Maths Class X Question Bank

Question 1 of 30

1. Question
1 point(s)
A tower stands vertically on the ground. From a point on the ground which is 60 m away from foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
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Hint

$\tan60=\dfrac{AB}{BC}\\ \\ \Rightarrow \sqrt3=\dfrac{h}{60}\\ \\ \Rightarrow h=60\sqrt3\ m$
Question 2 of 30

2. Question
1 point(s)
A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.
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Hint

$\sin30=\dfrac{AB}{AC}\\ \\ \Rightarrow \dfrac{1}{2}=\dfrac{AB}{15}\\ \\ \Rightarrow AB=\dfrac{15}{2}=7.5\ m$
Question 3 of 30

3. Question
1 point(s)
A tower stands vertically on the ground. From a point on the ground which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
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Hint

$\tan c=\dfrac{AB}{CB}\\ \\ \Rightarrow \tan60=\dfrac{H}{15}\\ \\ \Rightarrow H=15\sqrt3\ m$
Question 4 of 30

4. Question
1 point(s)
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
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Hint

$\cos C=\dfrac{BC}{AC}\\ \\ \Rightarrow \cos30=\dfrac{8}{AC}\Rightarrow AC=\dfrac{16}{\sqrt3}\\ \\ \sin30=\dfrac{AB}{AC}\\ \\ \Rightarrow \dfrac{1}{2}=\dfrac{AB}{\dfrac{16}{\sqrt3}}\\ \\ \Rightarrow AB=\dfrac{16}{2\sqrt3}=\dfrac{8}{\sqrt3}$
Height of tree $=AC+AB$
$=\dfrac{16}{\sqrt3}+\dfrac{8}{\sqrt3}=\dfrac{8}{\sqrt3}=\dfrac{24}{\sqrt3}=\dfrac{24\sqrt3}{3}\\ \\ =8\sqrt3\ m$
Question 5 of 30

5. Question
1 point(s)
A kite is flying at a height of 90 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°. Find the length of the string assuming that there is no slack in the string. [Take latex]\sqrt3=1.732 [/latex]]
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Hint

$\sin60=\dfrac{AB}{AC}\\ \\ \Rightarrow \dfrac{\sqrt3}{2}=\dfrac{90}{x}\\ \\ \Rightarrow x=\dfrac{180}{\sqrt3}=\dfrac{180\sqrt3}{3}=60\sqrt3\\ \\ =60\times1.732\\ \\ =103.92\ m$
Question 6 of 30

6. Question
1 point(s)
A player sitting on the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground as 60°. Find the distance between the foot of the tower and the ball.
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Hint

$\tan60=\dfrac{AB}{BC}\\ \\ \Rightarrow \sqrt3=\dfrac{20}{x}\ \Rightarrow x=\dfrac{20}{\sqrt3}\\ \\ =\dfrac{20\sqrt3}{3}\\ \\ =11.53\ m$
Question 7 of 30

7. Question
1 point(s)
The shadow of a tower is 30 m long, when the sun’s elevation is 30°. What is the length of the shadow, when sun’s elevation is 60° ?
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Hint

In $\triangle ABC$
$\dfrac{AB}{BC}=\tan30^o\\ \\ \Rightarrow \dfrac{x}{30}=\dfrac{1}{\sqrt3}\\ \\ \Rightarrow x=10\sqrt3$
In $\triangle ABD$ ,
$\dfrac{AB}{BD}=\tan60\\ \\ \Rightarrow \dfrac{10\sqrt3}{y}=\sqrt3\\ \\ \Rightarrow y=10\ m$
Length of shadow $=10\ m$
Question 8 of 30

8. Question
1 point(s)
From a point P on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flag is hosted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. [Take latex]\sqrt3=1.732 [/latex]]
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Hint

$\tan30=\dfrac{AB}{AD}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{10}{AP}\Rightarrow AP=10\sqrt3\ m\\ \\ \tan45=\dfrac{AD}{AP}\\ \\ \Rightarrow 1=\dfrac{x+10}{17.32}\\ \\ \Rightarrow x=17.32-10=7.32\ m$
Question 9 of 30

9. Question
1 point(s)
A person standing on the bank of the river observes that the angle subtended by a tree on the opposite bank is 60°, when he retreats 20 m from the bank, he finds the angle to be 30°, find the height of the tree.
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Hint

