MCQ 11 Mark
A car is moving away from the base of a 30 m high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is $10 \sqrt{3} m$ away from the base of the tower, is
- A
$30^{\circ}$
- B
$45^{\circ}$
- C
$90^{\circ}$
- ✓
$60^{\circ}$
AnswerCorrect option: D. $60^{\circ}$
View full question & answer→MCQ 21 Mark
If the angle of elevation of a tower from a distance of 100 metres from its foot is $60^{\circ}$, then the height of the tower is
- ✓
$100 \sqrt{3} m$
- B
$\frac{100}{\sqrt{3}} m$
- C
$50 \sqrt{3} m$
- D
$\frac{200}{\sqrt{3}} m$
AnswerCorrect option: A. $100 \sqrt{3} m$
View full question & answer→MCQ 31 Mark
At some time of the dayy, if the leng th of the shadow of a vertical pole is equal to its height, the angle of eleation of Sun's altitude is
- ✓
$45^{\circ}$
- B
$60^{\circ}$
- C
$30^{\circ}$
- D
$75^{\circ}$
AnswerCorrect option: A. $45^{\circ}$
(A)
Let $A B$ be a vertical pole of height $x m$ and let $A C$ be the length of its shadow when the angle of elevation of Sun is $\theta^{\circ}$. It is given that $A C=x m$. In $\triangle A C B$, we obtain
$
\tan \theta=\frac{A B}{A C}=\frac{x}{x}=1=\tan 45^{\circ} \Rightarrow \theta=45^{\circ}
$
Hence, Sun's altitude is $45^{\circ}$.
View full question & answer→MCQ 41 Mark
If a pole 6 m high casts a shadow $2 \sqrt{3} m$ long on the ground, then sun's elevation is
- A
$60^{\circ}$
- B
$45^{\circ}$
- ✓
$30^{\circ}$
- D
$90^{\circ}$
AnswerCorrect option: C. $30^{\circ}$
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
$2 \sqrt{2}$
- ✓
View full question & answer→MCQ 61 Mark
The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is $45^{\circ}$. Then the height of the tower (in metres) is
- A
$50 \sqrt{3}$
- ✓
- C
$\frac{50}{\sqrt{2}}$
- D
$\frac{50}{\sqrt{3}}$
View full question & answer→MCQ 71 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. $150 \sqrt{3}$
View full question & answer→MCQ 81 Mark
A ladder 15 m 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. $\frac{15}{2} m$
View full question & answer→MCQ 91 Mark
The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is $30^{\circ}$. The distance of the car from the base of the,tower (in metres) is
- A
$25 \sqrt{3}$
- B
$50 \sqrt{3}$
- ✓
$75 \sqrt{3}$
- D
AnswerCorrect option: C. $75 \sqrt{3}$
View full question & answer→MCQ 101 Mark
The length of shadow of a tower on the plane ground is $\sqrt{3}$ times the height of the tower. The angle of clevation of sum is
- A
$45^{\circ}$
- ✓
$30^{\circ}$
- C
$60^{\circ}$
- D
$90^{\circ}$
AnswerCorrect option: B. $30^{\circ}$
View full question & answer→MCQ 111 Mark
If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is
MCQ 121 Mark
The tops of two poles of height 16 m and 10 m are connected by a wire of length $l$ metres. If the wire makes an angle of $30^{\circ}$ with the horizontal, then $l=$
View full question & answer→MCQ 131 Mark
Two poles are ' $a$ ' metres apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the smaller is
AnswerCorrect option: B. $\frac{a}{2 \sqrt{2}}$ metres
View full question & answer→MCQ 141 Mark
The length of the shadow of a tower standing on level ground is found to be $2 x$ metres longer when the sun's elevation is $30^{\circ}$ than when it was $45^{\circ}$. The height of the tower in metres is
- ✓
$(\sqrt{3}+1) x$
- B
$(\sqrt{3}-1) x$
- C
$2 \sqrt{3} x$
- D
$3 \sqrt{2} x$
AnswerCorrect option: A. $(\sqrt{3}+1) x$
View full question & answer→MCQ 151 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
- A
$3 \sqrt{2} x$
- B
$2 \sqrt{3} x$
- ✓
$\frac{\sqrt{3}}{2} x$
- D
$\frac{2}{\sqrt{3}} x$
AnswerCorrect option: C. $\frac{\sqrt{3}}{2} x$
View full question & answer→MCQ 161 Mark
A tower subtends an angle of $30^{\circ}$ at a point on the same level as its foot. At a second point $h$ metres above the first, the depression of the foot of the tower is $60^{\circ}$. The height of the tower is
- A
$\frac{h}{2} m$
- B
$\sqrt{3} h m$
- ✓
$\frac{h}{3} m$
- D
$\frac{h}{\sqrt{3}} m$
AnswerCorrect option: C. $\frac{h}{3} m$
View full question & answer→MCQ 171 Mark
