How high above the ground is the balloon?

Question

a balloonist is directly above a straight road 1.5mi long that joins two villages. She finds that the town closer to her is at an of depression of 35 degrees and the farther town is at an angle of depression of 31 degrees.

Answer 1

h=xtan 35=(1.5-x)tan 31

x(tan 35+tan 31)=1.5tan 31

x=1.5tan 31/(tan 35+tan 31)

h=1.5*tan 31*tan 35/(tan 31+tan 35)

Answer 2

let x be the town closer to her
and 1.5-x = to the distance from the vertical height of balloon to another town

we all know that triangles has an angle equal to 180, so,

C=180-(31+35)
C=114 deg

using sine law,

1.5 / sin(114) = s / sin(35)
s=0.941786364

where s is the inclined distance of the 2nd town to the balloon (or we could say that it is hypotenuse)

cutting the triangle into 2, and using the triangle at the side of the second town, we obtain an angle of
S=90 deg, A=31 deg, B=59 deg

using again the sine law,

a/sin(A)=s/sin(S)
a/sin(31)=0.941786364/sin(90)

a=0.485055836 mi is the answer

in addition,

(1.5-x)^2=(0.941786364)^2 - (0.485055836)^2

x=0.848317609 is the distance of the town closer to her, to vertical height of the balloon

Answer 3

Let x and y the distances from the point P dropped directly from the balloon to the ground. x + y = 1.5 miles

The angles are 35 and 31 degrees respectively in the two right angled triangles.

Height / x = tan angle

h / x = tan 35 = 0.7

h/ y = tan 31 = 0.6

0.7 x = 0.6 y

0.7 x = 0.6 (1.5 - x)

0.7x = 0.9 - 0.6x

1.3x = 0.9

x = 0.9/1.3 = 0.692

y = 1.5 - 0.692 = 0.808

Height h = 0.7 x = 0.7 x 0.692 = 0.485 miles

Answer 4

Draw triangle ABC
A = 114°
B = 31°
C = 35°
BC = 1.5

1.5 /sin114° = AC / sin 31°
AC = 1.5 sin 31° / sin 114°
AC = 0.846
let AD = y = vertical height (D lies on BC)
sin 35° = y / 0.846
y = 0.846 sin 35°
y = 0.485
Height = 0.485 miles = 2562 ft