# math

posted by on .

a hot-air balloon is seen in the sky simultaneously by two observers standing a t two different points on the ground. (the 2 people are facing each other and the balloon is in between them in the air.)they are 1.75 miles apart on level ground. The angles of elevation are 33 and 37 respectively. how high above the ground is the balloon?

• math - ,

make a diagram
let the base be BC, where angleB= 37 and angle C = 33°
From the top vertex A drop an altitude to meet BC at D
let AD = h, BD = x, then DC = 1.75-x

you now have two right-angled triangles
from ABD,
tan 37 = h/x
h = xtan37

tan 33 = h/(1.75-x)
h = tan33(1.75-x)

then x tan37 = 1.75tan33 - xtan33
xtan37 - xtan33 = 1.75tan33
x(tan37-tan33) = 1.75tan33
x = 1.75tan33/(tan37-tan33)

once you have that evaluated, put that x value into
h = xtan37

you do all the button-pushing.

• math - ,

i got up to the very last step but when i plugged my x value into h=xtan37 i just got decimals, like .97131.. what is the answer supposed to be?