44116.The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree?
22 m
44 m
33 m
None of these
Explanation:
Let DC be the wall, AB be the tree.
Given that $\angle$DBC = 30°, $\angle$DAE = 60°, DC = 11 m
tan30°=$\dfrac{DC}{BC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{11}{BC}$
BC = 11√3 m
AE = BC =11√3 m ⋯(1)
tan60°=$\dfrac{ED}{AE}$
=>√3=$\dfrac{ED}{11\sqrt{3}}$ [∵ Substituted value of AE from (1)]
ED =11√3×√3=11×3=33
Height of the tree
= AB = EC = (ED + DC)
= (33 + 11) = 44 m
Let DC be the wall, AB be the tree.
Given that $\angle$DBC = 30°, $\angle$DAE = 60°, DC = 11 m
tan30°=$\dfrac{DC}{BC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{11}{BC}$
BC = 11√3 m
AE = BC =11√3 m ⋯(1)
tan60°=$\dfrac{ED}{AE}$
=>√3=$\dfrac{ED}{11\sqrt{3}}$ [∵ Substituted value of AE from (1)]
ED =11√3×√3=11×3=33
Height of the tree
= AB = EC = (ED + DC)
= (33 + 11) = 44 m
44122.A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 15 m away from the base of the pole, what is the height of the pole?
60√5m
15√5m
15√3 m
60√3m
Explanation:
Let CB be the pole and point D divides it such that BD : DC = 1 : 9
Given that AB = 15 m
Let the the two parts subtend equal angles at point A such that
$\angle$CAD =$\angle$BAD = θ
From "Angle Bisector Theorem", we have
$\dfrac{BD}{DC}=\dfrac{AB}{AC}$
⇒$\dfrac{1}{9}=\dfrac{15}{AC}$ [∵ BD : DC = 1 : 9 and AB = 15(given)]
=> AC = 15 × 9 m ...(eq: 1)
From the right $\triangle$ ABC,
CB=$\sqrt{AC^{2}-AB^{2}}$ (∵ Pythagorean theorem)
=$\sqrt{(15 \times 9)^{2}-15^{2}}$ (∵AC=15×9(eq:1) and AB=15 m(given))
=$\sqrt{15^{2} \times 9^{2}-15^{2}}$
=$\sqrt{15^{2} (9^{2}-1)}$=$\sqrt{15^{2} \times 80}$
=$\sqrt{15^{2} \times 16 \times 5}$ =15×4×√5
=60√5 m
Let CB be the pole and point D divides it such that BD : DC = 1 : 9
Given that AB = 15 m
Let the the two parts subtend equal angles at point A such that
$\angle$CAD =$\angle$BAD = θ
From "Angle Bisector Theorem", we have
$\dfrac{BD}{DC}=\dfrac{AB}{AC}$
⇒$\dfrac{1}{9}=\dfrac{15}{AC}$ [∵ BD : DC = 1 : 9 and AB = 15(given)]
=> AC = 15 × 9 m ...(eq: 1)
From the right $\triangle$ ABC,
CB=$\sqrt{AC^{2}-AB^{2}}$ (∵ Pythagorean theorem)
=$\sqrt{(15 \times 9)^{2}-15^{2}}$ (∵AC=15×9(eq:1) and AB=15 m(given))
=$\sqrt{15^{2} \times 9^{2}-15^{2}}$
=$\sqrt{15^{2} (9^{2}-1)}$=$\sqrt{15^{2} \times 80}$
=$\sqrt{15^{2} \times 16 \times 5}$ =15×4×√5
=60√5 m
44128.A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man s eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?
26.28 km/hr
32.42 km/hr
24.22 km/hr
31.25 km/hr
Explanation:
Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, $\angle$ACB = 45° , $\angle$ADC = 30°, BC = 100 m
tan45°=$\dfrac{AB}{BC}$
=>1=$\dfrac{AB}{100}$
=> AB = 100 ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{100}{BD}$ (∵ Substituted the value of AB from equation 1)
=> BD =100√3
CD = (BD - BC) =(100√3−100)=100(√3−1)
It is given that the distance CD is covered in 10 seconds.
i.e., the distance 100(√3−1) is covered in 10 seconds.
