MCQ 11 Mark
When the angle of elevation of the sun decreases from $60^{\circ}$ to $30^{\circ}$ the length of the shadow of the tower standing on the ground...........................
View full question & answer→MCQ 21 Mark
In the given figure $P$ and $Q$ represent............
- A
angle of elevation, line of vision
- B
angle of depression, horizontal line
- ✓
line of vision, angle of elevation
- D
horizontal line, angle of depression
AnswerCorrect option: C. line of vision, angle of elevation
line of vision, angle of elevation
View full question & answer→MCQ 31 Mark
In the given figure $\tan \theta=$..........
- ✓
$\frac{12}{5}$
- B
$\frac{5}{13}$
- C
$\frac{12}{3}$
- D
$\frac{5}{12}$
AnswerCorrect option: A. $\frac{12}{5}$
$\frac{12}{5}$
View full question & answer→MCQ 41 Mark
A ladder is placed along a wall such that its upper end is touching the wall at $4 m$ high. The length of the ladder is $5 m$. Then the foot of the ladder is ............... $m$ away from the ground.
View full question & answer→MCQ 51 Mark
There is a house, with height $x$ opposite to the hill. The height of the hill is $h$. The angle of elevation of the top of the hill from the bottom of the house is $\alpha$ and the angle of elevation of the top of the house from the bottom of the hill is $\beta$ then the value of $\frac{h}{x}$ is......................
- A
$1$
- ✓
more than $1$
- C
less than $1$
- D
AnswerCorrect option: B. more than $1$
more than $1$
View full question & answer→MCQ 61 Mark
The angle of elevation of the sun at 9 o'clock is $x$ and at $11$ o'clock is $y$. Then ...........
- ✓
$x$
- B
$x>y$
- C
$x \geq y$
- D
$x=y$
View full question & answer→MCQ 71 Mark
The decimal value of $\sqrt{3}$ is.................
- A
$0.73$
- B
$2.73$
- ✓
$1.73$
- D
$17.3$
AnswerCorrect option: C. $1.73$
$1.73$
View full question & answer→MCQ 81 Mark
The height of the tower is $50 m$ and the height of the house is $30 m$. From a point middle of the line segment joining them, the angle of elevation of the tower is $\alpha$ and the angle of elevation of the house is $\beta$ then.
- A
$\alpha=\beta$
- ✓
$\alpha>\beta$
- C
$\alpha \leq \beta$
- D
$\alpha<\beta$
AnswerCorrect option: B. $\alpha>\beta$
$\alpha>\beta$
View full question & answer→MCQ 91 Mark
As observed from the top of a lighthouse from the sealevel, the angle of depression of two ships $A$ and $B$ are $25^{\circ}$ and $40^{\circ}$ respectively. From the lighthouse...................
- A
$A$ and $B$ are at equal distance.
- B
the distance of $B$ is more than the distance of $A$.
- ✓
The distance of $A$ is more than the distance of $B$.
- D
AnswerCorrect option: C. The distance of $A$ is more than the distance of $B$.
The distance of $A$ is more than the distance of $B$.
View full question & answer→MCQ 101 Mark
From the bottom of the house $A$, the angle of elevation of the top of the house $B$ is $45^{\circ}$ and from the bottom of the house $B$, the angle of elevation of the top of the house $A$ is $65^{\circ}$. Then
- A
The height of $B$ is more than the height of $A$.
- B
The heights of $A$ and $B$ are equal.
- ✓
The height of $A$ is more than the height of $B$.
- D
AnswerCorrect option: C. The height of $A$ is more than the height of $B$.
The height of $A$ is more than the height of $B$.
View full question & answer→MCQ 111 Mark
As observed from the top of a lighthouse from the sealevel the angle of depression of two ships P and Q are $35^{\circ}$ and $50^{\circ}$ respectively. From the lighthouse,
- A
$P$ and $Q$ are at equal distances.
- B
The distance of $Q$ is more than that of $P$.
- ✓
The distance of $P$ is more than that of $Q$.
- D
AnswerCorrect option: C. The distance of $P$ is more than that of $Q$.
The distance of $P$ is more than that of $Q$.
