44114.A flagstaff is placed on top of a building. The flagstaff and building subtend equal angles at a point on level ground which is 200 m away from the foot of the building. If the height of the flagstaff is 50 m and the height of the building is h, which of the following is true?
h3 - 50h2 + (200)2h + (200)250 = 0
None of these
h3 + 50h2 + (200)2h - (200)250 = 0
h3 - 50h2 - (200)2h + (200)250 = 0
Explanation:
Let AD be the flagstaff and CD be the building.
Assume that the flagstaff and building subtend equal angles at point B.
Given that AD = 50 m, CD = h and BC = 200 m
Let $\angle$ABD = θ, $\angle$DBC = θ (∵ flagstaff and building subtend equal angles at a point on level ground).
Then, $\angle$ABC = 2θ
From the right $\triangle$ BCD,
tanθ=$\dfrac{DC}{BC}=\dfrac{h}{200}$ ⋯(eq:1)
From the right $\triangle$ BCA,
tan2θ=$\dfrac{AC}{BC}=\dfrac{AD + DC}{200}=\dfrac{50 + h}{200}$
⇒$\dfrac{2tanθ}{1−tan2θ}=\dfrac{50 + h}{200}$ (∵tan(2θ)=$\dfrac{2tanθ}{1−tan2θ}$)
⇒$\dfrac{2\left( \dfrac{h}{200}\right)}{1−\dfrac{h^{2}}{200^{2}}}=\dfrac{50 + h}{200}$(∵ substituted value of tan θ from eq:1)
⇒2h=$\left(1−\dfrac{h^{2}}{200^{2}}\right) $ (50 + h)
⇒2h=50+h−$\dfrac{50h^{2}}{200^{2}}−\dfrac{h^{3}}{200^{2}}$
⇒2(2002)h =50(200)2+h(200)2−50h2−h3
(∵ multiplied LHS and RHS by 2002)
⇒h3+50h2+(200)2h−50(200)2=0
Let AD be the flagstaff and CD be the building.
Assume that the flagstaff and building subtend equal angles at point B.
Given that AD = 50 m, CD = h and BC = 200 m
Let $\angle$ABD = θ, $\angle$DBC = θ (∵ flagstaff and building subtend equal angles at a point on level ground).
Then, $\angle$ABC = 2θ
From the right $\triangle$ BCD,
tanθ=$\dfrac{DC}{BC}=\dfrac{h}{200}$ ⋯(eq:1)
From the right $\triangle$ BCA,
tan2θ=$\dfrac{AC}{BC}=\dfrac{AD + DC}{200}=\dfrac{50 + h}{200}$
⇒$\dfrac{2tanθ}{1−tan2θ}=\dfrac{50 + h}{200}$ (∵tan(2θ)=$\dfrac{2tanθ}{1−tan2θ}$)
⇒$\dfrac{2\left( \dfrac{h}{200}\right)}{1−\dfrac{h^{2}}{200^{2}}}=\dfrac{50 + h}{200}$(∵ substituted value of tan θ from eq:1)
⇒2h=$\left(1−\dfrac{h^{2}}{200^{2}}\right) $ (50 + h)
⇒2h=50+h−$\dfrac{50h^{2}}{200^{2}}−\dfrac{h^{3}}{200^{2}}$
⇒2(2002)h =50(200)2+h(200)2−50h2−h3
(∵ multiplied LHS and RHS by 2002)
⇒h3+50h2+(200)2h−50(200)2=0
44117.Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles.
141 m and 282 m
70.5 m and 141 m
65 m and 130 m
130 m and 260 m
Explanation:
Let AB and CD be the poles with heights h and 2h respectively.
