Maths Class X Question Bank

Question 1 of 32

1. Question
1 point(s)
The angles of elevation of a cloud from a point h metres above a lake is 30° and the angle of depression of its reflection in the lake is 45°. If the height of the cloud be 200 m, find h.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\dfrac{200+h}{AC}=\tan45\\ \\ \Rightarrow AC=200+h\\ \\ \dfrac{BC}{AC}=\tan30\\ \\ \dfrac{200-h}{200+h}=\dfrac{1}{\sqrt3}\\ \\ \Rightarrow 200\sqrt3-h\sqrt3=200+h\\ \\ \Rightarrow \dfrac{200}{h}=\dfrac{\sqrt3+1}{\sqrt3-1}\\ \\ \Rightarrow h=\dfrac{200(\sqrt3-1)}{\sqrt3+1}$
Question 2 of 32

2. Question
1 point(s)
A bird is perched on the top of a tree 20 m high and its angle of elevation from a point on the ground 45°. The bird flies off horizontally straight away from the observer and in one second the angle of elevation of the bird reduces to 30°. Find the speed of the bird.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle OA_1B_1$
$\tan45=\dfrac{AB_1}{OB_1}\\ \\ \Rightarrow \dfrac{20}{OB_1}=1\Rightarrow OB_1=20$
In $\triangle OA_2B_2,\ \tan30=\dfrac{20}{OB_2}$
$\Rightarrow OB_2=20\sqrt3\\ \\ \Rightarrow B_1B_2+OB_1=20\sqrt3\\ \\ \Rightarrow B_1B_2=20\sqrt3-20=20(\sqrt3-1)$
speed $=\dfrac{D}{9}=\dfrac{20(\sqrt3-1)}{1}\\ \\ =20(\sqrt3-1)\ m/s$
Question 3 of 32

3. Question
1 point(s)
From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be $\alpha $ and $\beta (\alpha > \beta)$. If the distance between the objects is a metres, show that the height of the tower is $\dfrac{a\tan\alpha\tan\beta}{\tan\alpha-\tan\beta} $.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABC$
$\tan\alpha=\dfrac{h}{BC}\\ \\ \Rightarrow x=\dfrac{h}{\tan\alpha}$ …(1)
In $\tan\beta=\dfrac{b}{a+x}\\ \\ \Rightarrow h=\tan\beta(a+x)\\ \\ \Rightarrow h=a\tan\beta+x\tan\beta\\ \\ \Rightarrow h=\alpha\tan\beta+\dfrac{h\tan\beta}{\tan\alpha}\\ \\ \Rightarrow h-\dfrac{h\tan\beta}{\tan\alpha}=\alpha\tan\beta\\ \\ \Rightarrow \dfrac{h(\tan\alpha-\tan\beta)}{\tan\alpha}=a\tan\beta\\ \\ \Rightarrow h=\dfrac{a\tan\beta\tan\alpha}{\tan\alpha-\tan\beta}$ (Proved)
Question 4 of 32

4. Question
1 point(s)
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are $\alpha $ and $ \beta$ respectively. Prove that the height of the tower is $\dfrac{h\tan\alpha}{\tan\beta-\tan\alpha} $
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle OAB$
$\tan\alpha=\dfrac{AB}{OA}\\ \\ \Rightarrow y=x\tan\alpha$
$\Rightarrow x=\dfrac{y}{\tan\alpha}$ …(1)
In $\triangle OAC$
$\tan\beta=\dfrac{y+h}{x}\Rightarrow x=\dfrac{y+h}{\tan\beta}$ …(2)
$\dfrac{y}{\tan\alpha}=\dfrac{y+h}{\tan\beta}\\ \\ \Rightarrow y(\tan\beta-\tan\alpha)=h\tan\alpha\\ \\ \Rightarrow y=\dfrac{h\tan\alpha}{\tan\beta-\tan\alpha}$
Question 5 of 32

5. Question
1 point(s)
A tree standing on a horizontal plane is leaning towards east. At two points situated at distance $\alpha $ and $\beta $ exactly due west of it, the angles of elevation of the top are $\alpha $ and $\beta $ respectively. Prove that the height of the top of the tree from the ground is $ \dfrac{(b-a)\tan\alpha\tan\beta}{\tan\alpha-\tan\beta}$
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$OA=a\\ \\ OB=b\\ \\ PL=h,\ OL=x$
In $\triangle ALP$
$\tan\alpha=\dfrac{PL}{OA+OL}=\dfrac{h}{a+x}\\ \\ \Rightarrow \dfrac{1}{\cot \alpha}=\dfrac{h}{a+x}\ \Rightarrow h\cot \alpha=a+x$ …(1)
In $\triangle BLP$
$\tan\beta=\dfrac{PL}{OB+OL}\\ \\ \Rightarrow \tan\beta=\dfrac{h}{b+x}\\ \\ \Rightarrow h\cot \beta=b+x$ …(2)
Subtracting eqn (1) from (2) we get
$h\cot \beta-h\cot \alpha=b-a\\ \\ \Rightarrow h=\dfrac{b-a}{\cot \beta-\cot \alpha}\\ \\ =\dfrac{(b-a)(\tan\alpha\ \tan\beta)}{\tan\alpha-\tan\beta}$
Question 6 of 32

6. Question
1 point(s)
At a point on a level plane, a tower subtends an angle $\alpha $ and a man a metres tall standing on its top subtends an angle $\beta $. Prove that the height of the tower is $\dfrac{a\sin\alpha\cos(\alpha+\beta)}{\sin\beta} $
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\tan\alpha=\dfrac{h}{d}\\ \\ \tan(\alpha+\beta)=\dfrac{h+a}{d}\\ \\ \dfrac{\tan\alpha}{\tan(\alpha+\beta)}=\dfrac{h}{h+a}\\ \\ \Rightarrow \dfrac{\tan(\alpha+\beta)}{\tan\alpha}=\dfrac{h+a}{h}\\ \\ \Rightarrow h=\dfrac{a\tan\alpha}{\tan(\alpha+\beta)-\tan\alpha}\\ \\ =\dfrac{a\sin\alpha}{\cos\alpha}\ .\ \cos\alpha\ \cos(\alpha+\beta)\\ \\ \sin(\alpha+\beta)\ \cos2-\cos(\alpha+\beta)\sin\alpha\\ \\ \Rightarrow h=\dfrac{a\sin\alpha\cos(\alpha+\beta)}{\sin\beta}$ (Proved)
Question 7 of 32

7. Question
1 point(s)
An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After a flight of 10 seconds, its angle of elevation is observed to be 30° from the same point on the ground. Find the speed of the aeroplane in km/ hour.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABC$ ,
$\dfrac{AC}{BC}=\tan60\\ \\ \Rightarrow \dfrac{1000}{BC}=\sqrt3\\ \\ \Rightarrow BC=\dfrac{1000}{\sqrt3}\ m$
In $\triangle BDE$ ,
$\dfrac{DE}{BE}=\tan30\Rightarrow BE=1000\sqrt3\\ \\ CE=BE-BC\\ \\ \Rightarrow 10x=1000\sqrt3-\dfrac{1000}{\sqrt3}\\ \\ \Rightarrow x-100\left(\sqrt3-\dfrac{1}{\sqrt3}\right)=100\times1.154\\ \\ =115.4\ m/s\\ \\ =415.67\ km/hr$
Question 8 of 32

8. Question
1 point(s)
The angle of elevation of the top of a tower from two points P and Q at distance of 4 m and 9 m respectively from the base of the tower and in the same straight line with it are 60° and 30°. Prove that the height of the tower is 6 m.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABP$
$\tan\alpha=\dfrac{AB}{BP}$ …(1)
In $\triangle ABQ$
$\tan P=\dfrac{AB}{BQ}\\ \\ \tan(90-\alpha)=\dfrac{AB}{BQ}\\ \\ \cot\alpha=\dfrac{AB}{BQ}\\ \\ \dfrac{1}{\tan\alpha}=\dfrac{AB}{BQ}\Rightarrow \tan\alpha=\dfrac{BQ}{AB}$ …(2)
From (1) and (2)
$\dfrac{AB}{BP}=\dfrac{BQ}{AB}\Rightarrow AB^2=BQ\times BP\\ \\ =4\times9=36\\ \\ \Rightarrow AB=6\ m$
Question 9 of 32