In $\triangle ACB$
$\tan60=\dfrac{AB}{BC}\\ \\ \Rightarrow \sqrt3=\dfrac{h}{x}\\ \\ \Rightarrow x=\dfrac{h}{\sqrt3}$
In $\triangle ADB$
$\tan30=\dfrac{AB}{BD}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+20}\\ \\ \Rightarrow \sqrt3 h=x+20\\ \\ \sqrt3 h=\dfrac{h}{\sqrt3}+20\\ \\ \Rightarrow \dfrac{3h-h}{\sqrt3}=20\ \Rightarrow \dfrac{2h}{\sqrt3}=20\\ \\ \Rightarrow h=\dfrac{20\sqrt3}{2}=10\sqrt3\\ \\ =10\times1.732=17.32\ m$
Question 10 of 30

10. Question
1 point(s)
The angle 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. Prove that the height of tower is $\sqrt{ab} $.
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Hint

In $\triangle OAT$
$\tan\theta=\dfrac{OT}{OA}=\dfrac{h}{a}\\ \\ \tan(90-2)=\dfrac{OT}{OB}=\dfrac{h}{b}\\ \\ \Rightarrow \cot\alpha=\dfrac{h}{b}$
Multiplying eqn
$\tan\alpha\cot\alpha=\dfrac{h}{a}\times \dfrac{h}{b}=\dfrac{h^2}{ab}\\ \\ \Rightarrow \tan\alpha\times\dfrac{1}{\tan\alpha}=\dfrac{h^2}{ab}\\ \\ \Rightarrow 1=\dfrac{h^2}{ab}\Rightarrow h^2=ab\Rightarrow h=\sqrt{ab}$
Question 11 of 30

11. Question
1 point(s)
A tower is 60 m high. From the top of it the angles of depression of the top and the bottom of a tree are found to be 30° and 60° respectively. Find the height of the tree and its distance from the tower.
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Hint

In $\triangle AED$
$\dfrac{60-h}{ED}=\tan30^o\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{60-h}{BC}\\ \\ \Rightarrow \sqrt3\ (60-h)=x$
In $\triangle ABC$ ,
$\dfrac{60}{x}=\tan60^o\\ \\ \Rightarrow 60=\sqrt3x$
Putting the value of $x$ from eqn
$60=\sqrt3\times\sqrt3\ (60-h)\\ \\ \Rightarrow 60=3(60-h)\\ \\ \Rightarrow 20=(60-h)\Rightarrow h=40\ m\\ \\ x=\dfrac{60}{\sqrt3}=\dfrac{60\sqrt3}{3}=20\sqrt3\ m$
So, height of the tower is 40 m and distance from the tower = $20\sqrt3\ m$
Question 12 of 30

12. Question
1 point(s)
A man standing on the top of a multi-storey building, which is 30 m high, observes the angle of elevation of the top of a tower as 60° and the angle of depression of the base of the tower as 30°. Find the horizontal distance between the building and the tower. Also, find the height of the tower.
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Hint

In $\triangle ABD$
$\tan30=\dfrac{AB}{BD}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{30}{x}\\ \\ \Rightarrow x=30\sqrt3\ m$
In $\triangle AEC$
$\tan60=\dfrac{CE}{AE}\\ \\ \Rightarrow \sqrt3=\dfrac{h-30}{x}\\ \\ \Rightarrow \sqrt3=\dfrac{h-30}{30\sqrt3}\\ \\ \Rightarrow 90=h-30\Rightarrow h=120\ m$
So the horizontal distance is $30\sqrt3\ m$ and height is $120\ m$
Question 13 of 30

13. Question
1 point(s)
An aeroplane, when 3000 m high, passes vertically above another plane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the two aeroplanes.
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Hint

In $\triangle ADC$
$\tan60=\dfrac{AC}{DC}\\ \\ \Rightarrow \sqrt3=\dfrac{3000}{DC}\\ \\ \Rightarrow DC=\dfrac{3000}{\sqrt3}=\dfrac{3000\sqrt3}{3}=1000\sqrt3\\ \\ =1732\ m$
In $\triangle BDC$
$\tan45^o=\dfrac{BC}{DC}\\ \\ \Rightarrow 1=\dfrac{x}{1732}\Rightarrow x=1732\ m$
Difference in height $=3000-x$
$=3000-1732=1268\ m$
The vertical distance $=1268\ m$
Question 14 of 30