Required speed
=$\dfrac{Distance}{Time}=\dfrac{100(\sqrt{3}−1)}{10}$=10(1.73−1)
= 7.3 meter/seconds
= 7.3 × $\dfrac{18}{5}$ km/hr = 26.28 km/hr
Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, $\angle$ACB = 45° , $\angle$ADC = 30°, BC = 100 m
tan45°=$\dfrac{AB}{BC}$
=>1=$\dfrac{AB}{100}$
=> AB = 100 ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{100}{BD}$ (∵ Substituted the value of AB from equation 1)
=> BD =100√3
CD = (BD - BC) =(100√3−100)=100(√3−1)
It is given that the distance CD is covered in 10 seconds.
i.e., the distance 100(√3−1) is covered in 10 seconds.
Required speed
=$\dfrac{Distance}{Time}=\dfrac{100(\sqrt{3}−1)}{10}$=10(1.73−1)
= 7.3 meter/seconds
= 7.3 × $\dfrac{18}{5}$ km/hr = 26.28 km/hr
44129.The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
5 metres
8 metres
10 metres
12 metres
Explanation:
Consider the diagram shown above. AC represents the tower and DE represents the pole.
Given that AC = 15 m, $\angle$ADB = 30°, $\angle$AEC = 60°
Let DE = h
Then, BC = DE = h,
AB = (15-h) (∵ AC=15 and BC = h),
BD = CE
tan60°=$\dfrac{AC}{CE}$
=>√3=$\dfrac{15}{CE}$
=>CE=$\dfrac{15}{\sqrt{3}}$ ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{15−h}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{15−h}{\left( \dfrac{15}{\sqrt{3}}\right)}$ (∵ BD = CE and substituted the value of CE from equation 1 )
=>(15−h)=$\dfrac{1}{\sqrt{3}}\times\dfrac{15}{\sqrt{3}}=\dfrac{15}{3}=5$
=>h=15−5=10 m
i.e., height of the electric pole = 10 m
Consider the diagram shown above. AC represents the tower and DE represents the pole.
Given that AC = 15 m, $\angle$ADB = 30°, $\angle$AEC = 60°
Let DE = h
Then, BC = DE = h,
AB = (15-h) (∵ AC=15 and BC = h),
BD = CE
tan60°=$\dfrac{AC}{CE}$
=>√3=$\dfrac{15}{CE}$
=>CE=$\dfrac{15}{\sqrt{3}}$ ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{15−h}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{15−h}{\left( \dfrac{15}{\sqrt{3}}\right)}$ (∵ BD = CE and substituted the value of CE from equation 1 )
=>(15−h)=$\dfrac{1}{\sqrt{3}}\times\dfrac{15}{\sqrt{3}}=\dfrac{15}{3}=5$
=>h=15−5=10 m
i.e., height of the electric pole = 10 m
44130.The angle of elevation of the top of a lighthouse 60 m high, from two points on the ground on its opposite sides are 45° and 60°. What is the distance between these two points?
45 m
30 m
103.8 m
94.6 m
Explanation:
Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
$\angle$ BAD = 45° , $\angle$ BCD = 60°
tan45°=$\dfrac{BD}{BA}$
=>1=$\dfrac{60}{BA}$
=>BA=60 m ⋯(1)
tan60°=$\dfrac{BD}{BC}$
=>√3=$\dfrac{60}{BC}$
=>BC=$\dfrac{60}{\sqrt{3}}=\dfrac{60 \times \sqrt{3} }{\sqrt{3} \times \sqrt{3}}=\dfrac{60\sqrt{3} }{3}$
=20√3=20×1.73=34.6 m ⋯(2)
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (1) and (2)]
= 94.6 m
Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
$\angle$ BAD = 45° , $\angle$ BCD = 60°
tan45°=$\dfrac{BD}{BA}$
=>1=$\dfrac{60}{BA}$
=>BA=60 m ⋯(1)
tan60°=$\dfrac{BD}{BC}$
=>√3=$\dfrac{60}{BC}$
=>BC=$\dfrac{60}{\sqrt{3}}=\dfrac{60 \times \sqrt{3} }{\sqrt{3} \times \sqrt{3}}=\dfrac{60\sqrt{3} }{3}$
=20√3=20×1.73=34.6 m ⋯(2)
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (1) and (2)]
= 94.6 m
44133.To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window?
