ANGLE OF ELEVATION & DEPRESSION
-
Upload
mirabel-isambero -
Category
Documents
-
view
52 -
download
4
description
Transcript of ANGLE OF ELEVATION & DEPRESSION
If the object is above the horizontal level of the eyes we raise our head
upwards to view the object. Our eyes move through an angle to view the
object. The angle between the horizontal line and the line of sight
is called the angle of elevation
θ
If the object is below the horizontal level of the eyes we
move down our eyes through an angle to view the object. The
angle between the horizontal line and the line of sight is called the
angle of depression.Ф
Steps in Solving problems To solve problems the first step is to
draw the correct labeled diagram based on the question.
Mark the sides and angles given in the question.
Take the unknown quantities to be
found out as ‘x’ , ‘y’ , ‘p’ etc , and mark in the diagram.
Identify the correct trigonometric ratio which can be formed with the given quantities and the unknown quantity to be found out.
Hill, far away from the measuring point Take two points A,B (say) the distance between them is
measurable(500m) The angle of elevation at A is (45°) the angle of elevation at B is (60°)
Hill height= h
BA 500m)60 )45
C
D
height= h
BA500m
)60o )45 o
Tan45°=h/AD
1=h/AD h=AD
(or) h=AB + BD = 500+BD---(1)
Also tan 60°=h/BD
3=h/BD
BD=h/ 3 -------(2)
From (1) & (2) , find ‘h’
h = 500+BD--- (1) BD = h/ 3 -------(2) Substitute (2) in (1) h = 500 + h/ 3 (or) 3h – h = 500 3 (or) h (3 – 1 ) = 500 3 (or) h = 500 3 (3 – 1 ) = 500 3 (3 + 1 ) (3 – 1 ) (3+1 ) = (1500-500
x1.732)/2Simplify & find the answer.
AB is light house C and D are
positions of boats A man in a boat
observes that angle of elevation changes from 60 to 45 in 2 minutes
B
A
CD45°
60°
150m
To Find Speed
AD \ AB = cot 45 ° = 1
AD \ 150 = 1AD = 150 m
Let AC = X meters CD = (150 -x)
AC \ AB = cot 60° = 1 \ 3 x \ 150 = 1 \ 3
A
45° 60°
B
45°CD A
60°
B
X = ( 150 \ \/3 )m = ( 50 3 ) m
CD = (150 - 50 3) m = 63.4 m
Boat covers 63.4 m in 2 minutes.
Speed of boat = ( 63.4 ÷120 ) m\s = 0.53 m\s
To Find Height
An observer is 1.6 m tall and is 45 meters away .
The angle of elevation from his eye to top of tower is 30
Determine height of tower.
30°
45m
1.6
m
Solution
Height of the tower = h+1.6m
h=45 tan30° h=45 X 1/3 h=45 X 3/3 =15X1.73 =25.95
Height=27.55 m
1.6
30°45m
h45m
From the top of a tower 50m high the angles of depression of the top and bottom of a pole are observed to be 45º and 60 º respectively. Find the height of the pole , and the tower stand in the same line.
tower
pole
50m
60°
45 °
EB=CD & EC =BD Let Height of the pole CD =h m
Take AEC & ABD to solve for “h”
50m
B
C
D60°
45 °
A
E
h
There is a small island in the middle of a 100m wide river and a tall tree stands on the island. P and Q are points directly opposite each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively30 º and 45º,find the height of the tree.
100 mP Q30° 45°
Tree height =h m (say)h/PR =tan30° =1/33 h =PR3 h =100-RQ………..
(1)h/RQ =tan45° =1h =RQ………..(2)
There is a small island in the middle of a 100m wide river and a tall tree stands on the island. P and Q are points directly opposite each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively30 º and 45º,find the height of the tree.
tree
P Q30 45
R
From (1) & (2)
3 h =100-h
h(3 -1) =100
H =100/(3 -1)
= 100 (3 +1) /(3 -1) /(3 +1)
100x2.732/2
27.32/2
13.66 m
Height of the tree =13.7 m
A
θθ
фф
h’
h
W P
From a window ( h meters high above the ground) of a house in a street ,the angle of elevation and depression of the top and the foot of another house on opposite side of the street are θ and ф respectively. show that the height of the opposite house ish(1+tan θ cot ф ).
street
house
Let W be the window and AB be the house on the opposite side. then WP is the width of the street. In ΔBPW, tan ф =PB/WP
h/WP = tan ф or WP = h cot ф In Δ AWP, tan θ = AP/WP or h’/WP = tan θ or
h’ = WP tan θ h’ = h cot ф tan θ Height of the house = h + h’ =h+ h tan θ cot ф =h(1+ tan θ cot ф )