Given that distance between the poles, BD = 200 m
Let E be the middle point of BD,
$\angle$AEB = θ
$\angle$CED = (90-θ) (∵ given that angular elevations are complementary)
Since E is the middle point of BD, we have BE = ED = 100 m
From the right $\triangle$ ABE,
tan$\theta$=$\dfrac{AB}{BE}$
tan$\theta$=$\dfrac{h}{100}$
h = 100tanθ ... (1)
From the right $\triangle$ EDC,
tan(90-θ)=$\dfrac{CD}{ED}$
cotθ=$\dfrac{2h}{100}$ [tan(90−θ)=cotθ]
2h =100cotθ ⋯(2)
(1)×(2)=>2h2=1002 [∵tanθ×cotθ=tanθ×$\dfrac{1}{tanθ}$=1
=>√2h=100
=>h=$\dfrac{100}{\sqrt{2}}=\dfrac{100×\sqrt{2}}{\sqrt{2}×\sqrt{2}}$
=50√2=50×1.41=70.5
=>2h=2×70.5=141
i.e., the height of the poles are 70.5 m and 141 m.
Let AB and CD be the poles with heights h and 2h respectively.
Given that distance between the poles, BD = 200 m
Let E be the middle point of BD,
$\angle$AEB = θ
$\angle$CED = (90-θ) (∵ given that angular elevations are complementary)
Since E is the middle point of BD, we have BE = ED = 100 m
From the right $\triangle$ ABE,
tan$\theta$=$\dfrac{AB}{BE}$
tan$\theta$=$\dfrac{h}{100}$
h = 100tanθ ... (1)
From the right $\triangle$ EDC,
tan(90-θ)=$\dfrac{CD}{ED}$
cotθ=$\dfrac{2h}{100}$ [tan(90−θ)=cotθ]
2h =100cotθ ⋯(2)
(1)×(2)=>2h2=1002 [∵tanθ×cotθ=tanθ×$\dfrac{1}{tanθ}$=1
=>√2h=100
=>h=$\dfrac{100}{\sqrt{2}}=\dfrac{100×\sqrt{2}}{\sqrt{2}×\sqrt{2}}$
=50√2=50×1.41=70.5
=>2h=2×70.5=141
i.e., the height of the poles are 70.5 m and 141 m.
44118.When the sun s altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m. What is the height of the tower?
35 m
140 m
60.6 m
20.2 m
Explanation:
Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m, $\angle$ABD = 30°, $\angle$ACD = 60°,
Let CD = x, AD = h
From the right $\triangle$ CDA,
tan60°=$\dfrac{AD}{CD}$
√3=$\dfrac{h}{x} $ ⋯(eq:1)
From the right $\triangle$ BDA,
tan30°=$\dfrac{AD}{BD}$
$\dfrac{1}{\sqrt{3}}=\dfrac{h}{70+x} $ ⋯(eq:2)
$\dfrac{eq:1}{eq:2}⇒\dfrac{\sqrt{3}}{\left( \dfrac{1}{\sqrt{3}} \right)}=\dfrac{\left( \dfrac{h}{x}\right)}{\left( \dfrac{h}{70+x}\right)}$
⇒3=$\dfrac{70+x}{x}$
⇒2x=70
⇒x=35
Substituting this value of x in eq:1, we have
√3=$\dfrac{h}{35}$
⇒h=35√3=35×1.73
=60.55≈60.6
Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m, $\angle$ABD = 30°, $\angle$ACD = 60°,
Let CD = x, AD = h
From the right $\triangle$ CDA,
tan60°=$\dfrac{AD}{CD}$
√3=$\dfrac{h}{x} $ ⋯(eq:1)
From the right $\triangle$ BDA,
tan30°=$\dfrac{AD}{BD}$
$\dfrac{1}{\sqrt{3}}=\dfrac{h}{70+x} $ ⋯(eq:2)
$\dfrac{eq:1}{eq:2}⇒\dfrac{\sqrt{3}}{\left( \dfrac{1}{\sqrt{3}} \right)}=\dfrac{\left( \dfrac{h}{x}\right)}{\left( \dfrac{h}{70+x}\right)}$
⇒3=$\dfrac{70+x}{x}$
⇒2x=70
⇒x=35
Substituting this value of x in eq:1, we have
√3=$\dfrac{h}{35}$
⇒h=35√3=35×1.73
=60.55≈60.6
44119.A person, standing exactly midway between two towers, observes the top of the two towers at angle of elevation of 22.5° and 67.5°. What is the ratio of the height of the taller tower to the height of the shorter tower? (Given that tan 22.5° = √2−1)
1−2√2 : 1
1+2√2 : 1
3+2√2 : 1
3−2√2 : 1
Explanation:
Let ED be the taller tower and AB be the shorter tower.
Let C be the point of observation.
Given that $\angle$ACB = 22.5° and $\angle$DCE = 67.5°
Given that C is the midpoint of BD.
Hence, BC = CD
From the right $\triangle$ ABC,
tan22.5°=$\dfrac{AB}{BC}$ ⋯(eq:1)
From the right $\triangle$ CDE,
tan67.5°=$\dfrac{ED}{CD} $ ⋯(eq:2)
$\dfrac{(eq:2)}{(eq:1)}⇒\dfrac{tan67.5°}{tan22.5°} =\dfrac{\left( \dfrac{ED}{CD} \right)}{\left( \dfrac{AB}{BC} \right)}$
=$\dfrac{ED}{AB}$(∵ CD=BC)
⇒$\dfrac{tan(90°−22.5°)}{tan22.5°}=\dfrac{ED}{AB}$
⇒$\dfrac{cot22.5°}{tan22.5°}=\dfrac{ED}{AB}$ [∵ tan(90-θ)=cot θ]
⇒$\dfrac{\left( \dfrac{1}{tan22.5°}\right)}{tan22.5°}=\dfrac{ED}{AB }$ [∵cotθ=$\dfrac{1}{tanθ}$]
⇒$\dfrac{ED}{AB}=\dfrac{1}{(tan22.5°)^{2}}$
=$\dfrac{1}{(\sqrt{2}−1)^{2}}=\left( \dfrac{1}{\sqrt{2}−1} \right)^{2}$
=$\left[\dfrac{(\sqrt{2}+1)}{(\sqrt{2}−1)(\sqrt{2}+1)}\right]^{2}$
=$\left[\dfrac{(\sqrt{2}+1)}{(2−1)}\right]^{2}=\left[\dfrac{(\sqrt{2}+1)}{1}\right]^{2}$
=(√2+1)2=(2+2√2+1)
=(3+2√2)
Required ratio = ED : AB =(3+2√2):1
Let ED be the taller tower and AB be the shorter tower.
Let C be the point of observation.
Given that $\angle$ACB = 22.5° and $\angle$DCE = 67.5°
Given that C is the midpoint of BD.
Hence, BC = CD
From the right $\triangle$ ABC,
tan22.5°=$\dfrac{AB}{BC}$ ⋯(eq:1)
From the right $\triangle$ CDE,
tan67.5°=$\dfrac{ED}{CD} $ ⋯(eq:2)
$\dfrac{(eq:2)}{(eq:1)}⇒\dfrac{tan67.5°}{tan22.5°} =\dfrac{\left( \dfrac{ED}{CD} \right)}{\left( \dfrac{AB}{BC} \right)}$
=$\dfrac{ED}{AB}$(∵ CD=BC)
⇒$\dfrac{tan(90°−22.5°)}{tan22.5°}=\dfrac{ED}{AB}$
⇒$\dfrac{cot22.5°}{tan22.5°}=\dfrac{ED}{AB}$ [∵ tan(90-θ)=cot θ]
⇒$\dfrac{\left( \dfrac{1}{tan22.5°}\right)}{tan22.5°}=\dfrac{ED}{AB }$ [∵cotθ=$\dfrac{1}{tanθ}$]
⇒$\dfrac{ED}{AB}=\dfrac{1}{(tan22.5°)^{2}}$
=$\dfrac{1}{(\sqrt{2}−1)^{2}}=\left( \dfrac{1}{\sqrt{2}−1} \right)^{2}$
=$\left[\dfrac{(\sqrt{2}+1)}{(\sqrt{2}−1)(\sqrt{2}+1)}\right]^{2}$
=$\left[\dfrac{(\sqrt{2}+1)}{(2−1)}\right]^{2}=\left[\dfrac{(\sqrt{2}+1)}{1}\right]^{2}$
=(√2+1)2=(2+2√2+1)
=(3+2√2)
Required ratio = ED : AB =(3+2√2):1
44120.21. The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°,the elevation changes to 60°. What is the approximate height of the mountain?