The angle of elevation of a cloud from a point $/ 2$ metre above a lake is $\theta$. The angle of depression of its reflection in the lake is $45^{\circ}$. The height of the cloud is
- ✓
$h \tan \left(45^{\circ}+\theta\right)$
- B
$h \cot \left(45^{\circ}-\theta\right)$
- C
$I \tan \left(45^{\circ}-\theta\right)$
- D
$h \cot \left(45^{\circ}+\theta\right)$
AnswerCorrect option: A. $h \tan \left(45^{\circ}+\theta\right)$
View full question & answer→MCQ 181 Mark
Two persons are a metres apart and the height of one is double that of the other. If from the middle point of the line joining their feet, an observer finds the angular elevation of their tops to be complementary, then the height of the shorter post is
- A
$\frac{a}{4}$
- B
$\frac{a}{\sqrt{2}}$
- C
$a \sqrt{2}$
- ✓
$\frac{a}{2 \sqrt{2}}$
AnswerCorrect option: D. $\frac{a}{2 \sqrt{2}}$
View full question & answer→MCQ 191 Mark
The height of a tower is 100 m . When the angle of elevation of the sun changes from $30^{\circ}$ to $45^{\circ}$, the shadow of the tower becomes $x$ metres less. The value of $x$ is
- A
- B
$100 \sqrt{3} m$
- ✓
$100(\sqrt{3}-1) m$
- D
$\frac{100}{\sqrt{3}} m$
AnswerCorrect option: C. $100(\sqrt{3}-1) m$
View full question & answer→MCQ 201 Mark
If the angle of elevation of a cloud from a point 200 m 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
View full question & answer→MCQ 211 Mark
The angles of depression of two ships from the top of a light house are $45^{\circ}$ and $30^{\circ}$ towards east. If the ships are 100 m apart, the height of the light house is
AnswerCorrect option: D. $50(\sqrt{3}+1) m$
View full question & answer→MCQ 221 Mark
From the top of a cliff 25 m high the angle of elevation 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
MCQ 231 Mark
The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of $30^{\circ}$ with horizontal, then the length of the wire is
View full question & answer→MCQ 241 Mark
The angle of elevation of the top of a tower standing on a horizontal plane from a point $A$ is $\alpha$. After walking a distance $d$ towards the foot of the tower the angle of elevation is found to be $\beta$ The height of the tower is
- A
$\frac{d}{\cot \alpha+\cot \beta}$
- ✓
$\frac{d}{\cot \alpha-\cot \beta}$
- C
$\frac{d}{\tan \beta-\tan \alpha}$
- D
$\frac{d}{\tan \beta+\tan \alpha}$
AnswerCorrect option: B. $\frac{d}{\cot \alpha-\cot \beta}$
View full question & answer→MCQ 251 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
AnswerCorrect option: A. $(\sqrt{3}+1) h m$
View full question & answer→MCQ 261 Mark
If the angles of elevation of the top of a tower from two points distant $a$ and $b$ from the base and in the same straight line with it are complementary, then the height of the tower is
- A
$a b$
- ✓
$\sqrt{a b}$
- C
$\frac{a}{b}$
- D
$\sqrt{\frac{a}{b}}$
AnswerCorrect option: B. $\sqrt{a b}$
View full question & answer→MCQ 271 Mark
If the angles of elevation of a tower from two points distant $a$ and $b(a>b)$ from its foot and in the same straight line from it are $30^{\circ}$ and $60^{\circ}$, then the height of the tower is
- A
$\sqrt{a+b}$
- ✓
$\sqrt{a b}$
- C
$\sqrt{a-b}$
- D
$\sqrt{\frac{a}{b}}$
AnswerCorrect option: B. $\sqrt{a b}$
View full question & answer→MCQ 281 Mark
If the altitude of the sun is at $60^{\circ}$, then the height of the vertical tower that will cast a shadow of length 30 m is
- ✓
$30 \sqrt{3} m$
- B
- C
$\frac{30}{\sqrt{3}} m$
- D
$15 \sqrt{2} m$
AnswerCorrect option: A. $30 \sqrt{3} m$
View full question & answer→MCQ 291 Mark
The ratio of the length of a rod and its shadow is $1: \sqrt{3}$. The angle of elevation of the sun is
- ✓
$30^{\circ}$
- B
$45^{\circ}$
- C
$60^{\circ}$
- D
$90^{\circ}$
AnswerCorrect option: A. $30^{\circ}$
View full question & answer→MCQ 301 Mark
A man is climbing a ladder which is inclined to the wall at an angle of $30^{\circ}$. If he ascends at the rate of $2 m / sec$, then he approaches the wall at the rate of
- A
$2 m / sec$
- B
$25 m / sec$
- ✓
$1 m / sec$
- D
$1.5 m / sec$
AnswerCorrect option: C. $1 m / sec$
(C)
