Q: 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.

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

Q: 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.

\$\textbf{Step1: Draw diagram according to language of question}\$ \$\text{Let \$AB\$ and \$CD\$ be the multi-storied building and the building respectively.}\$ \$AE=CD=8\,m\$ \$[ Given ]\$ \$BE=AB-AE=(h-8)\,m\$ \$and\$ \$AC=DE=x\,m\$ \$ [ Given ]\$\$\textbf{Step2: Apply formula of trigonometric ratios }\$ \$Now, \$\$in\$ \$\triangle ACB\$, \$\Rightarrow\$ \$\tan\,45^o=\dfrac{AB}{AC}\$ \$\Rightarrow\$ \$1=\dfrac{h}{x}\$ \$\therefore\$ \$x=h\$ \$---- ( 1 )\$ \$In \$\$\triangle BDE\$, \$\Rightarrow\$ \$\tan\,30^o=\dfrac{BE}{ED}\$ \$\Rightarrow\$ \$\dfrac{1}{\sqrt{3}}=\dfrac{h-8}{x}\$ \$\therefore\$ \$x=\sqrt{3}(h-8)\$ \$------ ( 2 )\$\$\textbf{Step3 : Simplify to evaluate unknown }\$ \$From\ eq^n (1)\ and\ (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}\times \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}\$ \$\therefore\$ \$h=(12+4\sqrt{3})\,m\$ \$\therefore\$ \$x=(12+4\sqrt{3})\,m\$ \$[ From ( 1 ) ]\$\$\textbf{ Hence, The height of the multi-storied building and the distance between the two buildings is}\$ \$(12+4\sqrt{3})\,m\$

Q: 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.

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\$

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: 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.

Let \$AB=10m,\$ cliff of a ship where man is standingLet \$CE\$ be the hill.\$CD=AB=10m\$In \$\triangle ADE\$ \$\tan 60°=\cfrac { h }{ x } \$\$\sqrt{3}x=h\$\$x=\cfrac{h}{\sqrt{3}}\longrightarrow (1)\$In \$\triangle ABC\$ \$\tan 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\$\$\therefore CE=CD+DE=10+30=40m\$

Q: 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.

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

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.

Q: 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.

Let \$D\$ and \$E\$ be the initial and final positions of the plane respectivelyConsider the \$\triangle{ABD}\$\$\tan{{60}^{\circ}}=\dfrac{BD}{AB}\$\$\Rightarrow \sqrt{3}=\dfrac{1500\sqrt{3}}{AB}\$\$\Rightarrow AB=1500\$Consider the \$\triangle{ACE}\$\$\tan{{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.\$\therefore\$, speed of the plane \$=\dfrac{3000}{15}=200\$m\s.

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

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

Q: 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.

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

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 $\triangle BDC$,$\Rightarrow$  $\tan\,60^o=\dfrac{CD}{BD}$$\Rightarrow$  $\sqrt{3}=\dfrac{50}{BD}$$\Rightarrow$  $BD=\dfrac{50}{\sqrt{3}}\,m$In right angled $\triangle ABD$,$\Rightarrow$  $\tan\,30^o=\dfrac{AB}{BD}$$\Rightarrow$  $\dfrac{1}{\sqrt{3}}=\dfrac{h}{\dfrac{50}{\sqrt{3}}}$$\therefore$  $\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}h}{50}$$\therefore$  $h=\dfrac{50}{3}=16.66\,m$$\therefore$  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

$\textbf{Step1: Draw diagram according to language of question}$                   $\text{Let $AB$ and $CD$ be the multi-storied building and the building respectively.}$             $AE=CD=8\,m$                 $[ Given ]$             $BE=AB-AE=(h-8)\,m$              $and$  $AC=DE=x\,m$     $ [  Given ]$$\textbf{Step2: Apply formula of trigonometric ratios }$             $Now, $$in$ $\triangle ACB$,             $\Rightarrow$  $\tan\,45^o=\dfrac{AB}{AC}$             $\Rightarrow$  $1=\dfrac{h}{x}$             $\therefore$  $x=h$          $---- ( 1 )$             $In $$\triangle BDE$,                          $\Rightarrow$  $\tan\,30^o=\dfrac{BE}{ED}$             $\Rightarrow$  $\dfrac{1}{\sqrt{3}}=\dfrac{h-8}{x}$             $\therefore$  $x=\sqrt{3}(h-8)$             $------ ( 2 )$$\textbf{Step3 :  Simplify to evaluate unknown }$             $From\ eq^n (1)\ and\ (2)$             $\Rightarrow$  $h=\sqrt{3}h-8\sqrt{3}$             $\Rightarrow$  $\sqrt{3}h-h=8\sqrt{3}$             $\Rightarrow$  $h(\sqrt{3}-1)=8\sqrt{3}$             $\Rightarrow$  $h=\dfrac{8\sqrt{3}}{\sqrt{3}-1}$             $\Rightarrow$  $h=\dfrac{8\sqrt{3}}{\sqrt{3}+1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}$             $\Rightarrow$  $h=\dfrac{8\sqrt{3}(\sqrt{3}+1)}{3-1}$             $\Rightarrow$  $h=\dfrac{8\sqrt{3}(\sqrt{3}+1)}{2}$             $\therefore$  $h=(12+4\sqrt{3})\,m$             $\therefore$  $x=(12+4\sqrt{3})\,m$        $[ From ( 1 ) ]$$\textbf{ Hence, The height of the multi-storied building and the distance between the two buildings is}$ $(12+4\sqrt{3})\,m$

Question 3

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 4

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,$\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 $=h+7=7(\sqrt 3+1)$

Question 5

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 $\triangle ADE$ $\tan 60&#176;=\cfrac { h }{ x } $$\sqrt{3}x=h$$x=\cfrac{h}{\sqrt{3}}\longrightarrow (1)$In $\triangle ABC$ $\tan 30&#176;=\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$$h=30m$$\therefore CE=CD+DE=10+30=40m$

Question 6

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 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$$\tan 30^o=\dfrac{DC}{AC}$$\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 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 $\triangle{ABD}$$\tan{{60}^{\circ}}=\dfrac{BD}{AB}$$\Rightarrow \sqrt{3}=\dfrac{1500\sqrt{3}}{AB}$$\Rightarrow AB=1500$Consider the $\triangle{ACE}$$\tan{{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.$\therefore$, 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 '$ \theta $ ' be elevation. of sum then$tan\theta $ =$\frac{AB}{BC}=\frac{6}{2\sqrt{3}}$$ tan\theta =\sqrt{3}$$ tan\theta =tan \frac{\pi }{3}$ then $ \theta  = \frac{\pi }{3}$$\therefore$ 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 &#916;BTP,tan 30&#176; = TP/BP 1/&#8730;3 = TP/BPBP = TP&#8730;3In &#916;GTR,tan 60&#176; = TR/GR&#8730;3 = TR/GRGR = TR/&#8730;3As BP = GRTP&#8730;3 = TR/&#8730;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/&#8730;3GR = 75/&#8730;3 m = 25&#8730;3 m