9. Question
1 point(s)
A person standing on the bank of a river observes that the angle of elevation of the top of the tree standing on the opposite bank is 60°. When he moves 30 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and the width of the river.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ACB$
$\tan60=\dfrac{AB}{BC}\\ \\ \Rightarrow \sqrt3=\dfrac{h}{x}\Rightarrow x=\dfrac{h}{\sqrt3}$ …(1)
In $\triangle ADB$
$\tan30=\dfrac{AB}{BD}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+30}\\ \\ \Rightarrow \sqrt3h=x+30$ …(2)
$\sqrt3h=\dfrac{h}{\sqrt3}+30\Rightarrow \dfrac{2h}{\sqrt3}=30\\ \\ \Rightarrow 2h=30\sqrt3\\ \\ \Rightarrow h=15\sqrt3\ m\\ \\ x=\dfrac{15\sqrt3}{\sqrt3}=15\ m$
So the height of tree is $15\sqrt3\ m$ and width is $15\ m$
Question 10 of 32

10. Question
1 point(s)
The angle of elevation of a cloud from a point 60 m above the lake is 30° and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud above the lake.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle OAB$
$\tan30=\dfrac{AO}{OD}=\dfrac{H}{OD}\\ \\ \Rightarrow OD=\sqrt3H$ …(1)
In $\triangle A'OD$
$\tan60=\dfrac{OA'}{OD}=\dfrac{OB+BA'}{OD}\\ \\ \Rightarrow \sqrt3=\dfrac{60+60+H}{\sqrt3}=\dfrac{120+H}{\sqrt3}\\ \\ \Rightarrow 2H=120\Rightarrow H=60\ m$
Height of the cloud from the surface of the lake $=AB+A'B$
$=60+60=120\ m$
Question 11 of 32

11. Question
1 point(s)
A straight highways leads to the foot of tower. A man standing at the top of tower observes a car at an angle of depression 30° which is approaching the foot of the tower with a uniform speed. Six second later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

Time taken to car y distance = 6 sec
$\tan30=\dfrac{h}{x+y}$ and $\tan60=\dfrac{h}{x}\\ \\ \Rightarrow x+y=\sqrt3h\ \Rightarrow x=\dfrac{h}{\sqrt3}\\ \\ y=\dfrac{2h}{\sqrt3}=2x\\ \\ \dfrac{x}{t}=\dfrac{x+y}{6+t}\ (\because\ y=2x)\\ \\ 6+t=3t\\ \\ 2t=6\Rightarrow t=3\sec$
Question 12 of 32

12. Question
1 point(s)
As observed from the top of a light house, 100 m high above sea-level, the angle of depression of a ship sailing directly towards it, changes from 30° to 60°. Detremine the distance travelled by the ship during the period of observation $(\sqrt3=1.732) $
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABC$
$\tan60=\dfrac{AB}{BD}\\ \\ \Rightarrow x=\dfrac{100}{\sqrt3}$
In $\triangle ABD$
$\tan30=\dfrac{AB}{BD}\Rightarrow \dfrac{100}{y}=\dfrac{1}{\sqrt3}\Rightarrow y=100\sqrt3\\ \\ BD=BC+CD\\ \\ y=x+CD\\ \\ \Rightarrow CD=y-x=100\sqrt3-\dfrac{100}{\sqrt3}=\dfrac{100\times(3-1)}{\sqrt3}\\ \\ =\dfrac{100\times2}{\sqrt3}\\ \\ =\dfrac{200}{\sqrt3}=\dfrac{200\sqrt3}{3}\ m$
Question 13 of 32

13. Question
1 point(s)
A man standing on the deck of the ship which is 10 m above the sea level, observes the angle of elevation of the top of the cloud as $60^o $ and angle of depression of its reflection in the sea was found to be $ 30^o$. Find the height of the cloud and also the distance of the cloud from the ship
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ADE$
$\tan60=\dfrac{ED}{DA}\\ \\ \Rightarrow \sqrt3=\dfrac{h}{x}\\ \\ \Rightarrow h=x\sqrt3$ …(1)
In $\triangle ABC$
$\tan30=\dfrac{AB}{BC}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{10}{BC}\Rightarrow x=10\sqrt3$ …(2)
Substitute the value of $x$ from eqn (2) in eqn (1)
$h=(10\sqrt3)\sqrt3=30\ m\\ \\ CE=h+30=40\ m$
So the height of an will is 40 m and distance is $10\sqrt3\ m$
Question 14 of 32

