) An observer (O) is located 500 feet from a school (S). The observer notices a bird (B) flying at a 39° angle of elevation from his line of sight. How high is the bird flying over the school? You must show all work and calculations to receive full credit. (I would really appreciate it, thanks!)

 404.89 feet  

Step-by-step explanation:

Hi, since the situation forms a right triangle (see attachment) we have to apply the next trigonometric function.  

Tan α = opposite side / adjacent side

Where α is the angle of elevation of the bird (B), the opposite side (x) is the length of the bird above the ground, and the adjacent side is the distance between the school (S) and the observer (O).

Replacing with the values given:  

Tan 39 = x/ 50 0

Solving for x  

tan39 (500) =x  

x= 404.89 feet  

Feel free to ask for more if needed or if you did not understand something.  

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

jeff jeff

step-by-step explanation: