Can sombody solve this math problem?

Question

A plane flies a straight course. On the ground directly below
the flight path, observers 2 miles apart spot the plane at the
same time. The plane's angle of elevation is 46 degrees from
one observation point and 71 degrees from the other.
How high is the plane?

Answer 1

The angle opposite the two-mile side of the triangle is 63 degrees

Noe refer to the first answer above, and your text. Don't be so lazy..

Answer 2

EZ!!!
MUST USE TRIG+GEOMETRY
the 2 observers and plane create a triangle with angles 46 deg and 71 deg....now find out the 3rd angle
SUM of angles in triangle=180
180-46-71=63

APPLY SINE LAW:
a/sinA = b/sinB = c/sinC

2 miles / sin 63 = b / sin 71
b=2.122 miles
b is the diagonal distance from the observer of 71 deg to plane.

now we have a right angle triangle
the hypotneuse is b=2.122 miles and an angle of that triangle is 71. The vertical distance from the ground to the plane can be solved by this:

2.122miles(sin71)=
2.0067 miles

*draw a diagram to better understand the question*
*draw a vertical line from the plane down to a point on the ground

Answer 3

if both observers are viewing the plane from the same direction, then
height(1/tan46 + 1/tan71)=2

if the distance between the observers is y + z =2

then a triangle with base =y and height
has tan phi =height/y so y=height/tan phi
the other triangle has tan theta = height/z
so z=height/tan theta

since you know the angles, just plug it all in

j

Answer 4

The plane was 3,264 feet in the air.

Answer 5

Ain't nobody about to do your home work, do it yourself lazy.

Answer 6

i'm sure someone can; i draw a diagram or something to help you solve.

Answer 7

I am sure someone can but it isnt going to be me.

Answer 8

it is too difficult but hard to be solved