A weather balloon is directly between two tracking stations A and B. The angle of elevation from tracking station A is 55°. The angle of elevation from tracking B is 68°.

Question

If the tracking stations are 20 km apart, what is the height of the weather balloon? Include a diagram

Answer 1

Draw a triangle with Pt. A and Pt. B the corners of the base. Draw a line from the 3rd point, Pt. C, down to the base line at Pt D and perpendicular to it. Now we have 2 triangles with a common side, the height of the balloon, H. On the one side you have the distance from Pt. A to Pt. D we will call x and an angle of elevation of 55⁰. On the other side we have the distance from Pt. D to Pt. B of 20 - x and an angle of elevation of 68⁰. Because the height, H, is common to both sides we can see that

tan55⁰ = H/x--->H = xtan55⁰

tan68⁰ = H/(20 - x)--->H = (20 - x)tan68⁰

xtan55⁰ = 20tan68⁰ - xtan68⁰

xtan55⁰ + xtan68⁰ = 20tan68⁰

x(tan55⁰ + tan68⁰) = 20tan68⁰

x = 20tan68⁰/(tan55⁰ + tan68⁰)

x = 12.6822338

Putting this value into either equation and solving for H we get H = 18.1121 or 18.11km.

Hope this helps!

Answer 2

Draw and label a diagram

Drop of perpendicular from the balloon to the ground.

Label the point on the ground C

h represents the height of the balloon.

AC = 20 - CB

tan 55° = h / AC

AC tan 55° = h

(20 - CB) tan 55° = h

20 tan 55° - CB tan 55° = h

20 tan 55° - h = CB tan 55°

[20 tan 55° - h] / tan 55° = CB

tan 68° = h / CB

CB tan 68° = h

CB = h / tan 68°

[20 tan 55° - h] / tan 55° = h / tan 68°

Can you solve for h and answer?