Height and Distance - Aptitude Questions and Answers Part 2

6. A building is under construction. The top of the building forms 30° angle of elevation from a point on the adjoining plot that is 300 m. After a month, the angle of elevation formed by the top of the building from the same point increased to 60°. How much was the building constructed in this 1 month.  

a. 200/3 m
b. 1003 m
c. 2003 m
d. 3001
3

Answer: c. 2003 m

Explanation:

Tip:
Tan θ =P
B
Where, P = Perpendicular ; B = Base
Tan 30° =1; Tan 45° = 1;
3
Tan 60° = 3 ; Tan 90° = Not defined


height and distance ans-6

Original Building height = h = MQ
New building height = PQ
In △MQN, tan 30° =1=MQ
3NQ
∴ MQ =300
3
In △PQN, tan 60° = 3 =PQ
NQ
∴ PQ = 3003
Building grew = PQ - MQ
∴ Building grew = 3003 -300= 3002= 3 x 100 x2= 2003 m
333


7. Rohit was trying to identify a bird sitting at the top of a tree making a 45° angle of elevation. Since he could not identify the bird, he moved closer to the tree by 240 ft. and thus formed 60° angle of elevation. How far was he from the tree initially?

a.240ft
3 - 1
b.2403ft
3 - 1
c.240ft
3
d. 2403 ft

Answer: b.2403ft
3 - 1

Explanation:

Tip:
Tan θ =P
B
Where, P = Perpendicular ; B = Base
Tan 30° =1; Tan 45° = 1;
3
Tan 60° = 3 ; Tan 90° = Not defined


height and distance ans-7

Now, Tan 45° = 1 =PQ
NQ
∴ Tree height = PQ = NQ = (240 + MQ)
Tan 60° = 3 =PQ=240 + MQ
MQMQ
∴ 240 + MQ = 3 MQ
∴ MQ =240ft
3 - 1
∴ NQ = 240 + MQ =2403ft. = Rohit was this much far away initially
3 - 1


8. Two friends Ajay and Vijay formed sand castles of heights of height 8 cm and 15 cm respectively on the sea shore. The distance between the two castles was 24 cm. Find the distance between the tops of two castles.

a. 24cm
b. 24.5cm
c. 25cm
d. 31cm

Answer: c. 25cm

Explanation:

height and distance ans-8

Let MN = Ajay's castle & PQ = Vijay's castle
From the diagram we can see that
MR = 24cm   &   PR = 15 - 8 = 7cm
By Pythagoras Theorem,
Hypotenuse2 = (side1)2 + (side2)2
∴ MP = 242 + 72
∴ MP = 25cm


9. When looking at the top of a pole from two different points m and n on the ground, the angles of elevation formed are 60° and 45° respectively. Find the height of the pole.

a. question 9 option a
b. question 9 option b
c. mn units
d. 3mn units

Answer: a. question 9 option a

Explanation:

Tip:
Tan θ =P
B
Where, P = Perpendicular ; B = Base
Tan 30° =1; Tan 45° = 1;
3
Tan 60° = 3 ; Tan 90° = Not defined


height and distance ans-9

Height of pole = PQ
Tan 60° = 3 =PQ
m
Tan 45° = 1 =PQ
n
Multiply both equations
3 x 1 =PQxPQ
mn
∴ PQ = answer


10. Standing at one corner of a square piece of land, a boy looks in the diagonally opposite direction and sees an object that looks like a scare crow. At the moment his angle of elevation formed is 60°. He starts walking back home in a straight line and after 80 ft realizes that the angle of elevation of the object now is only 30°. Find the area of the land he was standing on.

a. 40/2 sq.ft.
b. 40sq.ft.
c. 800sq.ft.
d. 1600sq.ft.

Answer: c. 800sq.ft.

Explanation:

Tip:
Tan θ =P
B
Where, P = Perpendicular ; B = Base
Tan 30° =1; Tan 45° = 1;
3
Tan 60° = 3 ; Tan 90° = Not defined


height and distance ans-10

Tan 60° = 3 =PQ
QR
∴ PQ = 3 QR
Tan 30° =1=PQ=PQ
3SQ80+QR
∴ 80 + QR =  3 PQ
∴ 80 + QR = 3QR
∴ QR = 40 ft.
If we read carefully, we see that the boy (Point R) and the scarecrow (Point Q) are in diagonally opposite corners.
So QR is a diagonal of the square farm.
Diagonal of square = side x 2
∴ 40 = side x 2
∴ side = 40/2
∴ Area = (side)2 = (40/2)2 = (1600/2) = 800sq.ft.

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