An airplane is sighted at the same time by two ground observers who are 3 miles apart and in front of the airplane. They report the angles of elevation as 13 and 24 . How high is the airplane?
2007-11-23 01:14:33 · 5 answers · asked by labelapark 6 in Science & Mathematics ➔ Mathematics
Draw triangle ABC right angled at B :-
AB is vertical (height h miles)
BC is horizontal (x miles)
CD is horizontal (3 miles)
DB = x + 3 miles
tan 24° = h / x
x tan 24° = h
tan 13° = h / (x + 3)
(x + 3) tan 13° = h
x tan 24° = (x + 3) tan 13°
x ( tan 24° - tan 13° ) = 3 tan 13°
x = 3.23 miles
tan 13° = h / 6.23
h = 6.23 tan 13°
h = 1.44 miles
2007-11-28 02:40:31 · answer #1 · answered by Como 7 · 3⤊ 1⤋
The angle of elevation is 13 and 24 respectively. The angle of elevation of the person B further away from the airplane is 13 and the elevation is for the other person A is 24.Let the distance of the A from the perpendicular of the airplane be x then the point where the perpendiculor from the airplane meets the ground so distance of B from the point would be x+3
Tan13=Perpendicular/x+3
Tan24=Perpendicular/x
So Tan13/Tan24=x/x+3=0.2308/0.4452
So x/x+3=0.5184
x=(x+3)(0.5184)
x=(0.5184) x+3(0.5184)
x-(0.5184)x=1.5552
(0.4816)x=1.5552
x=1.5552/0.4816
x=3.2288miles
So A is standing 3.2288miles away from the point where the perpendicular meets the ground.
So the tan24=Perpendiculor/3.2288=0.4452
So
Perpenculor=(0.4452)*(3.2288)
Perpendiculor=1.4374 miles
So the plane is 1.4374 miles above the ground.
2007-11-23 10:01:06 · answer #2 · answered by Vaibhav Dwivedi 4 · 1⤊ 0⤋
vlee 1225 gave you a good way to solve the problem, however if there are 3 miles between the viewers and we let x = the distance from the viewer with 24 degrees elevation to the point below the airplane 3-x = the other distance.
The equations would be changed to:
tan 13 = h /(3-x)
and
tan 24 = h/x
solving for x we get:
x = 1.024
so tan 24 = h/1.024
the height is only :
.456 miles
2007-11-23 10:02:25 · answer #3 · answered by Peter m 5 · 0⤊ 1⤋
Let h be the height the plane is above graound.
Let x be the horizontal distance the first person is from the plane
then
tan (24) = h / x (all in miles)
the second person is at a horizontal dist of x + 3 miles away
tan (13) = h / (x + 3)
tan(24) = 0.445 = h/x
tan(13) = 0.231 = h/(x+3)
so
h = 0.445x = 0.231(x+3)
(0.445 - 0.231)x = 0.231(3) = 0.693
0.214x = 0.693
x = 0.693/0.241 = 2.88
h = 0.445x = 0.445(2.88) = 1.28
Ans the plane is 1.28 miles high
2007-11-23 09:41:11 · answer #4 · answered by vlee1225 6 · 1⤊ 1⤋
firstly, find the distance between the observer sitin at angle of 13 to the rocket by the sine rule:
a/SinA = b/ SinB
3/Sin143 = b/Sin 24
b = 1.7miles
now use the sin rule where the opposite side is the height and the hypothenuse is the 1.7miles distance between viewer observing at 13 to the rocket
sin13 = o/1.7
0 = 1.7 * Sin13
distance = 0.38miles
hope you get it and give me 10 points!!
2007-11-23 09:46:25 · answer #5 · answered by **PiNoY YFC** 7 · 1⤊ 1⤋