Maths Class X Question Bank

Question 1 of 31

1. Question
1 point(s)
The angles of elevation of the top of a tower from two points distance a and b from the base and in the same straight line with it, are complementary then the height of the tower is :
- $ab $
- $\sqrt{ab} $
- $\dfrac{a}{b} $
- $\sqrt{\dfrac{a}{b}} $
Correct

Incorrect

Hint

Let $\angle ACB=\theta$
$\angle ADB=90-\theta$
$\tan\theta=\dfrac{AB}{BC}=\dfrac{h}{a}$ …(1)
$\tan(90-\theta)=\dfrac{AB}{BD}=\dfrac{h}{b}$ …(2)
$\tan\theta\times \cot\theta=\dfrac{h}{a}\times \dfrac{h}{b}=\dfrac{h^2}{ab}\\ \\ \tan\theta\times \dfrac{1}{\tan\theta}=\dfrac{h^2}{ab}\\ \\ h^2=ab\Rightarrow h=\sqrt{ab}$
Question 2 of 31

2. Question
1 point(s)
The ratio of the length of a rod and its shadow is $1:\sqrt3 $. The altitude of the sun is :
- $30^o $
- $45^o $
- $60^o $
- $90^o $
Correct

Incorrect

Hint

$\tan\theta=\dfrac{OA}{OB}=\dfrac{1}{\sqrt3}\\ \\ \tan\theta=\dfrac{1}{\sqrt3}\Rightarrow \theta=30^o$
Question 3 of 31

3. Question
1 point(s)
The angle of elevation of a cloud from a point h metres above a lake is $\theta $. The angle of depression of its reflection in the lake is 45°. The height of the cloud is :
- $\dfrac{h(1+\tan\theta)}{(1-\tan\theta)} $
- $\dfrac{h(1-\tan\theta)}{(1+\tan\theta)} $
- $\dfrac{h(1+\tan\theta)}{\tan\theta} $
- None of these
Correct

Incorrect

Hint

$H=\dfrac{h(\tan\beta+\tan\alpha)}{\tan\beta-\tan\alpha}=\dfrac{h(1+\tan\theta)}{1-\tan\theta}$
Question 4 of 31

4. Question
1 point(s)
The height of a tower is 50m. When the sun’s altitude changes from $30^o $ to $45^o $, the shadow of the tower becomes x metres less. The value of x is :
- $50\ m $
- $50\sqrt3\ m $
- $50(\sqrt3-1)\ m $
- $\dfrac{50}{\sqrt3}\ m $
Correct

Incorrect

Hint

$\tan45=\dfrac{50}{DC}\\ \\ \Rightarrow DC=50\ m\\ \\ \tan30=\dfrac{AC}{BC}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{50}{x+50}\\ \\ \Rightarrow x+50=50\sqrt3\\ \\ \Rightarrow x=50(\sqrt3-1)$
Question 5 of 31

5. Question
1 point(s)
If the angle of elevation of a building from a distance of 100 m from its foot is $60^o $, then the height of the building is :
- $100\sqrt3\ m $
- $\dfrac{100}{\sqrt3}\ m $
- $50\sqrt3\ m $
- $\dfrac{200}{\sqrt3}\ m $
Correct

Incorrect

Hint

$60=\dfrac{h}{100}\\ \\ \Rightarrow \sqrt3=\dfrac{h}{100}\Rightarrow h=100\sqrt3 \ m$
Question 6 of 31

6. Question
1 point(s)
From the top of a cliff 50 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 :
- 50 m
- 100 m
- 125 m
- 150 m
Correct

Incorrect

Hint

$CD=EG=50\ m\\ \\ \tan\theta=\dfrac{EG}{DG}$
Also $\tan\theta=\dfrac{FG}{DG}$
So $\dfrac{EG}{DG}=\dfrac{FG}{DG}\Rightarrow EG=FG=50\ m$
Height of the tower $=50+50=100\ m$
Question 7 of 31

7. Question
1 point(s)
The altitude of the sun is $60^o $. The height of a tower which casts a shadow of length 30 m is :
- $30\sqrt3\ m $
- $15\ m $
- $\dfrac{30}{\sqrt3}\ m $
- $15\sqrt2\ m $
Correct

Incorrect

Hint

$\tan c=\dfrac{P}{h}=\dfrac{h}{30}\\ \\ \sqrt3=\dfrac{h}{30}\Rightarrow h=30\sqrt3\ m$
Question 8 of 31

