Question

GIVING 100 POINTS!!! An observer (O) is located 900 feet from a building (B). The observer notices a helicopter (H) flying at a 49° angle of elevation from his line of sight. How high is the helicopter flying over the building? You must show all work and calculations to receive full credit.

Answer 1

Answer:

1035.33 feet

Step-by-step explanation:

OBH is a triangle and the sum of the angles of a triangle = 180

∠OBH = 90°

∠OHB = 180 -(49 + 90) = 41°

The law of sines that the ratio of each side of a triangle to the sin of the angle opposite it is the same

Thus

A/sinα  = B/sinβ = C/sinγ

where A, B and C are the three sides of the triangle and α, β, γ corresponding respectively to the angles opposite them.

Using the law of sines  

$\frac{OB}{sin 41} = \frac{BH}{sin49}$

Therefore BH which is the height the helicopter is flying at is

$BH = OB \frac{sin49}{sin41}$

≈   1035.33 feet

Answer 2

$\boxed{\sf tan \Theta=\dfrac{Perpendicular}{Base}}$

So

• tan49=H/900
• H=900tan49
• 1035.3ft