triangle $\text{DBC},$
$\tan30^\circ=\frac{\frac{\text{h}}{2}}{\text{x}}$
$\Rightarrow\text{x}=\frac{\text{h}\sqrt3}{2}.....(\text{i})$
Again, in triangle $\text{ABC},$
$\tan\theta=\frac{\text{h}+\frac{\text{h}}{2}}{\text{x}}$
$\Rightarrow\tan\theta=\frac{3\text{h}}{2\text{x}}$
$\Rightarrow\tan\theta=\frac{3\text{h}\times2}{2\times\text{h}\sqrt3} \ [$From eq.$(i)] x$
$\Rightarrow\tan\theta=\frac{3}{\sqrt3}$
$\Rightarrow\tan\theta=\sqrt3$
$\Rightarrow\tan\theta=\tan60^\circ$
$\Rightarrow\theta=60^\circ$ View full question & answer→MCQ 361 Mark
lf the shadow of a tower is $\sqrt{3}$ times of its height, the altitude of the sun is :
- A
$15^\circ$
- ✓
$30^\circ$
- C
$45^\circ$
- D
$60^\circ$
AnswerCorrect option: B. $30^\circ$
Let the height of tower be $h.$
$\text{tan}\theta =\frac{\text{h}}{\sqrt{\text{3h}}}$
$\text{tan}\theta =\frac{\text{1}}{\sqrt{\text{3h}}}$
$\Rightarrow \theta = 30^\circ $
View full question & answer→MCQ 371 Mark
The angles of elevation of top of a pole from two points $A$ and $B$ on the horizontal line lying on opposite side of the pole are observed to be $30^\circ$ and $60^\circ$ If $AB = 100m,$ then height of the pole is :
- A
$20\sqrt{3}\text{m}$
- B
$15\sqrt{3}\text{m}$
- C
$10\sqrt{3}\text{m}$
- ✓
$25\sqrt{3}\text{m}$
AnswerCorrect option: D. $25\sqrt{3}\text{m}$
$25\sqrt{3}\text{m}$
View full question & answer→MCQ 381 Mark
The tops of two poles of height $16m$ and $10m$ are connected by a wire of length $l$ metres. If the wire makes an angle of $30^\circ$ with the horizontal, then $l =$
AnswerLet $AB$ and $CD$ be the poles such that $AB = 16m$ and $CD = 10m.$
The given information can be represented as,
Here $, AC$ is the length of wire which is $l$.
Also $, AE = AB - BE $
$= 16m - 10m = 6m$
We have to find the length of wire $l$.
So we use trigonometric ratios.
In triangle $\text{ACE},$
$\sin\text{C}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\ \sin30^\circ=\frac{6}{\text{l}}$
$\Rightarrow\ \frac{1}{2}=\frac{6}{\text{l}}$
$\Rightarrow\ \text{l}=12\text{m}$ View full question & answer→MCQ 391 Mark
The horizontal distance between two towers is $60m$ and angular depression of the top of the first as seen from the second, which is $150m$ in height, is $30^\circ $. The height of the first tower is :
- A
$\big(150+20\sqrt{3}\big)\text{m}$
- B
$\big(150+15\sqrt{3}\big)\text{m}$
- C
$\big(150-20\sqrt{5}\big)\text{m}$
- ✓
$\big(150-20\sqrt{3}\big)\text{m}$
AnswerCorrect option: D. $\big(150-20\sqrt{3}\big)\text{m}$
$\big(150-20\sqrt{3}\big)\text{m}$
View full question & answer→MCQ 401 Mark
The $...........$ of an object is the angle formed by the line of sight with the horizontal when the object is above the horizontal level.
AnswerThe angle of elevation of an object is the angle formed by the line of sight with the horizontal when the object is above the horizontal level.