View full question & answer→MCQ 121 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 meters) is
- A
$\frac{4}{\sqrt{3}}$
- B
$4 \sqrt{3}$
- C
$2 \sqrt{2}$
- ✓
Answer(d): Let $M N$ be the length of the ladder.
In right-angled triangle $M N B$,
$
\cos 60^{\circ}=\frac{B N}{M N} \Rightarrow \frac{1}{2}=\frac{2}{M N}
$
$
\Rightarrow M N=2 \times 2=4 m
$
Therefore, the length of the ladder is $4 m$. View full question & answer→MCQ 131 Mark
From a point on the ground, which is $15 m$ away from the foot of a vertical tower, the angle of elevation of the top of the tower, is found to be $60^{\circ}$. The height of the tower (in metres) is
- A
$5 \sqrt{3}$
- ✓
$15 \sqrt{3}$
- C
- D
AnswerCorrect option: B. $15 \sqrt{3}$
(b) : Let height of tower $A B=h$ metres
In $\triangle A P B$,
$
\tan 60^{\circ}=\frac{A B}{P B}=\frac{h}{15}
$
$\Rightarrow \sqrt{3}=\frac{h}{15}$
or $h=15 \sqrt{3} m$
$\therefore$ Height of the tower $=15 \sqrt{3} m$ View full question & answer→MCQ 141 Mark
A lamp post $5 \sqrt{3} m$ high casts a shadow $5 m$ long on the ground. The Sun's elevation at this moment is
- A
$30^{\circ}$
- B
$45^{\circ}$
- ✓
$60^{\circ}$
- D
$90^{\circ}$
AnswerCorrect option: C. $60^{\circ}$
(c) : Let $A B$ be the lamp post and $B C$ be the shadow of it. Let Sun's elevation be $\theta$.
In $\triangle A B C$,
$
\tan \theta=\frac{A B}{B C}=\frac{5 \sqrt{3}}{5}=\sqrt{3}
$
$\Rightarrow \tan \theta=\tan 60^{\circ}$$
\Rightarrow \theta=60^{\circ}
$
Hence, Sun's elevation at the moment is $60^{\circ}$. View full question & answer→MCQ 151 Mark
The figure shows the $...........$
observation of point $C$ from point $A$. The angle of depression of $A$ is - ✓
$30^{\circ}$
- B
$45^{\circ}$
- C
$50^{\circ}$
- D
$75^{\circ}$
AnswerCorrect option: A. $30^{\circ}$
In $\triangle A B C, \angle B=90^{\circ}$.
Let $\angle A C B=\theta$
$\therefore \tan \theta=\frac{A B}{B C}=\frac{4}{4 \sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow \tan \theta=\tan 30^{\circ}$
$\Rightarrow \theta=30^{\circ}$
Hence, the angle of depression from $A$ is $30^{\circ}$.
View full question & answer→MCQ 161 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. $60^{\circ}$
(b) : Let $B C=h$ be the length of shadow of vertical pole and $A B=h \sqrt{3}$ be the height of pole.
In $\triangle A B C, \tan \theta=\frac{h \sqrt{3}}{h}=\frac{\sqrt{3}}{1}$
$
\Rightarrow \theta=60^{\circ}
$ View full question & answer→MCQ 171 Mark
At certain time of the day, the length of the shadow of a tower is equal to its height. Then the Sun's altitude at that time is
- A
$30^{\circ}$
- B
$60^{\circ}$
- C
$90^{\circ}$
- ✓
$45^{\circ}$
AnswerCorrect option: D. $45^{\circ}$
(d) : Let $A B$ be the tower of height $h$ units and altitude of Sun at a certain time is $\theta$.
We have, $A B=B C=h$ (say)
In $\triangle A B C$,
$
\Rightarrow \tan \theta=\tan 45^{\circ} \Rightarrow \theta=45^{\circ}
$ View full question & answer→MCQ 181 Mark
A man sitting on the top of a tower of height $30 m$ observes the angle of depression of a dog sitting on the ground as $60^{\circ}$. Find the distance between the foot of the tower and the dog. $($Use $\sqrt{3}=1.732)$
- A
$5 \sqrt{3} m$
- ✓
$10 \sqrt{3} m$
- C
$8 \sqrt{3} m$
- D
$12 \sqrt{3} m$
AnswerCorrect option: B. $10 \sqrt{3} m$
Let $A B$ be the tower of height $30 m$ and the dog is sitting on the ground at point $C$.