1.2 km
0.6 km
1.4 km
2.7 km
Explanation:
Let A be the foot and C be the summit of a mountain.
Given that $\angle$CAB = 45°
From the diagram, CB is the height of the mountain. Let CB = x
Let D be the point after ascending 2 km towards the mountain such that AD = 2 km and given that $\angle$DAY = 30°
It is also given that from the point D, the elevation is 60°
i.e., $\angle$CDE = 60°
From the right $\triangle$ ABC,
tan45°=$\dfrac{CB}{AB}$
=>1=$\dfrac{x}{AB}$ [∵ CB = x(the height of the mountain)]
AB = x ⋯(1)
From the right $\triangle$ AYD,
sin30°=$\dfrac{DY}{AD}$
=>$\dfrac{1}{2}=\dfrac{DY}{2}$ (∵ Given that AD = 2)
=> DY = 1 ... (2)
cos30°=$\dfrac{AY}{AD}$
=> $\dfrac{\sqrt{3}}{2}=\dfrac{AY}{2}$ (∵ Given that AD = 2)
=> AY
=√3 ... (3)
From the right $\triangle$ CED,
tan60°=$\dfrac{CE}{DE}$
⇒tan60°=$\dfrac{CE}{DE}$
=>tan60°=$\dfrac{(CB-EB)}{YB}$ [∵ CE=(CB-EB) and DE=YB)]
=>tan60°=$\dfrac{(CB-DY)}{(AB-AY)}$ [∵ EB=DY and YB=(AB-AY)]
=>tan60°=$\dfrac{(x - 1)}{(x−\sqrt{3})}$ [∵ CB=x, DY=1(eq:2), AB=x(eq:1) and AY = √3(eq:3)]
⇒√3=$\dfrac{(x - 1)}{(x−\sqrt{3})}$
⇒x√3−3=x−1
⇒x(√3−1)=2
⇒0.73x=2
⇒x=$\dfrac{2}{0.73}$=2.7
i.e., the height of the mountain = 2.7 km
Let A be the foot and C be the summit of a mountain.
Given that $\angle$CAB = 45°
From the diagram, CB is the height of the mountain. Let CB = x
Let D be the point after ascending 2 km towards the mountain such that AD = 2 km and given that $\angle$DAY = 30°
It is also given that from the point D, the elevation is 60°
i.e., $\angle$CDE = 60°
From the right $\triangle$ ABC,
tan45°=$\dfrac{CB}{AB}$
=>1=$\dfrac{x}{AB}$ [∵ CB = x(the height of the mountain)]
AB = x ⋯(1)
From the right $\triangle$ AYD,
sin30°=$\dfrac{DY}{AD}$
=>$\dfrac{1}{2}=\dfrac{DY}{2}$ (∵ Given that AD = 2)
=> DY = 1 ... (2)
cos30°=$\dfrac{AY}{AD}$
=> $\dfrac{\sqrt{3}}{2}=\dfrac{AY}{2}$ (∵ Given that AD = 2)
=> AY
=√3 ... (3)
From the right $\triangle$ CED,
tan60°=$\dfrac{CE}{DE}$
⇒tan60°=$\dfrac{CE}{DE}$
=>tan60°=$\dfrac{(CB-EB)}{YB}$ [∵ CE=(CB-EB) and DE=YB)]
=>tan60°=$\dfrac{(CB-DY)}{(AB-AY)}$ [∵ EB=DY and YB=(AB-AY)]
=>tan60°=$\dfrac{(x - 1)}{(x−\sqrt{3})}$ [∵ CB=x, DY=1(eq:2), AB=x(eq:1) and AY = √3(eq:3)]
⇒√3=$\dfrac{(x - 1)}{(x−\sqrt{3})}$
⇒x√3−3=x−1