14. Question
1 point(s)
A circus artist is climbing a rope 12 m long which is tightly stretched and tied from the top of a vertical pole to the ground. find the height of the pole if the angle made by the rope with the ground is 30°.
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Hint

In $\triangle ABC$
$\sin30=\dfrac{AB}{AC}=\dfrac{h}{12}\\ \\ \Rightarrow \dfrac{1}{2}=\dfrac{h}{12}\Rightarrow h=6$
This height of the pole is 6m
Question 15 of 30

15. Question
1 point(s)
The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun’s altitude is 30° than when it was 45°. Prove that the height of tower is $(\sqrt3+1) $ x metres
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Hint

In $\triangle ABC$
$\tan45=\dfrac{h}{y}$
$\Rightarrow h=y$ …(1)
In $\triangle ABD$
$\tan30=\dfrac{AB}{BD}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{2x+y}\\ \\ \Rightarrow \sqrt3h=2x+y\\ \\ \Rightarrow \sqrt3h=2x+h\\ \\ \Rightarrow \sqrt3h-h=2x\\ \\ \Rightarrow h=\dfrac{2x}{\sqrt3-1}=\dfrac{2x}{\sqrt3-1}\times\dfrac{\sqrt3+1}{\sqrt3+1}\\ \\ \Rightarrow h=\dfrac{2x(\sqrt3+1)}{2}\\ \\ =x(\sqrt3+1)\ m$
Question 16 of 30

16. Question
1 point(s)
From the top of a 10 m tall tower the angle of depression of a point on a ground was found to be 60°. How far is the point from the base of the tower?
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Hint

In $\triangle ABC$
$\tan60=\dfrac{P}{b}\\ \\ \Rightarrow \sqrt3=\dfrac{10}{b}\\ \\ \Rightarrow b=\dfrac{10}{\sqrt3}=\dfrac{10\sqrt3}{3}=5.78\ m$
Question 17 of 30

17. Question
1 point(s)
An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from his eyes is 45°. What is the height of the tower ?
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Hint

In $\triangle ACB$
$\tan\theta=\dfrac{BC}{AC}\\ \\ \Rightarrow \tan45=\dfrac{BC}{28.5}\\ \\ \Rightarrow BC=28.5\ m$
$BD=28.5+1.5=30\ m$
Question 18 of 30

18. Question
1 point(s)
Two poles of equal heights are standing opposite to each other on either side of a road, which is 100 metres wide. From a point between them on the road, the angles of elevation of their tops are 30° and 60°. Find the position of the point and also, the heights of the poles.
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Hint

In $\triangle PAB$
$\tan60=\dfrac{AB}{AP}\\ \\ \Rightarrow \sqrt3=\dfrac{h}{2}\Rightarrow h=\sqrt3x$ …(1)
In $\triangle PCD$
$\tan30=\dfrac{CD}{PC}\\ \\ \Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{100-x}$ …(2)
From (1) and (2) we get
$3x=100-x\Rightarrow x=25\\ \\ h=25\sqrt3$
so the height of the poles is $25\sqrt3\ m$
$x=\dfrac{h}{\sqrt3}=\dfrac{25\sqrt3}{\sqrt3}=25\ m$
The position of the point is 25 m
Question 19 of 30

19. Question
1 point(s)
A tree breaks due to the storm and the broken part bends so that the top of the tree touches ground making an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground, is 10 metres. Find the height of the tree.
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Hint

In $\triangle ABC$
$\cos c=\dfrac{BC}{AC}\\ \\ \Rightarrow \cos30=\dfrac{10}{AC}\\ \\ \Rightarrow \dfrac{\sqrt3}{2}=\dfrac{10}{AC}\Rightarrow AC=\dfrac{20}{\sqrt3}$
In $\triangle ABC$
$\sin30=\dfrac{AB}{AC}\Rightarrow \dfrac{1}{2}=\dfrac{AB}{\dfrac{20}{\sqrt3}}\\ \\ \Rightarrow AB=\dfrac{20}{2\sqrt3}=\dfrac{10}{\sqrt3}$
Height of the tree $=AC+AB$
$=\dfrac{20}{\sqrt3}+\dfrac{10}{\sqrt3}=\dfrac{30}{\sqrt3}=\dfrac{30\sqrt3}{3}\\ \\ =10\sqrt3\ m$
Question 20 of 30

