Q: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60^o and the angle of depression of its foot is 45^o . The height of the tower in metre is

In \Delta ABE, \tan 60^o=\displaystyle\frac{h}{AB} h=AB\sqrt 3 ....(i)In \Delta ABC, \tan 45^o=\displaystyle\frac{7}{AB} AB=7 ...(ii)By equation (i) and equation (ii) h=7\sqrt 3 So, height of tower is =h+7=7(\sqrt 3+1)

Q: The angle of elevation of the top of a tower from a point A on the ground is 30. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60. Find the height of the tower and the distance of the tower from thy. point A.

In \Delta ADC \tan 30^o=\dfrac{DC}{AC} \dfrac{1}{\sqrt{3}}=\dfrac{h}{d} ........ (1) d=\sqrt{3}h \rightarrow d=\sqrt{3}\times 10\sqrt{3} =30 mIn \Delta BCD \tan 60^o=\dfrac{DC}{BC} \sqrt{3}=\dfrac{h}{d-20} ....... (2) on solving equation (1) & (2) \sqrt{3}=\dfrac{h}{\sqrt{3}h-20} \Rightarrow 3h-20\sqrt{3}=h 2h=20\sqrt{3} h=10\sqrt{3}m height of tower.

Question 1

The angle of elevation of the top of the building from the foot of the tower is  30  and the angle 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.

Accepted Answer

Given height of tower  CD = 50 m Let the height of the building,  AB = h In right angled  	riangle BDC , \Rightarrow    	an\,60^o=\dfrac{CD}{BD} \Rightarrow    \sqrt{3}=\dfrac{50}{BD} \Rightarrow    BD=\dfrac{50}{\sqrt{3}}\,m In right angled  	riangle ABD , \Rightarrow    	an\,30^o=\dfrac{AB}{BD} \Rightarrow    \dfrac{1}{\sqrt{3}}=\dfrac{h}{\dfrac{50}{\sqrt{3}}} 	herefore    \dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}h}{50} 	herefore    h=\dfrac{50}{3}=16.66\,m 	herefore   The height of the building is  16.66\,m .

Question 2

The angles of depression of the top and bottom of  8  m tall building from the top of a multi-storeyed building are  30^0  and  45^0  respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Accepted Answer

Let  AB  and  CD  be the multi-storied building and the building respectively.Let the height of the multi-storied building be  h\,m  and  the\ distance\ between\ the\ two\ building\ be\ x\,m AE=CD=8\,m                  [ Given ] BE=AB-AE=(h-8)\,m  and  AC=DE=x\,m       [  Given ]Now, in  	riangle ACB , \Rightarrow    	an\,45^o=\dfrac{AB}{AC} \Rightarrow    1=\dfrac{h}{x} 	herefore    x=h           ---- ( 1 )In  	riangle BDE , \Rightarrow    	an\,30^o=\dfrac{BE}{ED} \Rightarrow    \dfrac{1}{\sqrt{3}}=\dfrac{h-8}{x} 	herefore    x=\sqrt{3}(h-8)              ------ ( 2 ) \Rightarrow    h=\sqrt{3}h-8\sqrt{3} \Rightarrow    \sqrt{3}h-h=8\sqrt{3} \Rightarrow    h(\sqrt{3}-1)=8\sqrt{3} \Rightarrow    h=\dfrac{8\sqrt{3}}{\sqrt{3}-1} \Rightarrow    h=\dfrac{8\sqrt{3}}{\sqrt{3}+1}	imes \dfrac{\sqrt{3}+1}{\sqrt{3}+1} \Rightarrow    h=\dfrac{8\sqrt{3}(\sqrt{3}+1)}{3-1} \Rightarrow    h=\dfrac{8\sqrt{3}(\sqrt{3}+1)}{2} 	herefore    h=(12+4\sqrt{3})\,m 	herefore    x=(12+4\sqrt{3})\,m         [ From ( 1 ) ] 	herefore   The height of the multi-storied building and the distance between the two buildings is  (12+4\sqrt{3})\,m

Question 3

A man is standing on the deck of a ship which is 10m above the water level. He observes the angle of elevation of the top of hill as  { 60 }^{ \circ  }  and the angle of the base of hill as  { 30 }^{ \circ  } . Find the height of the hill from the base.

Accepted Answer

Let  AB=10m,  cliff of a ship where man is standingLet  CE  be the hill. CD=AB=10m In  	riangle ADE   	an 60°=\cfrac { h }{ x }  \sqrt{3}x=h x=\cfrac{h}{\sqrt{3}}\longrightarrow (1) In  	riangle ABC   	an 30°=\cfrac { 10 }{ x }  \cfrac { 1 }{ \sqrt { 3 }  } =\cfrac { 10 }{ x } x = 10  \sqrt{3} \longrightarrow (2) We have two equations  (1)  &  (2) , Multiplying them we get, x^2 =10h h=30m 	herefore CE=CD+DE=10+30=40m

Question 4

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. Find the height of the wall.

