L7 : Examples-6 - Application of Trigonometry , Maths, Class 10 Video Lecture

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applications of trigonometry part 7 is
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please make sure that you have watched
part 1 to part 6
we'll take an example a straight highway
leads to the front of the tower a man
standing in the top of the tower
observes a car at an angle of depression
for degree which is approaching the
front of the power to the uniform speed
six seconds later the angle of
dispersion of the car is 2 degree you
have to find the time taken by the car
to reach of the tower so let's draw this
power this is a tower man is standing on
the top of the tower so it must have a
man standing at the door tower this is a
car and this guy observes the car the
angle of discussion is 30 DB not that
this this prophecy underlined this angle
is 30 degree
now a car is approaching towards the
tower with a uniform speed this car will
move from here to this star so let's go
the star move from this point to this
point so this is T is equal to zero and
this is T is equal to 6 seconds the car
note from here to here now the angle of
dispersion became 60 so this young
became 60
the question is at what time is you can
decay what time or how much time in the
car will take to reach this point the
top put over top so let's first draw the
triangle this is my time
so once I have my triangle if we name is
triangle II we see we see this point is
s this D so this this angle is 60 so
Senden loves me come
Steve this angle is 30 this angle also
begun okay we have told that on this
point to this point into six seconds so
we are defined how much time it takes
from this thing this to find that
firstly to find the distance between BH
correct only then we can find the
distance between B and C this is Rishi
there only we can tell that from this
point recently took this per second so
from this point this point is correct so
what we can do now so let's assume the
height of the tower means and now let's
find C D and BC in compromise what do I
do with leg first this triangle triangle
a B C in triangle ABC same thing we'll
do have sees the frame of reference
angle so this is height this is
perpendicular this is base so drilling
is perpendicular and base now so we'll
use this funded by 3 pressure on array
array gold T and V P and B stands away
looking for tank she will say x 60
degree printing P by V here P is nothing
but Navy and base is nothing but VC this
is nothing but X by VC and times X is
nothing but lutely so we'll get VC is
nothing but X by this is the value of BC
that is X 5 all right now same thing in
trying a bigger triangle regular V D
which you don't see it is a ve now so we
have 30 degree here same thing P and V
so we will stand here so you say 1020
degree if we could do P by V so in this
case perpendicular is a B but the base
isn't it of VC to VD correct
and that is something but a B is X by D
D D D be this one and time that is
nothing but 1 by root 3 so we will get
VD is something but XQ cut so we got VD
s X mu T we have got D steam equation is
the equation month on this is the
equation the distance travelled by the
car is d c so DC is nothing but this is
nothing but BD minus VZ that is X 3
minus X by root 3 so we saw that this
becomes X into root 3 minus 1 by root 3
solve this this becomes X into root 3
into root 3 is 3 minus 1 by 3 that is 2
X value G this is the distance traveled
by car
C so to travel so as a travel or we can
find the speed of a car see this since
tab it is you expiratory time taken is
how much 6 seconds speed is equal to
distance by time that is X by root 3 by
6 this is meter per second in this video
the car speed of about is 2 X by 3 D 1
by 6 meter per second now here the speed
of the car you have to find how much
time it will take to travel BC right so
distance is going to speak into time for
time is equal to distance by time is
equal to distance by speed so you have
to find the time it takes strong to
reach from C to B so here the distance
is X by moving
and the speed is this much too weights
dream six exits cancel huge utensil UN
six become this becomes three seconds
it take three seconds for the car to
reach the foot of that simple what we
had done we have found this this since I
won't buy the car we have a way we were
told that a time it took the six events
we found the speed of a cup we have the
speed of a car we know the distance from
this current position to the tip of the
tower is acutely we know distance we
have speed before the time intake that
is few seconds let's take a more last
problem the angle of elevation of the
top of the tower from two points at a
distance four and nine meters from the
base of the tower in the same high time
it are complemented to the height where
tower is 6 meter this is the power they
are two points for n nine meters from
the base the tower
and this the angle of elevation actually
this angle and this angle if you see if
this angle is Theta this angle is 90
minus the question says the sum of the
angle is 90 degree and we to prove that
the height of the tower is 6 meter let
us draw the final bus this is my
triangle let the height VX include the
xv6 meter will same let height of tower
leaves now the distance here is this is
4 and this is my national big forest
night seriously B C so we are told that
in first case if you see the tan theta
is equal to X by Oh
because in triangle ABC the sites we are
dealing with this this will be
perpendicular because this is the theta
remembrance perpendicular in base
so under puzzle Prashad rahuri born
perpendicular invades
tan theta so the same tan theta is equal
to P by V and here P is a V and V is BC
that is X by 4 so I got time T does not
wait for the next equation I get in time
then a B the same thing we're dealing
with height X and this one base I'll say
tan ah she type angle a is 90 minus
theta tan of 90 minus theta is equal to
e by me here P is VB and B is VD that is
X by 9 and time 90 minus theta is cot
hiccupping we have Nautilus should I
have to be patient
question one and pitch into equation
once a times it is going to expect over
the equation du is odd theta is equal to
expand I will multiply these two will
get time to tie into cot theta tan theta
is X by 4 in particles will be
explaining 2010 particles will do 1 we
know is equal to X square by 56 or we
get X is equal to root 36 that is plus
minus 6 since we are talking about
distance we can equal negative part X is
equal to positive 6 to the height of the
turbulence since correct / Pio down we
found tan theta we found cot theta and
then need for use this relation time
here in the cottage circle 2 1 and we
got the hence thank you
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