MCQ Test: Height and Distance- 2 - Bank Exams MCQ

Given, BC = 150 m
∠ACB = 30°
and, ∠DCB = 45°
Then, AD = ?
In ΔABC, tan 30° = AB / BC
1 / √3 = AB / 150
∴ AB = 150 / √3 = 86.6m
In ΔDBC, tan 45° = DB / BC
1 = DB / 150
DB = 150
AD + AB = 150
[∵ DB = AD + AB]
∴ AD = 150 – AB
= 150 – 86.6 = 63.4m
Hence, option D is correct.

Given,
∠ ACB = 45°
∠ ADB = 30°
and distance between two ships, i.e.,
CD = 200 m
Then, AB = ?
Let BC = x m
In ΔABC,
tan 45º = AB /BC
(∵ tan 45° = 1)
1 = AB / x
∴ AB = x m ....(i)
In ΔABD, tan30º = AB / BD
∴ 1 / √3 = AB / (x + 200)
(∵ tan 30° = 1/√3 )
x = √3, AB – 200 ....(ii)
From Eqs. (i) and (ii),
AB = √3AB – 200
√3 AB – AB = 200
0.732 AB = 200
(∵ √3 = 1.732)
AB = 200 / 0.732 = 273 .22
= 273 m
Hence, option D is  correct.

Let AB be tower and BC be its shadow
∴ Let AB = x


∴ Angle of elevation of the sun = 30^∘

Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
∠BAD = 45°, ∠BCD = 60°


Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (i) and (ii)]
= 94.6 m

Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m, ∠DAC = 45°, ∠DBC = 60°

AC = 600.............................(ii)
Distance between the objects

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 200 m, ∠BAD = 30°, ∠BCD = 45°

Distance between the two ships
 

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have ∠DAC = 30°, ∠DBC = 45°
Let DC = h

We know that,
AB = (AC − BC)
⇒ 40 = (AC − BC)

= 54.6 m

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.
Given that AD = 18 m, ∠ ABC = 60°, ∠ DBC = 30°
Let DC be h


i.e., the height of the tower = 9 m

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m, ∠ ABD = 30°, ∠ ACD = 60°,
Let CD = x, AD = h
From the right ΔCDA


Substituting this value of x in eq : 1, we have

Let DC be the wall, AB be the tree.
Given that ∠DBC = 30°, ∠DAE = 60°, DC = 11 m

Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, ∠ ACB = 45°, ∠ ADC = 30°, BC = 100 m

(∵ Substituted the value of AB from equation 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

Consider the diagram shown above where QR represents the tree and PQ represents its shadow
We have, QR = PQ
Let ∠QPR = θ

i.e., required angle of elevation = 45°

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, ∠ ADC = 30° , ∠ ACB = 45°
Let AB = h, BC = x, CD = y


⇒ 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

Let AC be the tower and B be the position of the bus.
Then BC = the distance of the bus from the foot of the tower.
Given that height of the tower, AC = 80 m and the angle of depression, ∠DAB = 30°
∠ABC = ∠DAB = 30° (because DA || BC)

i.e., Distance of the bus from the foot of the tower = 138.4 m

Let the tree break at height h cm from ground at point M.
The broken part makes angle of 30°
∴ Broken Part MP = MN = 24 - h
in ΔMCQ

∴ 24 - h = 2h
∴ h = 8 cm = Tree breaks at this height

In ∠ABR, ∠ARB = 90° and ∠BAR = 45°
Sum of angles of a triangle = 180°
So ∠ABR = 180 - 90 - 45 = 45°
∴ BR = AR
AS = RQ = 10m
Also, BR = BQ - RQ = 25 - 10 = 15m
∴ AR = 15m = Distance between houses

In ΔSTR, by Pythagoras theorem
RT² = ST² + RS²
∴ ST² = 15² − 9² = 144
∴ ST = 12m
In ΔTQP, by Pythagoras theorem
PT² = TQ² + PQ²
∴ TQ² = 15² − 12² = 81
∴ TQ = 9m
Distance between houses
⇒ SQ = ST + TQ
⇒ SQ = 12 + 9
⇒ SQ = 21 m

Let MN = Ramesh's fort & PQ = Suresh's fort
From the diagram we can see that
MR = 24 cm & PR = 15 - 8 = 7 cm
By Pythagoras theorem,
Hypotenuse² = (side1)² + (side2)²

Both the man and pole are near each other and are illuminated by same sun from same direction.

So angle of elevation for sun is same for both.
So ration of object to shadow will be same for all objects. (Proportionality Rule)

∴ H = 54 ft = Height of pole