In $\triangle ABC$,
$\tan 60^{\circ}=\frac{A B}{B C}$
$\Rightarrow \sqrt{3}=\frac{30}{B C}$
$\Rightarrow B C=\frac{30}{\sqrt{3}}$
$=\frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$
$=10 \sqrt{3} m$
Hence, the distance between foot of the tower and the $\operatorname{dog}$ is $10 \sqrt{3} m$. View full question & answer→MCQ 191 Mark
From the given figure, the angle of depression of point $C$ from the point $P$ is
- A
$45^{\circ}$
- B
$90^{\circ}$
- C
$75^{\circ}$
- ✓
$30^{\circ}$
AnswerCorrect option: D. $30^{\circ}$
(d) : From figure, $\angle B P Q=\angle A B P$
[Alternate interior angles]
$
\Rightarrow \angle C P Q+\angle B P C=60^{\circ} \Rightarrow \angle C P Q+30^{\circ}=60^{\circ}
$
$
\Rightarrow \angle C P Q=30^{\circ}
$
Hence, angle of depression of point $C$ from the point $P$ is $30^{\circ}$.
View full question & answer→MCQ 201 Mark
A kite is flying at a height of $45 m$ above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^{\circ}$. Find the length of the string assuming that there is no slack in the string. $($Use $\sqrt{3}=1.732)$
- A
$45.35 m$
- B
$72.96 m$
- ✓
$51.96 m$
- D
$50 m$
AnswerCorrect option: C. $51.96 m$
Let $C$ be the position of kite.
Now, in $\triangle A B C$,
$\sin 60^{\circ}=\frac{B C}{A C}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{45}{A C}$
$\Rightarrow A C=\frac{45 \times 2}{\sqrt{3}}$
$=\frac{90}{\sqrt{3}}$
$=30 \sqrt{3}$
$=51.96 m$
Thus, the length of the string is $51.96 m$. View full question & answer→MCQ 211 Mark
The ratio of the length of a vertical rod and the length of its shadow is $1: \sqrt{3}$. Find the angle of elevation of the Sun at that moment.
- ✓
$30^{\circ}$
- B
$60^{\circ}$
- C
$45^{\circ}$
- D
AnswerCorrect option: A. $30^{\circ}$
(a) : Let $A C$ be the length of vertical rod, $A B$ be the length of its shadow and $\theta$ be the angle of elevation of the Sun.
In $\triangle A B C, \tan \theta=\frac{A C}{A B}$
$\Rightarrow \tan \theta=\frac{1}{\sqrt{3}}$ (Given)
$\Rightarrow \tan \theta=\tan 30^{\circ}$
$\Rightarrow \theta=30^{\circ}$ View full question & answer→MCQ 221 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. $30^{\circ}$
(b): Let $A B$ be the tower and $B C$ be the length of the shadow of the tower.
Let $\theta$ be the angle of elevation of Sun.
Given, $B C=\sqrt{3} A B$
In $\triangle A B C, \tan \theta=\frac{A B}{B C}=\frac{A B}{\sqrt{3} A B}$
$\Rightarrow \tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ}$
Thus, the angle of elevation of the Sun is $30^{\circ}$. View full question & answer→MCQ 231 Mark
The angle of elevation of the top of a tower from a point on the ground, which is $30 m$ away from the foot of the tower is $45^{\circ}$. The height of the tower (in metres) is
- A
- ✓
- C
$30 \sqrt{3}$
- D
$10 \sqrt{3}$
Answer(b) : Let $A B$ be the tower and $P$ be the point on the ground.
In $\triangle A B P, \frac{A B}{B P}=\tan 45^{\circ}$
$\Rightarrow \frac{A B}{30}=1 \Rightarrow A B=30 m$ View full question & answer→MCQ 241 Mark
The angle of elevation of the top of a pillar from a point on the ground is $15^{\circ}$. On walking $100 m$ towards the pillar, the angle of elevation becomes $30^{\circ}$. Find the height of the pillar.
