An observer(o) spots a plane(p) taking off from a local airport and flying at a 29 degree angle horizontal to her line of sight and located directly above a tower(t). The observer also notices a bird circling directly above her. If the distance from the plane(p) to the tower(t) is 6,000ft., how far is the bird(b) from the plane(p).

Answer:Based on the diagram, we can say that the answer is 10824 feet. We know this when we apply the formula BP=OT because it is a rectangle. so we need to find OT using trigonometry using the formula tan=OPP/ADJ like this

tan 29° = 6000 / x

x = 6000 / tan 29° = 10824

Step-by-step explanation:

I Hope this helps

Answer Link

windyyork windyyork · Answer 2 · 2019-07-04T09:36:59+02:00

windyyork windyyork

04-07-2019

Answer: Hence, the bird is 10824 feet away from the plane.

Step-by-step explanation:

Since we have given that

Angle of elevation = 29°

Length of plane to the tower = 6000 feet

So, it forms a right angled triangle.

So, we will use "Tangent":

Consider, ΔABC,

[tex]\tan 29^\circ=\dfrac{AB}{CB}\\\\0.55=\dfrac{6000}{BC}\\\\BC=\dfrac{6000}{0.554}\\\\BC=10824\ ft[/tex]

Hence, the bird is 10824 feet away from the plane.

Answer Link