Let $P Q$ be the wall and $Q R$ be the ladder. It is given that the ladder makes an angle of $30^{\circ}$ with the wall. Let the man take $t$ seconds to reach to the point $Q$ with the speed of $2 m / sec$. Then, $P Q=2!\ln \triangle Q P R$, we obtain
$\cos 60^{\circ}=\frac{P R}{Q R} \Rightarrow P R=t$
Thus, the man covers distance $P R=t$ metres in $t$ seconds Hence, the rate at which he approaches the wall is $\frac{t}{t}=1 m / sec$
View full question & answer→MCQ 311 Mark
Two friends Rohit and Mohit are standing on the opposite sides of a tower of height 60 metres. If their angles of depression seen from the top of the tower are $30^{\circ}$ and $45^{\circ}$ respectively, then the distance between the two friends is
- A
$60(\sqrt{3}-1) m$
- ✓
$60(\sqrt{3}+1) m$
- C
$30(\sqrt{3}-1) m$
- D
$30(\sqrt{3}+1) m$
AnswerCorrect option: B. $60(\sqrt{3}+1) m$
(B)
Let $A B$ be a tower of height 60 metres. In right triangles $P A B$ and $Q A B$, we obtain
$
\begin{array}{ll}
& \tan 30^{\circ}=\frac{A B}{A P} \text { and, } \tan 45^{\circ}=\frac{A B}{A Q} \\
\Rightarrow & \frac{1}{\sqrt{3}}=\frac{60}{A P} \text { and } 1=\frac{60}{A Q} \\
\Rightarrow \quad & A P=60 \sqrt{3} \text { and } A Q=60
\end{array}
$
Hence, $P Q=P A+A Q=60 \sqrt{3}+60=60(\sqrt{3}+1) m$.
View full question & answer→MCQ 321 Mark
If the height of a flagstaff is twice the height of the tower on which it is fixed 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 flag staff as seen from the same point is
- A
$45^{\circ}$
- B
$30^{\circ}$
- ✓
$60^{\circ}$
- D
$90^{\circ}$
AnswerCorrect option: C. $60^{\circ}$
(C)
Let $A B$ be a tower of height $h$ metres and $B C$ be a flagstaff of a height $2 h$ metres. The angle of elevation of the top $B$ of the tower as seen from a point $P$ on the ground is $30^{\circ}$. Let the angle of elevation of the top of the flagstaff as seen from point $P$ be $\theta$. In right triangles $P A B$ and $P A C$, we obtain
$\begin{array}{ll} & \tan 30^{\circ}=\frac{A B}{A P} \text { and } \tan \theta=\frac{A C}{A P} \\ \Rightarrow & \frac{1}{\sqrt{3}}=\frac{h}{A P} \text { and } \tan \theta=\frac{3 h}{A P} \\ \Rightarrow & A P=\sqrt{3} h \text { and } A P=\frac{3 h}{\tan \theta} \\ \Rightarrow & \sqrt{3} h=\frac{3 h}{\tan \theta} \Rightarrow \tan \theta=\sqrt{3} \Rightarrow \theta=60^{\circ}\end{array}$
View full question & answer→MCQ 331 Mark
A man on the top of a cliff ' $x$ ' metres high observes that the angles of elevation of the top of a tower is equal to the angle of depression of the foot of the tower. The height of the tower in metres is
- A
$2 \sqrt{2} x$
- ✓
$2 x$
- C
$\sqrt{2} x$
- D
$\frac{x}{2}$
Answer(B)
Let $A B$ be a Cliff of height $x$ metres and $P Q$ be a tower of height $h$ metres. Let $\theta$ be the angle of elevation of the top $Q$ of the tower observed from the top of the cliff and $\theta$ be the angle of depression of the foot of the tower. In right triangles $B R Q$ and $P A B$, we obtain
$\begin{array}{ll} & \tan \theta=\frac{Q R}{B R} \text { and } \tan \theta=\frac{A B}{P A} \\ \Rightarrow \quad & \tan =\frac{h-x}{A P} \text { and } \tan \theta=\frac{x}{A P} \\ \Rightarrow \quad & \frac{h-x}{A P}=\frac{x}{A P} \Rightarrow h-x=x \Rightarrow h=2 x\end{array}$
View full question & answer→MCQ 341 Mark
If the angles of elevation of the top of a tower from two points at a distance of 4 m and 16 m from the base of a tower and in the same line are complementary, then the height of the tower is
Answer(C)
Let $P Q$ be a tower of height $h$ metre such that the angles of elevation of its top at two points $R$ and $S$ are complementary. Let $\theta$ be the angle of elevation at $R$. Then, the angle of elevation at $S$ is $90^{\circ}-\theta$. In right triangles $P Q R$ and $P Q S$, we obtain
$\begin{aligned} \tan \theta & =\frac{P Q}{Q R} \text { and } \tan \left(90^{\circ}-\theta\right)=\frac{P Q}{Q S} \\ \Rightarrow \quad \tan \theta & =\frac{h}{4} \text { and } \cot \theta=\frac{h}{16}\end{aligned}$
$\Rightarrow \quad \tan \theta \times \cot \theta=\frac{h}{4} \times \frac{h}{16} \Rightarrow 1=\frac{h^2}{64} \Rightarrow h=8 m$.