- A
$25 m$
- ✓
$50 m$
- C
$50 \sqrt{2} m ( d )$
- D
$25 \sqrt{2} m$
AnswerCorrect option: B. $50 m$
Let $A B$ be the pillar and $C$ and $D$ be the two points of observation.
$\angle A C B=\angle C D A+\angle C A D$
$\Rightarrow 30^{\circ}=15^{\circ}+\angle C A D[$ By exterior angle property $]$
$\Rightarrow \angle C A D=15^{\circ}=\angle A D C$
$\therefore C A=C D=100 m$
$[\because$ Sides opposite to equal angles are equal $]$
In $\triangle A B C$,
$\sin 30^{\circ}=\frac{A B}{A C}$
$\Rightarrow \frac{1}{2}=\frac{A B}{100}$
$\Rightarrow 2 A B=100$
$\Rightarrow A B=50 m$
Hence, the height of the pillar is $50 m$. View full question & answer→MCQ 251 Mark
The tops of two poles of height $18 m$ and $10 m$ are connected by a wire of length $l$. If the wire makes an angle of $30^{\circ}$ with the horizontal, then $l$ is equal to
- A
$26 m$
- ✓
$16 m$
- C
$12 m$
- D
$10 m$
AnswerCorrect option: B. $16 m$
Let $A B$ and $C D$ be the two poles of height $18 m$ and $10 m$ respectively, such that $A C=l$.
$\text { In } \triangle C D P, \sin 30^{\circ}$
$\Rightarrow \frac{1}{2}=\frac{10}{P C}$
$\Rightarrow P C=20 m$
In $\triangle A B P, \sin 30^{\circ}=\frac{A B}{P A}$
$\Rightarrow \frac{1}{2}=\frac{18}{P A}$
$\Rightarrow P A=36 m$
Now, $l=A C=P A-P C$
$=36-20$
$=16 m$ View full question & answer→MCQ 261 Mark
A person walking $20 m$ towards a chimney in a horizontal line through its base observes that its angle of elevation changes from $30^{\circ}$ to $45^{\circ}$. The height of chimney is
AnswerCorrect option: B. $\frac{20}{\sqrt{3}-1} m$
Suppose height of the chimney be $h$ metres. Let $A$ and $B$ be the points of observation and $B C$ be $x m$. In $\triangle A C D$,
$\tan 30^{\circ}=\frac{C D}{A C}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x}$
$\Rightarrow 20+x=h \sqrt{3}$
$\Rightarrow x=h \sqrt{3}-20$
Now, in $\triangle D B C, \tan 45^{\circ}=\frac{C D}{B C}$
$\Rightarrow 1=\frac{h}{x}$
$\Rightarrow x=h$
From $(i)$ and $(ii),$ we get
$h=h \sqrt{3}-20$
$\Rightarrow h \sqrt{3}-h=20$
$\therefore h=\frac{20}{\sqrt{3}-1} m$ View full question & answer→MCQ 271 Mark
The angle of elevation of the top of a tower from a point on the ground, which is $15 m$ away from the foot of the tower is $60^{\circ}$. The height of the tower (in metres) is
- A
- ✓
- C
$30 \sqrt{3}$
- D
$15 \sqrt{3}$
Answer(b): Let $A B$ be the tower and $P$ be the point on the ground.
In $\triangle A B P, \frac{A B}{B P}=\tan 60^{\circ}$
$
\Rightarrow \frac{A B}{15}=\sqrt{3} \Rightarrow A B=15 \sqrt{3} m
$ View full question & answer→MCQ 281 Mark
The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of $100 m$ from its base is $45^{\circ}$. If the angle of elevation of the top of the complete pillar at the same point is to be $60^{\circ}$, then the height of the incomplete pillar is to be increased by
- A
$100(\sqrt{3}+1) m$
- B
$100 m$
- C
$100 \sqrt{3} m$
- ✓
$100(\sqrt{3}-1) m$
AnswerCorrect option: D. $100(\sqrt{3}-1) m$
(d) : Let $A B$ be the incomplete pillar and $A D$ is the height to be increased.