8. Question
1 point(s)
A pole subtends an angle of $30^o $ 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 pole is $60^o $. The height of the pole is :
- $\dfrac{h}{2}\ m $
- $\sqrt3\ m $
- $\dfrac{h}{3}\ m $
- $\dfrac{h}{\sqrt3}\ m $
Correct

Incorrect

Hint

$\tan60=\dfrac{BE}{EC}\\ \\ \Rightarrow \sqrt3=\dfrac{h}{\sqrt{EC}}\\ \\ \Rightarrow EC=\dfrac{h}{\sqrt3}=BD\\ \\ \tan30=\dfrac{AB}{BD}=\dfrac{1}{\sqrt3}\\ \\ \Rightarrow \dfrac{H}{\dfrac{h}{\sqrt3}}=\dfrac{1}{\sqrt3}\\ \\ H=\dfrac{\dfrac{h}{\sqrt3}}{\sqrt3}=\dfrac{h}{\sqrt3}\times\dfrac{1}{\sqrt3}=\dfrac{h}{3}$
Question 9 of 31

9. Question
1 point(s)
The angle of elevation of the top of a tower from a point P on the ground is $\alpha $. After walking a distance d towards the foot of the tower, angle of elevation is found to be $\beta $. The height of the tower is
- $\dfrac{d\sin\alpha\sin\beta}{1\sin(\beta-2)} $
- $\dfrac{d}{\cot\alpha-\cot\beta} $
- $\dfrac{d}{\tan\alpha+\tan\beta} $
- $\tan\beta-\tan\alpha $
Correct

Incorrect

Hint

$\dfrac{x+d}{h}=\cot 2\\ \\ \Rightarrow x=h\cot \alpha-d\ tx=td\Rightarrow P$
Also $\dfrac{x}{h}=\cot \beta\Rightarrow x=h\cot \beta$
So, $h\cot \alpha-d=h\cot \beta\\ \\ \Rightarrow h\left(\dfrac{\cos\alpha}{\sin \alpha}-\dfrac{\cos\beta}{\sin\beta}\right)=d\\ \\ \Rightarrow h\left(\dfrac{\sin\beta\cos\alpha-\cos \beta\sin\alpha}{\sin \alpha\ \sin\beta}\right)=d\\ \\ \Rightarrow h=\dfrac{d\sin\alpha\sin\beta}{\sin(\beta-\alpha)}$
Question 10 of 31

10. Question
1 point(s)
The length of the shadow of a tower standing on level plane is found to be 2x metres longer when the sun’s altitude is $30^o $ than when it was $45^o $. The height of the tower is :
- $x(\sqrt3-1)\ m $
- $x(\sqrt3+1)\ m $
- $\dfrac{x}{\sqrt3-1}m $
- $\dfrac{\sqrt3-1}{x}m $
Correct

Incorrect

Hint

$\tan45=\dfrac{AB}{BD}\\ \\ \Rightarrow h=BD\\ \\ \tan30=\dfrac{AB}{BC}\\ \\ \Rightarrow \dfrac{h}{BC}=\dfrac{1}{\sqrt3}\Rightarrow BC=\sqrt3h\\ \\ CD+h=\sqrt3h\\ \\ \Rightarrow 2x=(\sqrt3-1)\ h\\ \\ h=\dfrac{2x}{\sqrt3-1}=\dfrac{2x(\sqrt3+1)}{3-1}=x(\sqrt3+1)$
Question 11 of 31

11. Question
1 point(s)
As observed from the top of a 75 m high light– house from the sea–level, the angles of depression of two ships are $30^o $ and $45^o $. One ship is exactly behind the other on the same side of the light–house. The distance between the two ships is :
- $75\ m $
- $75(\sqrt3+1)m $
- $75\ (\sqrt3)m $
- $75\ (\sqrt3-1)m $
Correct

Incorrect

Hint

$\tan C=\dfrac{AD}{CD}\\ \\ \tan45=\dfrac{75}{CD}\Rightarrow CD=75\ m\\ \\ \tan30=\dfrac{AD}{BD}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{75}{BD}\Rightarrow BD=75\sqrt3\\ \\ BC+CD=75\sqrt3\\ \\ \Rightarrow BC=75\sqrt3-75=75(\sqrt3-1)m$
Question 12 of 31

12. Question
1 point(s)
The 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 short person is :
- $\dfrac{a}{4} $
- $\dfrac{a}{\sqrt2} $
- $a\sqrt2 $
- $\dfrac{a}{2\sqrt2} $
Correct