View full question & answer→MCQ 351 Mark
When the Sun's elevation is $30^{\circ}$, the shadow of a tower is 30 m long, if the Sun's elevation is $60^{\circ}$, then the length of the shadow is
Answer(C)
Let $Q P$ be a tower of height $h$ metre and let $Q S=30 m$ be the length of the shadow of the tower when Sun's elevation is $30^{\circ}$. Let $Q R$ be the shadow when Sun's elevation is $60^{\circ}$.
In triangles $P Q S$ and $P Q R$, we obtain
$
\begin{array}{ll}
& \tan 30^{\circ}=\frac{P Q}{Q S} \text { and } \tan 60^{\circ}=\frac{P Q}{Q R} \\
\Rightarrow & \frac{1}{\sqrt{3}}=\frac{h}{30} \text { and } \sqrt{3}=\frac{h}{x} \\
\Rightarrow & h=\frac{30}{\sqrt{3}} \text { and } x=\frac{h}{\sqrt{3}} \\
\Rightarrow & x=\frac{1}{\sqrt{3}} \times \frac{30}{\sqrt{3}}=10 m
\end{array}
$
Hence, the length of the shadow is 10 m .
View full question & answer→MCQ 361 Mark
If the angle of elevation of the top of a tower from a point on the ground, 100 m away from the foot of the tower is $30^{\circ}$, then the height of the tower is
- A
- B
$100 \sqrt{3} m$
- ✓
$\frac{100}{\sqrt{3}} m$
- D
$75 \sqrt{3} m$
AnswerCorrect option: C. $\frac{100}{\sqrt{3}} m$
(C)
Let $P Q$ be a tower of height $h$ metre such that the angle of elevation of its top $Q$ at a point $R, 100 m$ away from the foot $P$ of the tower, is $30^{\circ}$.
In right triangle $R P Q$, we obtain
$
\tan 30^{\circ}=\frac{P Q}{P R} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{100} \Rightarrow \frac{100}{\sqrt{3}} m
$
View full question & answer→MCQ 371 Mark
The measure of the angle of elevation of the top of the tower $75 \sqrt{3} m$ high from a point at a distance of 75 m from the foot of the tower in a horizontal plane is
- A
$30^{\circ}$
- ✓
$60^{\circ}$
- C
$90^{\circ}$
- D
$45^{\circ}$
AnswerCorrect option: B. $60^{\circ}$
(B)
Let $P Q$ be a tower of height $75 \sqrt{3} m$ and let $\theta$ be the angle of elevation of its top $Q$ from a point $R$ at a distance of 75 m from the tower. In right triangle $R P Q$, we obtain
$\tan \theta=\frac{P Q}{P R} \Rightarrow \tan \theta=\frac{75 \sqrt{3}}{75}=\sqrt{3} \Rightarrow \tan \theta=\tan 60^{\circ} \Rightarrow \theta=60^{\circ}$
Hence, the angle of elevation is $60^{\circ}$.
View full question & answer→MCQ 381 Mark
Figure shows the observation of point C from point $A$. The angle of depression from $A$ is
- A
$60^{\circ}$
- ✓
$30^{\circ}$
- C
$45^{\circ}$
- D
$75^{\circ}$
AnswerCorrect option: B. $30^{\circ}$
(B)
Let $\theta$ be the angle of depression of point $C$ from $A$. In right triangle $A B C$, we obtain
$
\begin{array}{ll}
& \tan \theta=\frac{A B}{B C} \\
\Rightarrow \quad & \tan \theta=\frac{2}{2 \sqrt{3}}=\frac{1}{\sqrt{3}} \\
\Rightarrow \quad & \tan \theta=\tan 30^{\circ} \Rightarrow \theta=30^{\circ}
\end{array}
$
Hence, the angle of depression is $30^{\circ}$.
View full question & answer→