8.65 m
2 m
2.5 m
3.65 m
Explanation:
Let AB be the man and CD be the window
Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
$\angle$DAF = 45° , $\angle$CAF = 60°
From the diagram, AF = BE = 5 m
From the right $\triangle$ AFD,
tan45°=$\dfrac{DF}{AF}$
=>1=$\dfrac{DF}{5}$
DF = 5 ⋯(1)
From the right $\triangle$ AFC,
tan60°=$\dfrac{CF}{AF}$
=>√3=$\dfrac{CF}{15}$
CF =5√3 ...(2)
Length of the window
= CD = (CF - DF)
=5√3−5 [∵ Substituted the value of CF and DF from (1) and (2)]
=5(√3−1)=5(1.73−1)
=5×0.73=3.65 m
Let AB be the man and CD be the window
Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
$\angle$DAF = 45° , $\angle$CAF = 60°
From the diagram, AF = BE = 5 m
From the right $\triangle$ AFD,
tan45°=$\dfrac{DF}{AF}$
=>1=$\dfrac{DF}{5}$
DF = 5 ⋯(1)
From the right $\triangle$ AFC,
tan60°=$\dfrac{CF}{AF}$
=>√3=$\dfrac{CF}{15}$
CF =5√3 ...(2)
Length of the window
= CD = (CF - DF)
=5√3−5 [∵ Substituted the value of CF and DF from (1) and (2)]
=5(√3−1)=5(1.73−1)
=5×0.73=3.65 m
44135.An aeroplane when 900 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 60° and 45° respectively. Approximately, how many meters higher is the one than the other?
381 m
169 m
254 m
211 m
Explanation:
Let C and D be the position of the aeroplanes.
Given that CB = 900 m, $\angle$CAB = 60°, $\angle$DAB = 45°
From the right $\triangle$ ABC,
tan60°=$\dfrac{CB}{AB}$
√3=$\dfrac{900}{AB} $
AB=$\dfrac{900}{\sqrt{3}} $
=$\dfrac{900 \times \sqrt{3} }{\sqrt{3} \times \sqrt{3}} =\dfrac{900 \sqrt{3} }{3} $=300√3
rom the right $\triangle$ ABD,
tan45°=$\dfrac{DB}{AB}$
1=$\dfrac{DB}{AB}$
DB=AB=300√3
Required height
= CD = (CB-DB)
=(900−300√3)=(900−300×1.73)
=(900−519)=381 m
Let C and D be the position of the aeroplanes.
Given that CB = 900 m, $\angle$CAB = 60°, $\angle$DAB = 45°
From the right $\triangle$ ABC,
tan60°=$\dfrac{CB}{AB}$
√3=$\dfrac{900}{AB} $
AB=$\dfrac{900}{\sqrt{3}} $
=$\dfrac{900 \times \sqrt{3} }{\sqrt{3} \times \sqrt{3}} =\dfrac{900 \sqrt{3} }{3} $=300√3
rom the right $\triangle$ ABD,
tan45°=$\dfrac{DB}{AB}$
1=$\dfrac{DB}{AB}$
DB=AB=300√3
Required height
= CD = (CB-DB)
=(900−300√3)=(900−300×1.73)
=(900−519)=381 m
44136.Two persons are on either sides of a tower of height 50 m. The persons observers the top of the tower at an angle of elevation of 30° and 60°. If a car crosses these two persons in 10 seconds, what is the speed of the car?
24√3 km/hr
None of these
24/√3km/hr
20√3/3 km/hr
Explanation:
Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, $\angle$ BAD = 30°, $\angle$BCD = 60°
From the right $\triangle$ ABD,
tan30°=$\dfrac{BD}{BA}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{50}{BA}$
⇒BA=50√3
From the right $\triangle$ CBD,
tan60°=$\dfrac{BD}{BC}$
=>√3=$\dfrac{50}{BC}$
=>BC=$\dfrac{50}{\sqrt{3}}=\dfrac{50 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\dfrac{50\sqrt{3}}{3}$
Distance between the two persons
= AC = BA + BC
=50√3+$\dfrac{50\sqrt{3}}{3}$
=√3 $\left( 50+\dfrac{50}{3} \right) $
=$\dfrac{200\sqrt{3}}{3}$ m
i.e., the distance travelled by the car in 10 seconds =$\dfrac{200\sqrt{3}}{3}$ m
Speed of the car =$\dfrac{Distance}{Time}$
=$\dfrac{\left( \dfrac{200\sqrt{3}}{3}\right)}{10}= \dfrac{20\sqrt{3}}{3}$ meter/second
= $\dfrac{20\sqrt{3}}{3} \times \dfrac{18}{5}$km/hr==24√3 km/hr
Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, $\angle$ BAD = 30°, $\angle$BCD = 60°
From the right $\triangle$ ABD,
tan30°=$\dfrac{BD}{BA}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{50}{BA}$
⇒BA=50√3
From the right $\triangle$ CBD,
tan60°=$\dfrac{BD}{BC}$
=>√3=$\dfrac{50}{BC}$
=>BC=$\dfrac{50}{\sqrt{3}}=\dfrac{50 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\dfrac{50\sqrt{3}}{3}$
Distance between the two persons
= AC = BA + BC
=50√3+$\dfrac{50\sqrt{3}}{3}$
=√3 $\left( 50+\dfrac{50}{3} \right) $
=$\dfrac{200\sqrt{3}}{3}$ m
i.e., the distance travelled by the car in 10 seconds =$\dfrac{200\sqrt{3}}{3}$ m
Speed of the car =$\dfrac{Distance}{Time}$
=$\dfrac{\left( \dfrac{200\sqrt{3}}{3}\right)}{10}= \dfrac{20\sqrt{3}}{3}$ meter/second
= $\dfrac{20\sqrt{3}}{3} \times \dfrac{18}{5}$km/hr==24√3 km/hr
44138.A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower?