Incorrect

Hint

$\tan(90-\theta)=\dfrac{2h}{\dfrac{a}{2}}\\ \\ \Rightarrow \cot\theta=\dfrac{4h}{a}\\ \\ \tan\theta=\dfrac{h}{\dfrac{a}{2}}\\ \\ \Rightarrow \dfrac{1}{\cot\theta}=\dfrac{2h}{a}\Rightarrow \dfrac{a}{4h}=\dfrac{2h}{a}\\ \\ \Rightarrow h=\dfrac{a}{2\sqrt2}$
Question 13 of 31

13. Question
1 point(s)
If the altitude of the sun is $60^o $, the height of the tower which casts a shadow of length 30 m is :
- $30\sqrt3\ m $
- $\dfrac{30}{3}\sqrt3\ m $
- $15\sqrt3\ m $
- $15\ m $
Correct

Incorrect

Hint

$\tan C=\dfrac{AB}{BC}\\ \\ \Rightarrow \sqrt2=\dfrac{h}{30}\Rightarrow h=30\sqrt3$
Question 14 of 31

14. Question
1 point(s)
The tops of two poles of heights 20 m and 14 m are connected by a wire. If the wire makes an angle of $ 30^o$ with the horizontal, then the length of the wire is :
- 34 m
- 12 m
- 6 m
- 17 m
Correct

Incorrect

Hint

$\sin B=\dfrac{AC}{AD}\\ \\ \Rightarrow \sin30=\dfrac{6}{h}\\ \\ \Rightarrow h=12m$
Question 15 of 31

15. Question
1 point(s)
The length of the string of a kite flying at 100 m above the ground with the elevation of $60^o $ is :
- $100\ m $
- $100\sqrt2\ m $
- $\dfrac{200}{\sqrt3} $
- $200\ m $
Correct

Incorrect

Hint

Let both of string be $x$
$\sin60=\dfrac{100}{x}\\ \\ \dfrac{\sqrt3}{2}=\dfrac{100}{x}\\ \\ x=\dfrac{200}{\sqrt3}$
Question 16 of 31

16. Question
1 point(s)
A ladder of 10 m length touches a wall at height of 5 m. The angle $\theta $ made by it with the horizontal is :
- $90^o $
- $60^o $
- $45^o $
- $30^o $
Correct

Incorrect

Hint

$\sin\theta=\dfrac{5}{10}\\ \\ \Rightarrow \sin\theta=\dfrac{1}{2}\Rightarrow \theta=30$
Question 17 of 31

17. Question
1 point(s)
The measure of angle of elevation of top of a tower $75\sqrt3 $ m high from a point at a distance of 75 m from foot of the tower in a horizontal plane is :
- $30^o $
- $60^o $
- $90^o $
- $45^o $
Correct

Incorrect

Hint

Let the angle of elevation be $\theta$
$\tan\theta=\dfrac{75\sqrt3}{75}=\sqrt3=60^o$
Question 18 of 31

18. Question
1 point(s)
A pole 10 m high casts a shadow 10 m long on the ground, then the sun’s elevation is :
- $60^o $
- $45^o $
- $30^o $
- $90^o $
Correct

Incorrect

Hint

$\tan x=\dfrac{10}{10}=1\\ \\ x=45^o$
Question 19 of 31

19. Question
1 point(s)
The angle of depression from the top of a tower 12 m high, at the point on the ground is $30^o $. The distance of the point from the top of the tower is
- $12\ m $
- $6\ m $
- $12\sqrt3\ m $
- $24\ m $
Correct

Incorrect

Hint

$\dfrac{AB}{BC}=\tan30\\ \\ \Rightarrow \dfrac{12}{BC}=\dfrac{1}{\sqrt3} \\ \\ BC=12\sqrt3\ m$
Question 20 of 31

20. Question
1 point(s)
A tree casts a shadow 4 m long on the ground, when the angle of elevation of the sun is $45^o $. The height of the tree (in metres) is :
- 3
- 4
- 4.5
- 5.2
Correct

Incorrect

Hint

In a triangle with me angle $90^o$ and other $45^o$ , the third angle is bound to be $45^o$ , So height of the tree is 4m
Question 21 of 31

21. Question
1 point(s)
If sun’s elevation is $60^o $, then a pole of height $6\ m $ will cast a shadow of length
- $6\sqrt3\ m $
- $\sqrt3\ m $
- $2\sqrt3\ m $
- $3\sqrt2\ m $
Correct