14. Question
1 point(s)
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height ‘h’. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are 45° and 60° respectively. Find the height of the tower.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

Let BC = height of tower = H
In $\triangle BCD$
$\tan45=\dfrac{BC}{CD}\\ \\ \Rightarrow 1=\dfrac{H}{CD}\Rightarrow H=CD$ …(1)
In $\triangle ACD$
$\tan60=\dfrac{AC}{CD}\\ \\ \sqrt3=\dfrac{H+h}{CD}\\ \\ \Rightarrow \sqrt3H=H+h\\ \\ \Rightarrow (\sqrt3-1)H=h\\ \\ \Rightarrow H=\dfrac{h}{\sqrt3-1}=\dfrac{h(\sqrt3+1)}{2}$
Question 15 of 32

15. Question
1 point(s)
From the top of a light house the angle of depression of a ship sailing towards it was found to be 30°. After 10 seconds the angle of depression changes to 60°. Assuming that the ship is sailing at uniform speed, find how much time it will take to reach the light house.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABC$
$\tan60=\dfrac{h}{y}\Rightarrow \sqrt3=\dfrac{h}{y}\\ \\ \Rightarrow h=\sqrt3y$ …(1)
In $\triangle ADC$
$\tan30=\dfrac{h}{x+y}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+y}\\ \\ \Rightarrow x+y=3y\Rightarrow x=2y$
Time takesn $=\dfrac{10}{2}=5$ second
Question 16 of 32

16. Question
1 point(s)
From a point on the ground the angle of elevation of the bottom and the top of a flagstaff situated on the top of a 120 m tall house, was found to be 30° and 45° respectively. Find the height of the flagstaff.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABC$
$\tan30=\dfrac{BC}{AB}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{120}{AB}\\ \\ \Rightarrow AB=120\sqrt3$ …(1)
In $\triangle ABD$
$\tan45=\dfrac{h+120}{120\sqrt3}\\ \\ \Rightarrow 1=\dfrac{h+120}{120\sqrt3}\\ \\ \Rightarrow 120\sqrt3-120=h\\ \\ \Rightarrow 120(\sqrt3-1)=h$
Question 17 of 32

17. Question
1 point(s)
A vertical tower is surmounted by a flag staff of height 5 metres. At a point on the ground, the angles of elevation of bottom and top of flag staff are 45° and 60° respectively. Find the height of the tower.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABC$
$\tan60=\dfrac{5+h}{x}\\ \\ \Rightarrow 5+h=\sqrt3x$ …(1)
In $\triangle DBL$
$\tan45=\dfrac{h}{x}\Rightarrow h=x$
Substituting in eqn (1) we get
$5+h=\sqrt3\ h\\ \\ \Rightarrow 5=h(\sqrt3-1)\\ \\ \Rightarrow 5=h\times 0.732\\ \\ \Rightarrow h=\dfrac{5}{0.732}=6.84\ m$
Question 18 of 32

18. Question
1 point(s)
A man on the top of a vertical tower observes a car moving towards the tower. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this the car will reach the tower ?
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABD$
$\tan30=\dfrac{AB}{BD}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{AB}{BD}\\ \\ \Rightarrow \dfrac{BD}{\sqrt3}=AB\Rightarrow \dfrac{BC+CD}{\sqrt3}=AB$ ….(1)
In $\triangle ABC$
$\tan45=\dfrac{AB}{BC}\\ \\ \Rightarrow 1=\dfrac{AB}{BC}\Rightarrow AB=BC$ …(2)
$\dfrac{AB+CD}{\sqrt3}=AB\\ \\ \Rightarrow AB+CD=\sqrt3\ AB\\ \\ \Rightarrow AB(\sqrt3-1)=CD$ …(3)
Let m/min be the speed of car
$CD=12v$ ….(4)
$AB(\sqrt3-D)=12v\\ \\ \Rightarrow \dfrac{AB(\sqrt3-1)}{12}=v$ …(5)
$Time =\dfrac{BC}{v}=\dfrac{12\ BC}{AB(\sqrt3-1)}\\ \\ =\dfrac{12\ AB}{AB\ (\sqrt3-1)}$
$Time =\dfrac{12}{\sqrt3-1}=16.4\ min$
Question 19 of 32

