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
Original Building height = h = MQ
New building height = PQ
∴ PQ = 3003
Building grew = PQ - MQ
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?
d. 2403 ft
Explanation:
∴ Tree height = PQ = NQ = (240 + MQ)
∴ 240 + MQ = 3 MQ
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
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.
b.
c. mn units
d. 3mn units
Height of pole = PQ
Multiply both equations
∴ PQ = 
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.
∴ PQ = 3 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.
a. 200/3 m
b. 1003 m
c. 2003 m
| d. 300 | 1 |
| 3 |
Answer: c. 2003 m
Explanation:
Tip:
Where, P = Perpendicular ; B = Base
Tan 60° = 3 ; Tan 90° = Not defined
| Tan θ = | P |
| B |
| Tan 30° = | 1 | ; Tan 45° = 1; |
| 3 |
Original Building height = h = MQ
New building height = PQ
| In △MQN, tan 30° = | 1 | = | MQ |
| 3 | NQ |
| ∴ MQ = | 300 |
| 3 |
| In △PQN, tan 60° = 3 = | PQ |
| NQ |
Building grew = PQ - MQ
| ∴ Building grew = 3003 - | 300 | = 300 | 2 | = 3 x 100 x | 2 | = 2003 m |
| 3 | 3 | 3 |
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. | 240 | ft |
| 3 - 1 |
| b. | 2403 | ft |
| 3 - 1 |
| c. | 240 | ft |
| 3 |
| Answer: b. | 2403 | ft |
| 3 - 1 |
Explanation:
Tip:
Where, P = Perpendicular ; B = Base
Tan 60° = 3 ; Tan 90° = Not defined
| Tan θ = | P |
| B |
| Tan 30° = | 1 | ; Tan 45° = 1; |
| 3 |
| Now, Tan 45° = 1 = | PQ |
| NQ |
| Tan 60° = 3 = | PQ | = | 240 + MQ |
| MQ | MQ |
| ∴ MQ = | 240 | ft |
| 3 - 1 |
| ∴ NQ = 240 + MQ = | 2403 | ft. = 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:
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.
b.
c. mn units
d. 3mn units
Answer: a.
Explanation:
Tip:
Where, P = Perpendicular ; B = Base
Tan 60° = 3 ; Tan 90° = Not defined
| Tan θ = | P |
| B |
| Tan 30° = | 1 | ; Tan 45° = 1; |
| 3 |
Height of pole = PQ
| Tan 60° = 3 = | PQ |
| m |
| Tan 45° = 1 = | PQ |
| n |
| 3 x 1 = | PQ | x | PQ |
| m | n |
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:
Where, P = Perpendicular ; B = Base
Tan 60° = 3 ; Tan 90° = Not defined
| Tan θ = | P |
| B |
| Tan 30° = | 1 | ; Tan 45° = 1; |
| 3 |
| Tan 60° = 3 = | PQ |
| QR |
| Tan 30° = | 1 | = | PQ | = | PQ |
| 3 | SQ | 80+QR |
∴ 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.