Incorrect

Hint

$\tan60=\dfrac{AB}{BC}\\ \\ \Rightarrow \sqrt3=\dfrac{6}{BC}\\ \\ \Rightarrow BC=\dfrac{6}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{6\sqrt3}{3}\\ \\ BC=2\sqrt3\ m$
Question 22 of 31

22. Question
1 point(s)
The angle fo elevation of the top of a building 50 m high, from a point on the ground is $45^o $. The distance of the point from the foot of the building is :
- 100 m
- 50 m
- 45 m
- 60 m
Correct

Incorrect

Hint

$\tan45=\dfrac{height}{distance}\\ \\ \Rightarrow 1=\dfrac{50}{distance}\\ \\ \Rightarrow distance=50\ m$
Question 23 of 31

23. Question
1 point(s)
A tree 6 m tall casts a 4 m long shadow. At the same time a pole casts a shadow 10 m long. The height of the pole is :
- 40 m
- 20 m
- 15 m
- 10 m
Correct

Incorrect

Hint

$\dfrac{6}{4}\times10=15\ m$
Question 24 of 31

24. Question
1 point(s)
If the angle of elevation of top of a tower from a point at a distance of $100\ m $ from its foot is $ 60^o$, then the height of the tower is :
- $50\sqrt3\ m $
- $\dfrac{200}{\sqrt3}\ m $
- $\dfrac{100}{\sqrt3}\ m $
- $100\sqrt3\ m $
Correct

Incorrect

Hint

$\tan60=\dfrac{h}{100}\\ \\ \Rightarrow \sqrt3=\dfrac{h}{100}\Rightarrow h=100\sqrt3\ m$
Question 25 of 31

25. Question
1 point(s)
The angle formed by the line of sight with the horizontal, when the point being viewed is above the horizontal level is called :
- vertical angle
- angle of depression
- angle of elevation
- obtuse angle
Correct

Incorrect

Hint

Angle of elevation.
Question 26 of 31

26. Question
1 point(s)
If the ratio of height of a tower and the length of its shadow on the ground is $\sqrt3:1 $, then the angle of elevation of the sun is :
- $60^o $
- $45^o $
- $30^o $
- $90^o $
Correct

Incorrect

Hint

$\dfrac{AB}{BC}=\dfrac{\sqrt3}{1}\\ \\ \Rightarrow \tan\theta=\dfrac{AB}{BC}=\sqrt3\\ \\ \theta=60^o$
Question 27 of 31

27. Question
1 point(s)
If the height and length of the shadow of a man are the same, then the angle of elevation of the sun is :
- $30^o $
- $60^o $
- $45^o $
- $15^o $
Correct

Incorrect

Hint

$\tan\theta=1\\ \\ \theta=45^o$
Question 28 of 31

28. Question
1 point(s)
If AB = 4 m and AC = 8 m, then angle of observation of A as observed from C is :
- $60^o $
- $30^o $
- $45^o $
- Cannot be determined
Correct

Incorrect

Hint

$\cos\theta=\dfrac{4}{8}=\dfrac{1}{2}\Rightarrow \theta=60^o$
Question 29 of 31

29. Question
1 point(s)
If the angle of depression of an object from a $75\ m $ high tower is $30^o $, then the distance of the object from the base of tower is
- $25\sqrt3\ m $
- $50\sqrt3\ m $
- $75\sqrt3\ m $
- $150\ m $
Correct

Incorrect

Hint

$\tan30=\dfrac{AB}{OB}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{75}{OB}\\ \\ \Rightarrow OB=75\sqrt3$
Question 30 of 31

30. Question
1 point(s)
The figure, shows the observation of point C from point A. The angle of depression from A is :
- $60^o $
- $30^o $
- $45^o $
- $75^o $
Correct

Incorrect

Hint

$\tan\theta=\dfrac{4}{4\sqrt3}=\dfrac{1}{\sqrt3}\\ \\ \theta=30^o$
Question 31 of 31

31. Question
1 point(s)
From the figure, the angle of depression of point C from the point P is :
- $90^o $
- $60^o $
- $30^o $
- $45^o $
Correct

Incorrect

Hint

$\angle A+\angle APB+\angle ABP=180^o\\ \\ 90+\angle APB+60=180^o\\ \\ \angle APB=30\\ \\ \angle APB+\angle BPC+\angle QPC=90\\ \\ 30+30+\theta=90\\ \\ Q=30^o$