⇒x(√3−1)=2
⇒0.73x=2
⇒x=$\dfrac{2}{0.73}$=2.7
i.e., the height of the mountain = 2.7 km
44124.On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 600 m, the distance between the objects is approximately equal to :
272 m
284 m
288 m
254 m
Explanation:
Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m , $\angle$DAC = 45°, $\angle$DBC = 60°
tan60°=$\dfrac{DC}{BC}$
=>√3=$\dfrac{600}{BC}$
=>BC=$\dfrac{600}{\sqrt{3}}$ ⋯(1)
tan45°=$\dfrac{DC}{AC}$
=>1=$\dfrac{600}{AC}$
=>AC=600 ⋯(2)
Distance between the objects
= AB = (AC - BC)
=600−$\dfrac{600}{\sqrt{3}} $ [∵ from (1) and (2)]
=600 $\left( 1-\dfrac{1}{\sqrt{3}} \right) $
=600 $\left( \dfrac{\sqrt{3}-1}{\sqrt{3}} \right) $
=600 $\left( \dfrac{\sqrt{3}-1}{\sqrt{3}} \right) \times \dfrac{\sqrt{3}}{\sqrt{3}}$
=$\dfrac{600\sqrt{3}(\sqrt{3}−1)}{3} $
=200√3(√3−1)
=200(3−√3)
=200(3−1.73)
=254 m
Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m , $\angle$DAC = 45°, $\angle$DBC = 60°
tan60°=$\dfrac{DC}{BC}$
=>√3=$\dfrac{600}{BC}$
=>BC=$\dfrac{600}{\sqrt{3}}$ ⋯(1)
tan45°=$\dfrac{DC}{AC}$
=>1=$\dfrac{600}{AC}$
=>AC=600 ⋯(2)
Distance between the objects
= AB = (AC - BC)
=600−$\dfrac{600}{\sqrt{3}} $ [∵ from (1) and (2)]
=600 $\left( 1-\dfrac{1}{\sqrt{3}} \right) $
=600 $\left( \dfrac{\sqrt{3}-1}{\sqrt{3}} \right) $
=600 $\left( \dfrac{\sqrt{3}-1}{\sqrt{3}} \right) \times \dfrac{\sqrt{3}}{\sqrt{3}}$
=$\dfrac{600\sqrt{3}(\sqrt{3}−1)}{3} $
=200√3(√3−1)
=200(3−√3)
=200(3−1.73)
=254 m
44127.A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
8 min 17 second
10 min 57 second
14 min 34 second
12 min 23 second
Explanation:
Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, $\angle$ADC = 30° , $\angle$ACB = 45°
Let AB = h, BC = x, CD = y
tan45°=$\dfrac{AB}{BC}=\dfrac{h}{x}$
=>1=$\dfrac{h}{x}$
=>h=x ⋯(1)
tan30°=$\dfrac{AB}{BD}=\dfrac{AB}{(BC+CD)}=\dfrac{h}{x+y}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{x+y}$
=>x + y = √3h
=>y = √3h - x
=>y = √3h−h (∵ Substituted the value of x from equation 1 )
=>y = h(√3−1)
Given that distance y is covered in 8 minutes.
i.e, distance h(√3−1) is covered in 8 minutes.
Time to travel distance x= Time to travel distance h (∵ Since x = h as per equation 1).
Let distance h is covered in t minutes.