19. Question
1 point(s)
An aeroplane at an altitude of 200 m observes the angles of depression of two opposite points on two banks of the river to be 45° and 60°. Find, in metres, the width of the river. (use $\sqrt3=1.732 $)
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ACD$
$\dfrac{CD}{AD}=\tan60\\ \\ \Rightarrow AD=\dfrac{200}{\sqrt3}=\dfrac{200\sqrt3}{3}$
In $\triangle BCD$
$\dfrac{CD}{DB}=\tan45\ \Rightarrow \dfrac{CD}{DB}=1\Rightarrow DB=200\ m$
Width of the river $=200+\dfrac{200\sqrt3}{3}$
$=200\left(1+\dfrac{\sqrt3}{3}\right)\\ \\ =\dfrac{200\times4.732}{3}=315.5\ m$
Question 20 of 32

20. Question
1 point(s)
From a window, 60 m high above the ground, of a house in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are 60° and 45° respectively. Show that the height of the opposite house is $60\ (1+\sqrt3) $ metres.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle\ CDE$
$\dfrac{DE}{CD}=\tan45\\ \\ \Rightarrow \dfrac{60}{y}=1\\ \\ \Rightarrow y=60\ m$
In $\triangle \ ADE$
$\dfrac{AB}{BE}=\tan60\Rightarrow \dfrac{x}{60} =\sqrt3\ \Rightarrow x=60\sqrt3$
Height $=x+60$
$=60+60\sqrt3\\ \\ =60(1+\sqrt3)\ m$
Question 21 of 32

21. Question
1 point(s)
Two pillars of equal heights are on either side of a road, which is 100 m wide. The angles of elevation of the top of the pillars are 60° and 30° at a point on the road between the pillars. Find the position of the point between the pillars on the road and the height of the pillars.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle PAB$
$\tan60=\dfrac{AB}{AP}\Rightarrow \sqrt3=\dfrac{h}{x}\Rightarrow h=\sqrt3x$ ….(1)
In $\triangle PCD$
$\tan30=\dfrac{CD}{PC}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{100-x}$ …(2)
From (1) and (2)
$3x=100-x\Rightarrow x=25\\ \\ h=25\sqrt3\ m$
so the height is $25\sqrt3$ and distance is $25\ m$
Question 22 of 32

22. Question
1 point(s)
From the top of a building 60 m high the angles of depression of the top and the bottom of a tower are observed to be 30° and 60° respectively. Find the height of the tower.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle AEC$
$\tan30=\dfrac{AE}{EC}\\ \\ \Rightarrow EC=\sqrt3\ AE\\ \\ \tan60=\dfrac{AB}{BD}\\ \\ \Rightarrow AB=\sqrt3\ BD=\sqrt3\ EC\\ \\ EC=\dfrac{AB}{\sqrt3}=\dfrac{60}{\sqrt3}=20\sqrt3\\ \\ AE=20\ m\\ \\ EB=AB-AE=60-20=40\ m$
Question 23 of 32

23. Question
1 point(s)
Find the height of a mountain if the elevation of its top at an unknown distance from the base is 60° and at a distance 10 km further off from the mountain, along the same line, the angle of elevation is 30°.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle CAB$
$\dfrac{h}{x}=\tan60=\sqrt3\\ \\ \Rightarrow h=\sqrt3x$ …(1)
In $\triangle DAB$
$\dfrac{h}{10+x}=\tan30=\dfrac{1}{\sqrt3}\\ \\ \Rightarrow h=\dfrac{10+x}{\sqrt3}\Rightarrow \sqrt3x=\dfrac{10+x}{\sqrt3}\\ \\ \Rightarrow 3x=10+x\Rightarrow x=5\\ \\ h=\sqrt3\times5=5\sqrt3$
Question 24 of 32