In $\triangle A B C, \frac{A B}{B C}=\tan 45^{\circ}$
$\Rightarrow \frac{A B}{100}=1 \Rightarrow A B=100 m$
In $\triangle D B C, \frac{D B}{B C}=\tan 60^{\circ}$
$\Rightarrow \quad \frac{A D+100}{100}=\sqrt{3}$
$\Rightarrow A D=100(\sqrt{3}-1) m$
Hence, the height of the incomplete pillar is to be increased by $100(\sqrt{3}-1) m$. View full question & answer→MCQ 291 Mark
The angle of depression of a car parked on the road from the top of a $150 m$ 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
AnswerCorrect option: B. $150 \sqrt{3}$
(b): Let, $A C=x m$ be the distance between tower and car and let $A B=150 m$ be height of tower.
In $\triangle A B C$,
$\tan 30^{\circ}=\frac{150}{x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{150}{x} \Rightarrow x=150 \sqrt{3} m$
Hence, distance between tower and car $=150 \sqrt{3} m$. View full question & answer→MCQ 301 Mark
A lamp post $8 m$ high casts a shadow $8 \sqrt{3} m$ long on the ground. The sun's elevation at this moment is
- ✓
$30^{\circ}$
- B
$45^{\circ}$
- C
$60^{\circ}$
- D
$90^{\circ}$
AnswerCorrect option: A. $30^{\circ}$
(a): Let $A B$ be the lamp post and $B C$ be the shadow of it. Let sun's elevation be $\theta$.
In $\triangle A B C$,
$\tan \theta=\frac{A B}{B C}=\frac{8}{8 \sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow \tan \theta=\tan 30^{\circ}$
$\Rightarrow \theta=30^{\circ}$
Hence, sun's elevation at the moment is $30^{\circ}$. View full question & answer→MCQ 311 Mark
The length of a string between a kite and a point on the ground is $85 m$. If the string makes an angle $\theta$ with level ground such that $\tan \theta=\frac{15}{8}$, then how high is the kite?
- ✓
$75 m$
- B
$78.05 m$
- C
$226 m$
- D
AnswerCorrect option: A. $75 m$
(a) : Length of the string of the kite $=A B=85 m$
$\tan \theta=\frac{15}{8}$
$\Rightarrow \cot \theta=\frac{8}{15} \Rightarrow \operatorname{cosec}^2 \theta-1=\frac{64}{225}$
$\Rightarrow \operatorname{cosec}^2 \theta=1+\frac{64}{225}=\frac{289}{225}$
$\Rightarrow \operatorname{cosec} \theta=\sqrt{\frac{289}{225}}=\frac{17}{15} \Rightarrow \sin \theta=\frac{15}{17}$
In $\triangle A B C, \sin \theta=\frac{B C}{A B} \Rightarrow \frac{15}{17}=\frac{B C}{85} \Rightarrow B C=75$
$\therefore \quad$ Height of the kite $=75 m$ View full question & answer→MCQ 321 Mark
Two men standing on opposite sides of a flagstaff measure the angles of elevation of the top of the flagstaff is $30^{\circ}$ and $60^{\circ}$. If the height of the flagstaff is $20 m$, then approximate distance between the men is $($Use $\sqrt{3}=1.732)$
- ✓
$46.19 m$
- B
$40 m$
- C
$50 m$
- D
$30 m$
AnswerCorrect option: A. $46.19 m$
Let $C$ and $D$ be the positions of the men and $A B$ be the height of flagstaff.
In $\triangle A B C, \tan 30^{\circ}=\frac{A B}{B C}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{20}{B C}$
$\Rightarrow B C=20 \sqrt{3}$
In $\triangle A B D, \tan 60^{\circ}=\frac{A B}{B D}$
$\Rightarrow \sqrt{3}=\frac{20}{B D}$
$\Rightarrow B D=\frac{20}{\sqrt{3}}$
Distance between the men, $C D=B C+B D$
$=20 \sqrt{3}+\frac{20}{\sqrt{3}}$
$=\frac{60+20}{\sqrt{3}}$
$=\frac{80}{\sqrt{3}}$
$=\frac{80 \sqrt{3}}{3}$
$=\frac{80 \times 1.732}{3}$
$=46.187 \approx 46.19 m$ View full question & answer→MCQ 331 Mark
There are two temples one on each bank of a river just opposite to each other. One temple is $40 m$ high. As observed from the top of this temple, the angle of depression of the top and foot of the other temple are $30^{\circ}$ and $60^{\circ}$ respectively. The width of river is
AnswerCorrect option: A. $\frac{40 \sqrt{3}}{3} m$
Let $A B$ be the temple of height $40 m$ and $P Q$ be the other temple.