since distance is proportional to the time when the speed is constant, we have
h(√3−1)∝8 ⋯(A)
h∝t ⋯(B)
$\dfrac{(A)}{(B)}=>\dfrac{h(\sqrt{3}−1)}{h}=\dfrac{8}{t}$
=>(√3−1)=$\dfrac{8}{t}$
=>t=$\dfrac{8}{(\sqrt{3}−1)}=\dfrac{8}{(1.73−1)}$
=$\dfrac{8}{.73}=\dfrac{800}{73}$minutes
=10$\dfrac{70}{73}$minutes
≈10 minutes 57 seconds
Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, $\angle$ADC = 30° , $\angle$ACB = 45°
Let AB = h, BC = x, CD = y
tan45°=$\dfrac{AB}{BC}=\dfrac{h}{x}$
=>1=$\dfrac{h}{x}$
=>h=x ⋯(1)
tan30°=$\dfrac{AB}{BD}=\dfrac{AB}{(BC+CD)}=\dfrac{h}{x+y}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{x+y}$
=>x + y = √3h
=>y = √3h - x
=>y = √3h−h (∵ Substituted the value of x from equation 1 )
=>y = h(√3−1)
Given that distance y is covered in 8 minutes.
i.e, distance h(√3−1) is covered in 8 minutes.
Time to travel distance x= Time to travel distance h (∵ Since x = h as per equation 1).
Let distance h is covered in t minutes.
since distance is proportional to the time when the speed is constant, we have
h(√3−1)∝8 ⋯(A)
h∝t ⋯(B)
$\dfrac{(A)}{(B)}=>\dfrac{h(\sqrt{3}−1)}{h}=\dfrac{8}{t}$
=>(√3−1)=$\dfrac{8}{t}$
=>t=$\dfrac{8}{(\sqrt{3}−1)}=\dfrac{8}{(1.73−1)}$
=$\dfrac{8}{.73}=\dfrac{800}{73}$minutes
=10$\dfrac{70}{73}$minutes
≈10 minutes 57 seconds
44131.The angle of elevation of the top of the tower from a point on the ground is sin−1$\left(\dfrac{3}{5}\right)$. If the point of observation is 20 meters away from the foot of the tower, what is the height of the tower?
9 m
18 m
15 m
12 m
Explanation:
Solution 1 :
Consider a right-angled triangle PQR as shown below.
Let QR = 3 and PR = 5 such that sinθ=$\dfrac{3}{5}$ [i.e., θ=sin−1 $\left( \dfrac{3}{5}\right)$]
PQ =$\sqrt{PR^{2}-QR^{2}}$ (∵ Pythagorean theorem)
i.e., when θ=sin−1$\left(\dfrac{3}{5}\right)$, PQ : QR = 4 : 3 ...(eq:1)
Now Let s solve the question. Let P be the point of observation and QR be the tower as shown in the below diagram.
Given that θ=sin−1$\left(\dfrac{3}{5}\right)$ and PQ = 20 m
We know that PQ : QR = 4 : 3 (from eq:1)
i.e., 20 : QR = 4 : 3
=> 20 × 3 = QR × 4
=> QR = 15 m
i.e., height of the tower = 15 m
Solution 2 :
Let P be the point of observation and QR be the tower.
Given that θ=sin−1$\left(\dfrac{3}{5}\right)$ and PQ = 20 m
Let the height of the tower, QR = h and PR = x
From the right $\triangle$ PQR,
sinθ=$\dfrac{QR}{PR}$
⇒$sin\left[sin^{−1}\left(\dfrac{3}{5}\right)\right]=\dfrac{h}{x}$
⇒$\dfrac{3}{5}=\dfrac{h}{x}$
⇒x=$\dfrac{5h}{3}$ ...(eq:1)
From Pythagorean theorem, we have
PQ2+QR2=PR2
202+h2=x2
202+h2=$\left(\dfrac{5h}{3}\right)^{2}$ (∵ Substituted the value of x from eq:1)
202+h2=$\left(\dfrac{25h^{2}}{9}\right)$
$\dfrac{25h^{2}}{9}$=202
$\dfrac{4h}{3}$=20
h=$\dfrac{3×20}{4}$=15 m
i.e., height of the tower = 15 m
Solution 1 :
Consider a right-angled triangle PQR as shown below.
Let QR = 3 and PR = 5 such that sinθ=$\dfrac{3}{5}$ [i.e., θ=sin−1 $\left( \dfrac{3}{5}\right)$]
PQ =$\sqrt{PR^{2}-QR^{2}}$ (∵ Pythagorean theorem)
i.e., when θ=sin−1$\left(\dfrac{3}{5}\right)$, PQ : QR = 4 : 3 ...(eq:1)
Now Let s solve the question. Let P be the point of observation and QR be the tower as shown in the below diagram.