9 m
10.40 m
15.57 m
12 m
Explanation:
Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.
Given that AD = 18 m, $\angle$ABC = 60°, $\angle$DBC = 30°
Let DC be h.
tan30°=$\dfrac{DC}{BC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{BC}$
=>h=$\dfrac{BC}{\sqrt{3}}$ ...(1)
tan60°=$\dfrac{AC}{BC}$
=>√3=$\dfrac{18+h}{BC}$
18+h=BC×√3 ⋯(2)
$ \dfrac{(1)}{(2)}=>\dfrac{h}{18+h}=\dfrac{\dfrac{BC}{\sqrt{3}}} {(BC×\sqrt{3})}=\dfrac{1}{3}$
=>3h=18+h
=>2h=18
=>h=9 m
i.e., the height of the tower = 9 m
Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.
Given that AD = 18 m, $\angle$ABC = 60°, $\angle$DBC = 30°
Let DC be h.
tan30°=$\dfrac{DC}{BC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{BC}$
=>h=$\dfrac{BC}{\sqrt{3}}$ ...(1)
tan60°=$\dfrac{AC}{BC}$
=>√3=$\dfrac{18+h}{BC}$
18+h=BC×√3 ⋯(2)
$ \dfrac{(1)}{(2)}=>\dfrac{h}{18+h}=\dfrac{\dfrac{BC}{\sqrt{3}}} {(BC×\sqrt{3})}=\dfrac{1}{3}$
=>3h=18+h
=>2h=18
=>h=9 m
i.e., the height of the tower = 9 m
44139.The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is:
64.2 m
2 m
52.2 m
54.6 m
Explanation:
Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have $\angle$DAC = 30°, $\angle$DBC = 45°
Let DC = h
tan30°=$\dfrac{DC}{AC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{AC}$
=>AC = h√3 ⋯(1)
tan45°=$\dfrac{DC}{BC}$
=>1=$\dfrac{h}{BC}$
=>BC=h ⋯(2)
We know that, AB = (AC - BC)
=> 40 = (AC - BC)
=> 40=(h√3−h)[∵ from (1) & (2)]
=>40=h(√3−1)
=>h=$\dfrac{40}{(\sqrt{3}−1)}$
=$\dfrac{40}{(\sqrt{3}−1)} \times \dfrac{(\sqrt{3}+1)}{(\sqrt{3}+1)} $
=$\dfrac{40(\sqrt{3}+1)}{(3−1)}$
=$\dfrac{40(\sqrt{3}+1)}{(2)}$
=20(√3+1)
=20(1.73+1)
=20×2.73
=54.6 m
Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have $\angle$DAC = 30°, $\angle$DBC = 45°
Let DC = h
tan30°=$\dfrac{DC}{AC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{AC}$
=>AC = h√3 ⋯(1)
tan45°=$\dfrac{DC}{BC}$
=>1=$\dfrac{h}{BC}$
=>BC=h ⋯(2)
We know that, AB = (AC - BC)
=> 40 = (AC - BC)
=> 40=(h√3−h)[∵ from (1) & (2)]
=>40=h(√3−1)
=>h=$\dfrac{40}{(\sqrt{3}−1)}$
=$\dfrac{40}{(\sqrt{3}−1)} \times \dfrac{(\sqrt{3}+1)}{(\sqrt{3}+1)} $
=$\dfrac{40(\sqrt{3}+1)}{(3−1)}$
=$\dfrac{40(\sqrt{3}+1)}{(2)}$
=20(√3+1)
=20(1.73+1)
=20×2.73
=54.6 m