24. Question
1 point(s)
From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive kilometre stones on opposite sides of the areoplane are observed to be 60° and 30°. Show that height (in metres) of aeroplane above the road is $\dfrac{\sqrt3}{4} $ km
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABD$ ,
$\tan\alpha=\dfrac{AD}{BD}\\ \\ \Rightarrow BD=\dfrac{h}{\tan\alpha}$ …(1)
In $\triangle ACD,\ \tan\beta=\dfrac{AC}{CD}$
$\Rightarrow \tan\beta=\dfrac{h}{CD}\\ \\ \Rightarrow CD=\dfrac{h}{\tan\beta}$ …(2)
$BC=\dfrac{h}{\tan\alpha}+\dfrac{h}{\tan\beta}=h\left(\dfrac{1}{\tan\alpha}+\dfrac{1}{\tan\beta}\right)\\ \\ \Rightarrow 1=h\left(\dfrac{\tan\alpha+\tan\beta}{\tan\alpha+\tan\beta}\right)\\ \\ \Rightarrow h=\dfrac{\tan\alpha\ \tan\beta}{\tan\alpha+\tan\beta}\\ \\ \dfrac{\tan60\ \tan30}{\tan60+\tan30}\\ \\ =\dfrac{\sqrt3\times\dfrac{1}{\sqrt3}}{\sqrt3+\dfrac{1}{\sqrt3}}=\dfrac{1}{\dfrac{3+1}{\sqrt3}}=\dfrac{\sqrt3}{4}\ km$
Question 25 of 32

25. Question
1 point(s)
From the top and foot of a tower 40 m high, the angle of elevation of the top of a light house are found to be 30° and 60° respectively. Find the height of the light house. Also, find the distance of the top of the light house from the foot of the tower.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle CDE$
$\tan30=\dfrac{DE}{CD}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x}\\ \\ \Rightarrow x=\sqrt3h$ …(1)
In $\triangle ABE$
$\tan60=\dfrac{BE}{AB}\ \Rightarrow \sqrt3=\dfrac{h+40}{x}\\ \\ \Rightarrow x=\dfrac{h+90}{\sqrt3}$ …(2)
Comparing (1) and (2)
$\sqrt3h=\dfrac{h+40}{\sqrt3}\Rightarrow h=20\ m$
Height of the length house $=40+20=60\ m$
In $\triangle ABE$
$\dfrac{\sqrt3}{2}=\dfrac{60}{AE}\Rightarrow AE=\dfrac{120}{\sqrt3}=\dfrac{120\sqrt3}{3}\\ \\ =40\sqrt3\ m$
Question 26 of 32

26. Question
1 point(s)
The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle\ ABC$
$\tan30-\dfrac{BC}{AB}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{AB}\Rightarrow AB=\sqrt3\ h$ …(1)
In $\triangle \ BAD$
$\tan60=\dfrac{AD}{AB}\ \Rightarrow \sqrt3=\dfrac{50}{AB}\ \Rightarrow AB=\dfrac{50}{\sqrt3}$
Comparing (1) and (2)
$\sqrt3h=\dfrac{50}{\sqrt3}\ \Rightarrow 3h=50\Rightarrow h=\dfrac{50}{3}\\ \\ =16\dfrac{2}{3}m$
Question 27 of 32

27. Question
1 point(s)
The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storied building are 30° and 45° respectively. Find the height of the multistoried building.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\tan30=\dfrac{DE}{AD}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{y}{x}$
In $\triangle EBC$
$\tan45=\dfrac{CE}{BC}\Rightarrow 1=\dfrac{8+y}{x}\\ \\ \Rightarrow x=8+y\\ \\ \Rightarrow \sqrt3y=8+y\\ \\ \Rightarrow y=4$
Height of the tower $=CE=8+y=8+4=12\ cm$
$x=\sqrt 3\times4=6.92\ m$
Question 28 of 32

28. Question
1 point(s)
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of 20 m high building are 45° and 60° respectively. Find the height of the transmission tower.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle ABC$
$\tan45=\dfrac{BC}{AB}\\ \\ \Rightarrow 1=\dfrac{20}{AB}\Rightarrow AB=20\ m$
In $\triangle \ ABD,\ \tan60=\dfrac{BD}{AB}\Rightarrow \sqrt3=\dfrac{h+20}{20}\\ \\ \Rightarrow h=20(\sqrt3-1)m$
Question 29 of 32