20. Question
1 point(s)
A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is standing on the roof of a 20 m high building and is flying his kite at an elevation of 45°. Both the boys are on the opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.
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Hint

Let the length of second string $=x\ m$
In $\triangle ABC$
$\sin30=\dfrac{AC}{AD}\Rightarrow \dfrac{1}{2}=\dfrac{AC}{100}\Rightarrow AC=50\ m$
In $\triangle AEF$
$\sin45=\dfrac{AF}{AE}\Rightarrow \dfrac{1}{\sqrt2}=\dfrac{AC-FC}{x}\\ \\ \Rightarrow \dfrac{1}{\sqrt2}=\dfrac{50-20}{x}\\ \\ \Rightarrow \dfrac{1}{\sqrt2}=\dfrac{30}{x}\\ \\ \Rightarrow x=30\sqrt2\ m$
so the lenght of the string $=30\sqrt2\ m$
Question 21 of 30

21. Question
1 point(s)
At the foot of a mountain the elevation of its summit is 45°. After ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.
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Hint

In $\triangle OCE$
$\sin30=\dfrac{CE}{OC}\Rightarrow \dfrac{1}{2}=\dfrac{CE}{1000}\\ \\ \Rightarrow CE=500\\ \\ \cos30=\dfrac{OE}{OC}\\ \\ \Rightarrow \dfrac{\sqrt3}{2}=\dfrac{OE}{1000}\\ \\ \Rightarrow OE=500\sqrt3$
In $\triangle AOB$ ,
$\tan45=\dfrac{AB}{OA}\Rightarrow OA=AB\\ \\ \therefore CD=EA=OA-OE=h-500\sqrt3\\ \\ BD=AB-AD=AB-CE=h-500$
In $\triangle BCD$ ,
$\tan60=\dfrac{BD}{CD}\Rightarrow \sqrt3=\dfrac{h-500}{h-500\sqrt3}\\ \\ \Rightarrow h=\dfrac{1000(\sqrt3+1)}{2}=500\times2.732=1366\ m$
Question 22 of 30

22. Question
1 point(s)
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of the tower PQ and the distance XQ.
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Hint

In $\triangle MYQ$ ,
$\tan45=\dfrac{MQ}{MY}\\ \\ \Rightarrow 1=\dfrac{h-40}{MY}\Rightarrow MY=h-40\\ \\ \Rightarrow PX=MY=h-40$ …(1)
In $\triangle QXP$ ,
$\tan60=\dfrac{PQ}{PX}\Rightarrow \sqrt3=\dfrac{h}{h-40}\\ \\ \Rightarrow h(\sqrt3-1)=40\sqrt3\\ \\ \Rightarrow h=\dfrac{40\sqrt3}{\sqrt3-1}\times\dfrac{\sqrt3+1}{\sqrt3+1}\\ \\ \Rightarrow h=\dfrac{40\sqrt3(\sqrt3+1)}{2}\\ \\ \Rightarrow h=20\sqrt3(\sqrt3+1)\\ \\ \Rightarrow h=60+20\sqrt3\\ \\ =60+34.6=94.6\\ \\ Px=h-40=94.6-40\\ \\ =54.6\\ \\ xQ=\sqrt{PQ^2+xP^2}\\ \\ =\sqrt{(94.6)^2+(54.6)^2}\\ \\ =\sqrt{8949.16+2981.6}\\ \\ =\sqrt{11930.32}\\ \\ =109.22=109\ m$
Question 23 of 30

23. Question
1 point(s)
A round ballon of radius r subtends an angle $\alpha $ at the eye of the observer while the angle of elevation of its centre is $\beta $. Prove that the height of the centre of the balloon is r sin $ \beta$ cosec $\dfrac{\alpha}{2} $.
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Hint