View full question & answer→MCQ 411 Mark
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height $5m$. From a point on the plane the angles of elevation of the bottom and top of the flagstaff are respectively $30^\circ$ and $60^\circ $. The height of the tower is :
AnswerCorrect option: B. $2.5m$
Here Height of the tower $= CD = h $ meters, height of the flagstaff $= AD = 5$ meters,
Angle of elevation of top of the tower $=\angle\text{DBC}=30^\circ$ and angle of elevation of the top of the flagstaff from ground $=\angle\text{ABC}=60^\circ$
Now, in triangle $\text{DBC},$
$\tan30^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\text{x}=\text{h}\sqrt3....(\text{i})$
And $\tan60^\circ=\frac{\text{h}+5}{\text{x}}$
$\Rightarrow\sqrt3=\frac{\text{h}+5}{\text{x}}$
$\Rightarrow\sqrt3=\frac{\text{h}+5}{\text{x}}....(\text{ii})$ View full question & answer→MCQ 421 Mark
If the height of a vertical pole is $\sqrt{3}$ times the length of its shadow on the ground, then the angle of elevation of the sun at that time is :
- A
$30^\circ$
- ✓
$60^\circ$
- C
$45^\circ$
- D
$75^\circ$
AnswerCorrect option: B. $60^\circ$
Let $AB$ be a vertical pole and let its shadow be $BC$
Let $BC = x m,$ then length of pole $=\sqrt{3}\text{x},$
$\theta$ be the angle of elevation
$\therefore\tan\theta=\frac{\text{AB}}{\text{BC}}$
$=\frac{\sqrt{3}\text{x}}{\text{x}}=\sqrt{3}$
$=\tan60^\circ$
$\therefore\theta=60^\circ$ View full question & answer→MCQ 431 Mark
The upper part of a tree broken by the windfalls to the ground without being detached. The top of the broken part touches the ground at an angle of $30^\circ$ at a point $8m$ from the foot of the tree. The original height of the tree is :
- A
$8\text{m}$
- ✓
$8\sqrt3\text{m}$
- C
$24\sqrt3\text{m}$
- D
$24\text{m}$
AnswerCorrect option: B. $8\sqrt3\text{m}$
In right triangle $\text{ABC},$
$\cos30^\circ=\frac{\text{BC}}{\text{AC}}$
$\Rightarrow\frac{\sqrt3}{2}=\frac{8}{\text{AC}}$
$\Rightarrow\text{AC}=\frac{16}{\sqrt3}\text{m}$
Again, $\tan30^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{\text{AB}}{8}$
$\Rightarrow\text{AB}=\frac{8}{\sqrt3}\text{m}$
$\therefore$ Height of the tree $=\text{AB}+\text{AC}$
$=\frac{8}{\sqrt3}+\frac{16}{\sqrt3}=\frac{24}{\sqrt3}$
$=\frac{24}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=8\sqrt3\text{m}$
The height of the tree is $8\sqrt3\text{m}.$ View full question & answer→MCQ 441 Mark
The $...........$ is the angle between the horizontal and the line of sight to an object when the object is below the horizontal level.
AnswerThe angle of depression is the angle between the horizontal and line of sight to an object when the object is below the horizontal level.
The angle of depression is formed when the observer is higher than the object he is looking at.
It is the angle between the horizontal line and the line joining observer's eye and the object.
It plays very important role in determining the heights and distances.
View full question & answer→MCQ 451 Mark
If the angle of depression of a car from a $100m$ high tower is $45^\circ,$ then the distance of the car from the tower is :
- A
$200\sqrt3\text{m}$
- B
$200\text{m}$
- C
$100\sqrt3\text{m}$
- ✓
$100\text{m}$
AnswerCorrect option: D. $100\text{m}$
Let the distance of the car from the tower be $x$ meters.
$\tan45^\circ=\frac{\text{AC}}{\text{BC}}$
$\Rightarrow1=\frac{100}{\text{x}}\text{m}$
$\text{x}=100\text{m}$
Therefore, the distance of the car from the tower is $100m.$ View full question & answer→MCQ 461 Mark
If the height of a tower and the distance of the point of observation from its foot, both, are increased by $10\%,$ then the angle of elevation of its top :
Answer
Let height of the tower be $h$ meters and distance of the point of observation from its foot be $x$ meters and angle of elevation be $\theta$