Accepted Answer

Let the length of the ladder be  =15 m(hypotenuse)The angle between the ladder and the wall  \angle{BCA}={60}^{\circ} Angle between ladder and the ground  \angle{CAB} ={90}^{\circ}-{60}^{\circ}={30}^{\circ} Height of the wall  = BC \sin{{30}^{\circ}}=\dfrac{BC}{15} \Rightarrow \dfrac{1}{2}=\dfrac{BC}{15} \Rightarrow BC=\dfrac{15}{2}=7.5 m

Question 5

From the top of a  7 m high building, the angle of elevation of the top of a cable tower is  60^o  and the angle of depression of its foot is  45^o . The height of the tower in metre is

Accepted Answer

In  \Delta ABE, 	an 60^o=\displaystyle\frac{h}{AB} h=AB\sqrt 3     ....(i)In  \Delta ABC, 	an 45^o=\displaystyle\frac{7}{AB} AB=7    ...(ii)By equation (i) and equation (ii) h=7\sqrt 3 So, height of tower is  =h+7=7(\sqrt 3+1)

Question 6

The angle of elevation of an airplane from a point on the ground is  {60^ \circ } . After a flight of 30 seconds, the angle of elevation become  {30^ \circ } . If the airplane is flying at a constant height of  3000\sqrt 3 \,m , find the speed of the airplane.

Accepted Answer

REF.Image tan 60^{\circ}=\frac{3000\sqrt{3}}{x} x=\frac{3000\sqrt{3}}{\sqrt{3}}= 3000 m tan 30^{\circ}=\frac{3000\sqrt{3}}{1/\sqrt{3}}=9000 m distance covered = 9000- 3000= 6000 m.speed =  \frac{6000}{30}= 200 m/s

Question 7

The angle of elevation of the top of a tower from a point A on the ground is 30. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60. Find the height of the tower and the distance of the tower from thy. point A.

Accepted Answer

In  \Delta ADC 	an 30^o=\dfrac{DC}{AC} \dfrac{1}{\sqrt{3}}=\dfrac{h}{d}  ........ (1) d=\sqrt{3}h ightarrow d=\sqrt{3}	imes 10\sqrt{3} =30 mIn  \Delta BCD 	an 60^o=\dfrac{DC}{BC} \sqrt{3}=\dfrac{h}{d-20}  ....... (2) on solving equation  (1)  &  (2) \sqrt{3}=\dfrac{h}{\sqrt{3}h-20} \Rightarrow 3h-20\sqrt{3}=h 2h=20\sqrt{3} h=10\sqrt{3}m height of tower.

Question 8

The angle of elevation of a jet plane from a point  A  on the ground is  {60}^{\circ} . After a flight of  15  second, the angle of elevation changes to  {30}^{\circ} . If the jet plane is flying at a constant height of   1500\sqrt{2} m.Find the speed of the jet plane.

Accepted Answer

Let  D  and  E  be the initial and final positions of the plane respectivelyConsider the  	riangle{ABD} 	an{{60}^{\circ}}=\dfrac{BD}{AB} \Rightarrow \sqrt{3}=\dfrac{1500\sqrt{3}}{AB} \Rightarrow AB=1500 Consider the  	riangle{ACE} 	an{{30}^{\circ}}=\dfrac{CE}{AC} \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{1500\sqrt{3}}{AC} \Rightarrow AC=4500 BC=AC-AB=4500-1500=3000 m DE=BC=3000 mi.e., the plane travels a distance of  3000  m in  15  seconds. 	herefore , speed of the plane  =\dfrac{3000}{15}=200 m\s.

Question 9

A pole  6m  high costs a shadow  2\sqrt {3}m  long on the ground then find the sun's elevation..

Accepted Answer

R.E.F image Given height of pole =6mlength of shadow = 2\sqrt{3}m Let '  	heta   ' be elevation. of sum then tan	heta   = \frac{AB}{BC}=\frac{6}{2\sqrt{3}}  tan	heta =\sqrt{3}  tan	heta =tan \frac{\pi }{3}  then   	heta  = \frac{\pi }{3} 	herefore  Angle of elevation of sum is  60^{\circ}.

Question 10

The angles of depression of the top and bottom of a building 50 meters high as 4 observed from the top of a tower are 30 and 60, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Accepted Answer

In ΔBTP,tan 30° = TP/BP 1/√3 = TP/BPBP = TP√3In ΔGTR,tan 60° = TR/GR√3 = TR/GRGR = TR/√3As BP = GRTP√3 = TR/√33 TP = TP + PR2 TP = BGTP = 50/2 m = 25 mNow, TR = TP + PRTR = (25 + 50) mHeight of tower = TR = 75 mDistance between building and tower = GR = TR/√3GR = 75/√3 m = 25√3 m