$\text { In } \triangle A B Q, \frac{A B}{B Q}=\tan 60^{\circ}$
$\Rightarrow \frac{40}{B Q}=\sqrt{3}$
$\Rightarrow B Q=\frac{40}{\sqrt{3}}$
So, width of the river $=\frac{40 \sqrt{3}}{3} m$ View full question & answer→MCQ 341 Mark
Suppose a straight vertical tree is broken at some point due to storm and the broken part is inclined at a certain distant from the foot of the tree. If the top of broken part of a tree touches the ground at a point whose distance from foot of the tree is equal to height of remaining part, then its angle of inclination is
- A
$30^{\circ}$
- B
$60^{\circ}$
- ✓
$45^{\circ}$
- D
AnswerCorrect option: C. $45^{\circ}$
(c) : Let $AB$ be the tree of height $h$.Let $B D=B C=x m$
$\therefore$ In $\triangle C B D$,
$
\begin{aligned}
\tan \theta & =\frac{B C}{B D}=1 \\
\Rightarrow \tan \theta & =\tan 45^{\circ} \Rightarrow \theta=45^{\circ}
\end{aligned}
$
View full question & answer→MCQ 351 Mark
If the ratio of the height of a tree and its shadow is $1: \frac{1}{\sqrt{3}}$, then the angle of the Sun's elevation is
- A
$30^{\circ}$
- B
$45^{\circ}$
- ✓
$60^{\circ}$
- D
$90^{\circ}$
AnswerCorrect option: C. $60^{\circ}$
(c) : Let $A B$ be the tree and $B C$ be its shadow such that $\frac{A B}{B C}=\frac{1}{\left(\frac{1}{\sqrt{3}}\right)}=\sqrt{3}$.
Let $\theta$ be the Sun's elevation.
$
\text { In } \triangle A B C, \tan \theta=\frac{A B}{B C}=\sqrt{3}=\tan 60^{\circ}
$
$
\Rightarrow \theta=60^{\circ}
$
Hence, the angle of the Sun's elevation is $60^{\circ}$. View full question & answer→MCQ 361 Mark
A steel pole is $30 m$ high. To keep the pole upright, one end of a steel wire is tied to the top of the pole while the other end has been fixed on the ground. If the steel wire makes an angle of $45^{\circ}$ with the horizontal through the base point of the pole, then find the length of the steel wire.
- ✓
$30 \sqrt{2} m$
- B
$30 \sqrt{3} m$
- C
$15 m$
- D
$15 \sqrt{2} m$
AnswerCorrect option: A. $30 \sqrt{2} m$
(a) : Let $A B$ be the pole and $A C$ be the steel wire fixed on the ground at $C$.
In $\triangle A B C, \sin 45^{\circ}=\frac{A B}{A C}$
$
\Rightarrow \frac{1}{\sqrt{2}}=\frac{30}{A C} \Rightarrow A C=30 \sqrt{2} m
$ View full question & answer→MCQ 371 Mark
A portion of a $45 m$ long tree is broken by tornado and the top struck up the ground making an angle of $30^{\circ}$ with the ground level. The height of the point where the tree is broken, is equal to
- A
$30 m$
- ✓
$15 m$
- C
$10 m$
- D
$20 m$
AnswerCorrect option: B. $15 m$
(b) : Let $A B$ be the tree which is broken at $C$.
Let $B C=x m$.