29. Question
1 point(s)
An aircraft is flying at a constant height with a speed of 360 km/hour. From a point on the ground, the angle of elevation at an instant was observed to be 45°. After 20 seconds, the angle of elevation was observed to be 30°. Determine the height at which the aircraft is flying. (use $\sqrt3=1.732 $)
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

Distance AB = Speed $\times$ times
$=360\times\dfrac{20}{60\times60}=20\ km\\ \\ DC=2\ km$
In $\triangle AOD$
$\tan45=\dfrac{h}{OD}\Rightarrow 1=\dfrac{h}{OD}\Rightarrow h=OD$
In $\triangle BOC$
$\tan30=\dfrac{h}{OD+2}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{h+2}\\ \\ \Rightarrow h=\dfrac{2}{\sqrt3-1}\times\dfrac{\sqrt3+1}{\sqrt3+1}\\ \\ =\sqrt3+1\\ \\ =2.732\ km$
Question 30 of 32

30. Question
1 point(s)
A boy 2 m tall is standing at some distance from a 20 m tall building. The angle of elevation from his eyes of the top of the building increases from $30^o $ to $60^o $ as he walks towards the building. Find the distance he walked towards the building.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle AGF$
$\tan60=\dfrac{AG}{FG}\\ \\ \Rightarrow \sqrt3=\dfrac{28}{FG}\\ \\ \Rightarrow FG=\dfrac{28}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{28\sqrt3}{3}$
In $\triangle AGE$
$\tan30=\dfrac{AG}{x+FG}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{28}{x+\dfrac{28\sqrt3}{3}}\\ \\ \Rightarrow x=\dfrac{56\sqrt3}{3}=18.67\sqrt3\ m$
Question 31 of 32

31. Question
1 point(s)
Two men on either side of a cliff, 60 m high, observe the angles of elevation of the top of the cliff to be 45° and 60° respectively. Find the distance between two men.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle BCD$
$\dfrac{CD}{DB}=\tan45\\ \\ \Rightarrow \dfrac{60}{y}=1\Rightarrow y=60\ m$
In $\triangle ACD$
$\dfrac{CD}{AD}=\tan60\Rightarrow \dfrac{60}{x}=\sqrt3\Rightarrow x=20\sqrt3$
Distance $=AB=x+y$
$=20\sqrt3+60\\ \\ =20(3+\sqrt3)m$
Question 32 of 32

32. Question
1 point(s)
If the angle of elevation of a cloud from a point h metres above a lake is $\alpha $ and the angle of depression of its reflection in the lake is $\beta $, prove that the distance of the cloud from the point of observation is $\dfrac{2h\sec\alpha}{\tan\beta-\tan\alpha} $
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle DCE$
$\tan\alpha=\dfrac{DE}{CE}\Rightarrow CE=\dfrac{H}{\tan\alpha}$ …(1)
In $\triangle CED'$
$CE=\dfrac{h+H+h}{\tan\beta}=\dfrac{2h+H}{\tan\beta}$
From (1) and (2)
$\dfrac{H}{\tan\alpha}=\dfrac{2h+H}{\tan\beta}\Rightarrow H\tan\beta=2h\tan\alpha+H\tan\alpha\\ \\ H\tan\beta-H\tan\alpha=2h\tan\alpha\\ \\ \Rightarrow H(\tan\beta-\tan\alpha)=2h\tan\alpha\\ \\ \Rightarrow H=\dfrac{2h\tan\alpha}{\tan\beta-\tan\alpha}$ …(3)
In $\triangle DCE$
$\sin\alpha=\dfrac{DE}{CD}\Rightarrow CD=\dfrac{DE}{\sin\alpha}=\dfrac{H}{\sin\alpha}$
Substitute is the value of H from eqn
$CD=2h\tan\alpha\\ \\ (\tan\beta-\tan\alpha) \ \sin\alpha$
$CD=\dfrac{2h\sec\alpha}{\tan\beta-\tan\alpha}$ (Proved)