Heights
and Distance
Trigonometric ratios
find many uses. Among them is their use in determining HEIGHTS and DISTANCE of
any object which has certain ANGLES and the measure of a related length.
Terminology used in this section
1– Angle of
Elevation:Let O be the eye of an observer and in the figure, object A is above
the horizontal line O B . Angle θ is called the Angle of Elevation.
2– Angle of Depression: Again, O be the eye of an
observer and object A is below the horizontal line O B .Angle θ is called Angle
of depression .
Now it is
clear from the adjoining figure (3) that the Angle
of elevation and Angle of depression
always be equal.
You know the values of standard angles e.g.,0°,30°,45°,60°,90°
What about the values like 40°,74°,35° 24’,42°41’……so
n.
Well there are some simple steps to know the values
of nonstandard angles.
How to read table for nonstandard angles →
Reading the value of sin 33° ,Sin 33°18’ ,33°27'.
To locate the value of 33°
Look at the extreme left column . you find the value
of 33°0’.value is .5446..
So,
Sin 33°= .5446
To locate the value of 33°18’
First you move to the right in the row of 33° till
you reach the column of 18’.
you find
5490 ,that is .5490. So Sin 33°18’ = 0.5490
To locate the value of Sin 33°27'
First move to the right in the row of 33 till you
reach the column of 24’.
you find 5505, that is .5505. So, Sin 33°27’ =
0.5505 + mean difference for 3’
= 0.5505 + 7 = 0.5512
This section illustrate the process of solving problems rel
ated to HEIGHT and DISTANCE in very easy way…
1 – A ladder rests against a vertical wall such that
the top of the ladder
reaches the top of the wall .The ladder is inclined at 60° with the
ground ,and the bottom of the ladder is 1.5 m away from the foot
of the wall. FIND (A) the length of the ladder, and (B) the
height of the wall.
SOLUTION → First draw a diagram according to question
(A) A O =length of ladder
A B =height of the wall, ⇒ B
O =1.5 m
Cos 60° = base / hypotanuse =BO/AO = 1.5 / AO ⇨ A O = 1.5 ⤬2
= 3 m
(B)
tan 60° = Altitude / base = AB / BO ⇨ √3 = AB / 1.5 ⇨ AB = √3 ⤬1.5
=1.732⤬1.5
=2.6 m (∵√3 =1.732)
(2) - vertical pillar and a tower 120 m
high are in the same horizontal plane .from the top of the tower , the angle of
depression of the top and the foot of the pillar are 45° and 60°. Find
(A) The
distance between the pillar and the tower. (B)The height of the pillar .
Solution→
Draw a diagram according
to question
AB = length of pillar
DC =length of tower
BD = distance between pillar and tower
∵ Angle of elevation = angle of depression
⊾CBD = 60°
, ⊾CAH = 45°
Tan 60°= Altitude / base = DC / BD ⇨ √ 3 = 120 /BD
BD
= 120 / √ 3 = 120 / 1.732 = 69.16
m (∵ √3 = 1.732 )
Again→
tan 45°= Altitude / base = CH / AH ⇨ 1 = DC - HD / BD ( ∵AH = BD )
1
= 120 - HD /69.16 ⇨ 69.16 = 120 –HD
HD = 120 – 69.16 = 50.84 m
AB
= HD = 50.84 m
NOTE: For any queries or suggestion please post a comment :) !
******************
No comments:
Post a Comment