In $\triangle B C D, \frac{B C}{D C}=\sin 30^{\circ}$
$\Rightarrow \quad \frac{x}{45-x}=\frac{1}{2}$
$\Rightarrow 2 x=45-x \Rightarrow x=15$
Hence, the height of the point where tree is broken is $15 m$. View full question & answer→MCQ 381 Mark
A man standing on the deck of a ship, which is $10 m$ above the 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 the hill as $30^{\circ}$. The distance of the hill from the ship is
- A
$40 m$
- ✓
$10 \sqrt{3} m$
- C
$10 m$
- D
$20 \sqrt{3} m$
AnswerCorrect option: B. $10 \sqrt{3} m$
Let $A B$ be the hill and $C D$ be the deck of the ship.
$C D=10 m ($Given$)$
In $\triangle C D B$,
$\tan 30^{\circ}=\frac{C D}{B D}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{10}{B D}$
$\Rightarrow B D=10 \sqrt{3} m$
Hence, the distance of the hill from the ship is $10 \sqrt{3} m$. View full question & answer→MCQ 391 Mark
If the length of shadow of a building, when the Sun's altitude is $60^{\circ}$, is $20 m$ less than when it was $45^{\circ}$, then the height of the building is (Use $\sqrt{3}=1.732$ )
- A
$54.48 m$
- B
$47.32 m$
- C
$64.32 m$
- D
$57.48 m$
View full question & answer→MCQ 401 Mark
A peacock sitting on the top of a tree observes a serpent on the ground making an angle of depression $30^{\circ}$. If the peacock with a speed of $300 m$ per minute catches the serpent in 12 seconds, then the height of the tree is
- A
$30 m$
- B
$30 \sqrt{3} m$
- C
$\frac{30}{\sqrt{3}} m$
- D
$15 m$
View full question & answer→MCQ 411 Mark
A bridge across a river makes an angle of $45^{\circ}$ with the river bank. If the length of the bridge across the river is $50 m$, then what is the width of the river?
- A
$20 \sqrt{2} m$
- B
$50 \sqrt{2} m$
- ✓
$25 \sqrt{2} m$
- D
$10 \sqrt{2} m$
AnswerCorrect option: C. $25 \sqrt{2} m$
From figure, in $\triangle A B C$,
we have $\sin 45^{\circ}=\frac{B C}{A C}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{B C}{50}$
$\Rightarrow B C=\frac{50}{\sqrt{2}}=25 \sqrt{2} m$
Hence, the width of the river is $25 \sqrt{2} m$.
View full question & answer→MCQ 421 Mark
The angle of elevation is always
Answer(b) : The angle of elevation is always an acute angle.
View full question & answer→MCQ 431 Mark
A figure is given below : Jyoti and Ravi observed the figure and said the following :
Jyoti : $C D=14 m$
Ravi : $C D=15 m$
Which of them is/are correct? AnswerClearly, $B D=A B-A D$
$=(10 \sqrt{3}-2 \sqrt{3}) m =8 \sqrt{3} m$
Now, in $\triangle B C D$, we have
$\sin 60^{\circ}=\frac{B D}{D C}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{8 \sqrt{3}}{D C} $
$\Rightarrow D C=16 m$
View full question & answer→MCQ 441 Mark
A window is $6 m$ above the ground. A ladder is placed against the wall such that its top reaches the window. If angle made by the foot of ladder to the ground is $30^{\circ}$, then length of the ladder is
- A
$8 m$
- B
$10 m$
- ✓
$12 m$
- D
$14 m$
AnswerCorrect option: C. $12 m$
(c) : Let $A C$ be the length of the ladder.
In $\triangle A B C, \frac{B C}{A C}=\sin 30^{\circ}$
$
\Rightarrow \frac{6}{A C}=\frac{1}{2} \Rightarrow A C=12 m
$ View full question & answer→MCQ 451 Mark
If the height of the window is $8 m$ above the ground. A ladder is placed against the wall such that its top reaches the window. If angle ofelevation is observed to be $45^{\circ}$, then horizontal distance between the foot of ladder and wall is
Answer(d) : Let $A B$ be the horizontal distance between the foot of ladder and wall.
In $\triangle A B C, \frac{B C}{A B}=\tan 45^{\circ}$
$\Rightarrow \frac{8}{A B}=1 \Rightarrow A B=8 m$ View full question & answer→