In $\triangle OAP$
$\sin\dfrac{\alpha}{2}=\dfrac{OA}{OP}=\dfrac{r}{OP}\\ \\ \Rightarrow OP=r\ cosec\ \dfrac{\alpha}{2}$ …(1)
In $\triangle OPL$
$\sin\beta=\dfrac{OL}{OP}\Rightarrow OL=OP\sin\beta\\ \\ \Rightarrow OL=r\ cosec\ \dfrac{\alpha}{2}\sin\beta$
so the height of the balloon $r\sin\beta\ cosec\ \dfrac{\alpha}{2}$
Question 24 of 30

24. Question
1 point(s)
The angle of elevation of a cliff from a fixed point is $\theta $. After going up a distance of k metres towards the top of the cliff at an angle of $\phi $ , it is found that the angle of elevation is $ \alpha$. Show that the height of the cliff is $\dfrac{k(\cos\phi-\sin\phi\cot\alpha)}{\cot\theta-\cot\alpha} $ metres.
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Hint

In $\triangle EFC$
$\sin\theta=\dfrac{EF}{CE}\\ \\ =\dfrac{EF}{k}\ (\because CE=k)\\ \\ \Rightarrow EF=k\sin\theta\\ \\ \cos \theta=\dfrac{CF}{CE}\\ \\ \Rightarrow CF=k\cos\theta\\ \\ AD=AB-DB\\ \\ =h-k\cos\theta$
In $\triangle EFC$
$\tan\alpha=\dfrac{AD}{ED}=\dfrac{h-k\cos\theta}{ED}\\ \\ ED=(h-k\cos\theta)\cot\alpha\\ \\ BC=BF+CF\\ \\ =ED+CF\\ \\ =(h-k\cos\theta)\cot\alpha+\cos\theta$
In $\triangle ABC$
$\tan\theta=\dfrac{AB}{BC}=\dfrac{h}{(h-k\cos\theta)\cot\alpha+k\cos\theta}\\ \\ \Rightarrow \tan\theta[(h-k\cos\theta)\cot \alpha+k\cos\theta]-h\\ \\ \Rightarrow [(h\tan\theta.\cot\alpha-k\cos\theta\cot\alpha)+(k\tan\theta\ \cos\theta)]=h\\ \\ \Rightarrow h=\dfrac{k(\cos\theta-\sin\theta\cot\alpha)}{\cot\theta-\cot\alpha}$ (Proved)
Question 25 of 30

25. 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 from the same point is $\beta $ prove that the height of the cloud is $\dfrac{h(\tan\beta+\tan\alpha)}{(\tan\beta-\tan\alpha)} $
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Hint

In $\triangle CPM$
$\dfrac{CM}{PM}=\tan\alpha\\ \\ \Rightarrow PM=\dfrac{CM}{\tan\alpha}=\dfrac{H-h}{\tan\alpha}$ …(1)
In $\triangle PMC'$
$\dfrac{C'M}{PM}=\tan\beta\Rightarrow PM=\dfrac{C'M}{\tan\beta}\Rightarrow PM=\dfrac{H+h}{\tan\beta}$ …(2)
From (1) and (2)
$\dfrac{H-h}{\tan\alpha}=\dfrac{H+h}{\tan\beta}\ \Rightarrow H\tan\beta-H\tan\alpha=h\tan\alpha+h\tan\beta\\ \\ \Rightarrow H=\dfrac{h(\tan\alpha+\tan\beta)}{\tan\beta-\tan\alpha}$ (Proved)
Question 26 of 30

26. Question
1 point(s)
A man on a cliff observes a boat at an angle of depression of $30^o $ which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be $60^o $. Find the time taken by the boat to reach the shore.
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Hint

Let the speed of the boat is vm/min and time is t min
In $\triangle DBC$ ,
$\dfrac{BC}{DB}=\tan60\Rightarrow \dfrac{h}{vt}=\sqrt3$
$\Rightarrow h=\sqrt3\ vt$ ..(1)
In $\triangle ABC$ ,
$\dfrac{BC}{AB}=\tan30\Rightarrow \dfrac{h}{6v+vt}=\dfrac{1}{\sqrt3}\\ \\ \Rightarrow h=\dfrac{6v+vt}{\sqrt3}$ …(2)
From (1) and (2)
$\sqrt3\ vt=\dfrac{6v+vt}{\sqrt3}\Rightarrow 2vt=6v\\ \\ \Rightarrow t=3\ minutes$
Question 27 of 30