Given that θ=sin−1$\left(\dfrac{3}{5}\right)$ and PQ = 20 m
We know that PQ : QR = 4 : 3 (from eq:1)
i.e., 20 : QR = 4 : 3
=> 20 × 3 = QR × 4
=> QR = 15 m
i.e., height of the tower = 15 m
Solution 2 :
Let P be the point of observation and QR be the tower.
Given that θ=sin−1$\left(\dfrac{3}{5}\right)$ and PQ = 20 m
Let the height of the tower, QR = h and PR = x
From the right $\triangle$ PQR,
sinθ=$\dfrac{QR}{PR}$
⇒$sin\left[sin^{−1}\left(\dfrac{3}{5}\right)\right]=\dfrac{h}{x}$
⇒$\dfrac{3}{5}=\dfrac{h}{x}$
⇒x=$\dfrac{5h}{3}$ ...(eq:1)
From Pythagorean theorem, we have
PQ2+QR2=PR2
202+h2=x2
202+h2=$\left(\dfrac{5h}{3}\right)^{2}$ (∵ Substituted the value of x from eq:1)
202+h2=$\left(\dfrac{25h^{2}}{9}\right)$
$\dfrac{25h^{2}}{9}$=202
$\dfrac{4h}{3}$=20
h=$\dfrac{3×20}{4}$=15 m
i.e., height of the tower = 15 m
44132.From the foot and the top of a building of height 230 m, a person observes the top of a tower with angles of elevation of b and a respectively. What is the distance between the top of these buildings if tan a = 5/12 and tan b = 4/5
400 m
250 m
600 m
650 m
Explanation:
Let ED be the building and AC be the tower.
Given that ED = 230 m, $\angle$ADC = b, $\angle$AEB = a
Also given that tan a = 5/12 and tan b = 4/5
Let AC = h
Required Distance = Distance between the top of these buildings = AE
From the right $\triangle$ ABE,
tan(a)=$\dfrac{AB}{BE}$
=> $\dfrac{5}{12}=\dfrac{(h-230)}{BE}$ [∵ tan(a)=5/12(given), AB = (AC-BC) = (AC-ED) = (h-230)]
=> BE =$\dfrac{12(h-230)}{5}$ ⋯(eq:1)
From the right $\triangle$ ACD,
tan(b)=$\dfrac{AC}{CD}$
=> $\dfrac{4}{5}=\dfrac{h}{CD}$ [∵ tan(b) = 4/5(given), AC=h]
=> CD =$\dfrac{5}{h4}$ ⋯(eq:2)
From the diagram, BE = CD
⇒$\dfrac{12(h-230)}{5}=\dfrac{5h}{4}$ (from eq:1 & eq:2)
⇒48h−(4×12×230)=25h
⇒23h=(4×12×230)
⇒h=$\dfrac{(4×12×230)}{23}$=480 m ⋯(eq:3)
AB = (AC - BC)
= (480 - 230) [∵ Since AC=h=480(from eq:3) and BC=ED=230 m(given)]
= 250 m
In the triangle ABE, tan(a) = 5/12. Let s figure out the value of sin(a) now.
Consider a triangle with opposite side = 5 and adjacent side = 12 such that tan(a) = 5/12
hypotenuse =$\sqrt{5^{2}+12^{2}}$=13
i.e., sin(a) =$\dfrac{opposite}{ sidehypotenuse}=\dfrac{5}{13}$
We have seen that sin(a) = $\dfrac{5}{13}$
⇒$\dfrac{AB}{AE}=\dfrac{5}{13}$
⇒AE = AB×$\dfrac{13}{5}$
=250×$\dfrac{13}{5}$=650 m
i.e., Distance between the top of the buildings = 650 m
Let ED be the building and AC be the tower.