$\therefore\tan\theta=\frac{\text{h}}{\text{x}}......(\text{i})$
Now, new height $= h + 10\%$ of $\text{h}=\text{h}+\frac{10}{100}$
$\text{h}=\frac{11\text{h}}{10}$ And new distance $= x + 10\%$ of $\text{x}=\text{x}+\frac{10}{100}$
$\text{x}=\frac{11\text{x}}{10}$
$\therefore\tan\theta=\frac{\frac{11\text{h}}{10}}{\frac{11\text{x}}{10}}=\frac{\text{h}}{\text{x}}.....(\text{ii})$
From eq. $(i)$ and $(ii),$ it is clear that the angle of elevation is same
i.e., angle of elevation remains unchanged. View full question & answer→MCQ 471 Mark
It is found that on walking $x$ meters towards a chimney in a horizontal line through its base, the elevation of its top changes from $30^\circ$ to $60^\circ$ . The height of the chimney is :
AnswerCorrect option: C. $\frac{\sqrt{3}}{2}\text{x}$
In the figure $,AB$ is chimney and $CB$ and $DB$ are its shadow
$\tan60^\circ=\frac{\text{AB}}{\text{BC}}=\frac{\text{h}}{\text{BC}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}}{\text{BC}}$
$\Rightarrow\ \text{BC}=\frac{\text{h}}{\sqrt{3}}\ .....(\text{i})$
and $\tan30^\circ=\frac{\text{h}}{\text{DB}}=\frac{\text{h}}{\text{DB}+\text{BC}}$
$\frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}+\text{BC}}$
$\text{x}+\text{BC}=\text{h}\sqrt{3}$
$\Rightarrow\ \text{BC}=\text{h}\sqrt{3}-\text{x}\ .....(\text{ii})$
From $(i)$ and $(ii)$
$\frac{\text{h}}{\sqrt{3}}=\text{h}\sqrt{3}-\text{x}$
$\Rightarrow\ \frac{\text{h}}{\sqrt{3}}-\text{h}\sqrt{3}=-\text{x}$
$\text{x}=\text{h}\sqrt{3}-\frac{\text{h}}{\sqrt{3}}$
$=\text{h}\Big(\sqrt{3}-\frac{1}{\sqrt{3}}\Big)$
$\text{x}=\text{h}\frac{3-1}{\sqrt{3}}$
$\Rightarrow\ \text{x}=\frac{2\text{h}}{\sqrt{3}}$
$\therefore\ \text{h}=\frac{\sqrt{3}}{2}\text{x}$ View full question & answer→MCQ 481 Mark
If a pole $12m$ high casts a shadow $4\sqrt{3}\text{m}$ long on the ground, then the sun's elevation is :
- ✓
$60^\circ$
- B
$45^\circ$
- C
$30^\circ$
- D
$90^\circ$
AnswerCorrect option: A. $60^\circ$
Let $AB$ be the pole and $BC$ be its shadow and $\theta$ be the sun's elevation.
We have,
$AB = 12m$ and $\text{BC}=4\sqrt{3}\text{m}$
In $\triangle\text{ABC},$
$\tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\tan\theta=\frac{12}{4\sqrt{3}}$
$\Rightarrow\tan\theta=\frac{3}{\sqrt{3}}$
$\Rightarrow\tan\theta=\frac{3}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow\tan\theta=\frac{3\sqrt{3}}{3}$
$\Rightarrow\tan\theta=\sqrt{3}$
$\Rightarrow\tan\theta=\tan60^\circ$
$\therefore\ \theta=60^\circ$ View full question & answer→MCQ 491 Mark
A balloon moving in a straight line passes vertically above two points $A$ and $B$ on horizontal plane $1000$ ft apart. When above $A$ it has an altitude of $60^\circ$ as seen from $B$. When above $B$ it has an altitude of $45^\circ$ as seen from $A$. The distance of $B$ from the point at which it will touch the plane is :
AnswerCorrect option: A. $500(\sqrt{3}+1)\text{ft}$
$\tan60=\frac{\text{h}}{1000}$
$\text{h}=1000\sqrt{3}$
$\frac{1000\sqrt{3}}{1000+\text{x}}=\frac{1000}{4}$
$\text{x}^2=\frac{1000}{(\sqrt{3}-1)}$
$\text{x}^2=500(\sqrt{3}+1)$
View full question & answer→MCQ 501 Mark
The tops of two poles of height $16m$ and $10m$ are connected by a wire. If the wire makes an angle of $60^\circ$ with the horizontal, then the length of the wire is :
- ✓
$12\text{m}$
- B
$10\text{m}$
- C
$10\sqrt3\text{m}$
- D
$16\text{m}$
AnswerCorrect option: A. $12\text{m}$
Given : Two poles $BC = 16m$ and $AD = 10m$
And $\angle\text{CDE}=30^\circ$
To find : Lenght of wire $CD = x$
$\therefore$ In triangle $\text{CDE},$
$\sin30^\circ=\frac{\text{CE}}{\text{CD}}$
$\Rightarrow\frac{1}{2}=\frac{\text{BC}-\text{BE}}{\text{CD}}$
$\Rightarrow\frac{1}{2}=\frac{16-10}{\text{x}}$
$\Rightarrow\frac{1}{2}=\frac{6}{\text{x}}$
$\Rightarrow\text{x}=12\text{m}$
Therefore, the lenght of the wire is $12m.$ View full question & answer→