27. Question
1 point(s)
A ladder rests agains a wall at an angle $\alpha $ to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides a distance b down the wall making an angle $\beta $ with the horizontal. Show that $\dfrac{a}{b}=\dfrac{\cos\alpha-\cos\beta}{\sin\beta-\sin\alpha} $
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Hint

Let $xy=wz=h$
In $\triangle oxy$
$\sin\alpha=\dfrac{oy}{xy}=\dfrac{oy}{h}\\ \\ \Rightarrow oy=h\sin\alpha\\ \\ \cos\alpha=\dfrac{ox}{xy}\Rightarrow \cos\alpha=\dfrac{ox}{h}\\ \\ \Rightarrow ox=h\cos\alpha$
In $\triangle owz$ ,
$\sin\beta=\dfrac{oz}{wz}=\dfrac{oz}{h}\Rightarrow oz=h\sin\beta\\ \\ \cos\beta=\dfrac{ow}{wz}=\dfrac{ow}{h}\Rightarrow ow=h\cos\beta\\ \\ a=ow-ox\\ \\ b=oy-oz\\ \\ \dfrac{a}{b}=\dfrac{ow-ox}{oy-oz}=\dfrac{\cos\beta-\cos\alpha}{\sin\alpha-\sin\beta}\\ \\ \Rightarrow \dfrac{a}{b}=\dfrac{\cos\alpha-\cos\beta}{\sin\beta-\sin\alpha}$ (Proved)
Question 28 of 30

28. Question
1 point(s)
The angles of elevation of the top of a hill at the city centres of two towns on either side of the hill are observed to be 30° and 60°. If the distance up-hill from the first city centre is 9 km, find in kilometres the distance up-hill from the other city centre.
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Hint

$\sin30=\dfrac{h}{9}=\dfrac{1}{2}\\ \\ h=4.5\\ \\ \dfrac{h}{y}=\sin60=\dfrac{\sqrt3}{2}\\ \\ y=\dfrac{2\times4.5}{\sqrt3}=\dfrac{9}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{9\sqrt3}{3}\\ \\ =3\sqrt3\\ \\ =3\times1.732\\ \\ =5.196\\ \\ =5.20\ km$
Question 29 of 30

29. Question
1 point(s)
An aeroplane when 3000 m high passes vertically above another aeroplane at an instance when their angles of elevation at the same observation point are 60° and 45° respectively. How many metres higher is the one than the other ?
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Hint

$x+h=3000\ m$
In $\triangle BCD$
$\tan45=\dfrac{BC}{CD}\\ \\ \Rightarrow 1=\dfrac{x}{y}\Rightarrow x=y$ …(1)
In $\triangle ACD$
$\tan60=\dfrac{AC}{CD}\Rightarrow \sqrt3=\dfrac{x+h}{y}\\ \\ \Rightarrow \sqrt3=\dfrac{3000}{y}\\ \\ \Rightarrow y=\dfrac{3000}{\sqrt3}$ …(2)
using (1)
$x=\dfrac{3000}{\sqrt3}=\dfrac{3000\sqrt3}{3}=1000\sqrt3\\ \\ =1730\ m$
Distance between the places $=3000-1730$
$=1270\ m$
Question 30 of 30

30. Question
1 point(s)
The distance between two vertical poles is 60 m. The height of one of the poles is double the height of the other. The angles of elevation of the tops of the poles from the middle point of the line segment joining their feet are complementary to each other. Find the heights of the poles.
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Hint

In $\triangle ABE$
$\tan\theta=\dfrac{AB}{BE}\\ \\ \Rightarrow \tan\theta=\dfrac{h}{30}\\ \\ \Rightarrow h=30\tan\theta$ …(1)
In $\triangle CDE$
$\tan(90-\theta)=\dfrac{2h}{36}\Rightarrow \cot\theta=\dfrac{h}{15}\Rightarrow h=15\cot \theta$ …(2)
By multiplying both the equations
$h^2=30\times15\times\tan\theta\times \cot\theta\\ \\ \Rightarrow h^2=30\times15\\ \\ \Rightarrow h=15\sqrt2\ m=AB\\ \\ 2h=30\sqrt2\ m=CD$