Given that ED = 230 m, $\angle$ADC = b, $\angle$AEB = a
Also given that tan a = 5/12 and tan b = 4/5
Let AC = h
Required Distance = Distance between the top of these buildings = AE
From the right $\triangle$ ABE,
tan(a)=$\dfrac{AB}{BE}$
=> $\dfrac{5}{12}=\dfrac{(h-230)}{BE}$ [∵ tan(a)=5/12(given), AB = (AC-BC) = (AC-ED) = (h-230)]
=> BE =$\dfrac{12(h-230)}{5}$ ⋯(eq:1)
From the right $\triangle$ ACD,
tan(b)=$\dfrac{AC}{CD}$
=> $\dfrac{4}{5}=\dfrac{h}{CD}$ [∵ tan(b) = 4/5(given), AC=h]
=> CD =$\dfrac{5}{h4}$ ⋯(eq:2)
From the diagram, BE = CD
⇒$\dfrac{12(h-230)}{5}=\dfrac{5h}{4}$ (from eq:1 & eq:2)
⇒48h−(4×12×230)=25h
⇒23h=(4×12×230)
⇒h=$\dfrac{(4×12×230)}{23}$=480 m ⋯(eq:3)
AB = (AC - BC)
= (480 - 230) [∵ Since AC=h=480(from eq:3) and BC=ED=230 m(given)]
= 250 m
In the triangle ABE, tan(a) = 5/12. Let s figure out the value of sin(a) now.
Consider a triangle with opposite side = 5 and adjacent side = 12 such that tan(a) = 5/12
hypotenuse =$\sqrt{5^{2}+12^{2}}$=13
i.e., sin(a) =$\dfrac{opposite}{ sidehypotenuse}=\dfrac{5}{13}$
We have seen that sin(a) = $\dfrac{5}{13}$
⇒$\dfrac{AB}{AE}=\dfrac{5}{13}$
⇒AE = AB×$\dfrac{13}{5}$
=250×$\dfrac{13}{5}$=650 m
i.e., Distance between the top of the buildings = 650 m
44137.From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole?
52 m
50 m
66.67 m
33.33 m
Explanation:
Consider the diagram shown above. AC represents the hill and DE represents the pole
Given that AC = 100 m
$\angle$XAD = $\angle$ADB = 30° (∵ AX || BD )
$\angle$XAE = $\angle$AEC = 60° (∵ AX || CE)
Let DE = h
Then, BC = DE = h,
AB = (100-h) (∵ AC=100 and BC = h),
BD = CE
tan60°=$\dfrac{AC}{CE}$
=>√3=$\dfrac{100}{CE}$
=>CE=$\dfrac{100}{\sqrt{3}} $ ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}} =\dfrac{100-h}{\left(\dfrac{100}{\sqrt{3}}\right)}$ (∵ BD = CE and substituted the value of CE from equation 1)
=>(100−h)=$\dfrac{1}{\sqrt{3}} \times \dfrac{100}{\sqrt{3}}$
=$\dfrac{100}{3}=33.33$
=>h=100−33.33=66.67 m
i.e., the height of the pole = 66.67 m
Consider the diagram shown above. AC represents the hill and DE represents the pole
Given that AC = 100 m
$\angle$XAD = $\angle$ADB = 30° (∵ AX || BD )
$\angle$XAE = $\angle$AEC = 60° (∵ AX || CE)
Let DE = h
Then, BC = DE = h,
AB = (100-h) (∵ AC=100 and BC = h),
BD = CE
tan60°=$\dfrac{AC}{CE}$
=>√3=$\dfrac{100}{CE}$
=>CE=$\dfrac{100}{\sqrt{3}} $ ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}} =\dfrac{100-h}{\left(\dfrac{100}{\sqrt{3}}\right)}$ (∵ BD = CE and substituted the value of CE from equation 1)
=>(100−h)=$\dfrac{1}{\sqrt{3}} \times \dfrac{100}{\sqrt{3}}$
=$\dfrac{100}{3}=33.33$
=>h=100−33.33=66.67 m
i.e., the height of the pole = 66.67 m