Answer:
see below
B.
Step-by-step explanation:
The correct answer for the question that is being presented above is this one: "C) ∠B = ∠B' = 33° and ∠C = ∠C' = 80° ."
Answer:
Step-by-step explanation:
Given
velocity of airplane is 360 mi/hr
inclination of climbing ![30^{\circ}](https://tex.z-dn.net/?f=30%5E%7B%5Ccirc%7D)
Plane velocity can be divided in to x & y component
Velocity in x component is ![360\cos30](https://tex.z-dn.net/?f=360%5Ccos30%20)
Velocity in y component is ![360\sin30](https://tex.z-dn.net/?f=360%5Csin30%20)
Thus distance traveled in x axis is ![=360\cos30\times \frac{1}{60}](https://tex.z-dn.net/?f=%3D360%5Ccos30%5Ctimes%20%5Cfrac%7B1%7D%7B60%7D)
=60cos30=51.96 miles
Distance travel in upward direction i.e. in Y axis
![=360\sin30\times \frac{1}{60}=30 miles](https://tex.z-dn.net/?f=%3D360%5Csin30%5Ctimes%20%5Cfrac%7B1%7D%7B60%7D%3D30%20miles%20)
so total altitude gain by plane is 30+4=34 miles
Horizontal distance traveled is 51.96
inclination w.r.t to point p
![tan\theta =\frac{34}{51.96}=0.654](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%5Cfrac%7B34%7D%7B51.96%7D%3D0.654)
![\theta =33.18^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D33.18%5E%7B%5Ccirc%7D)
Initial inclination is ![90^{\circ}](https://tex.z-dn.net/?f=90%5E%7B%5Ccirc%7D)
therefore rate at which its distance is changing is ![360\cos30=311.76 miles /hour](https://tex.z-dn.net/?f=360%5Ccos30%3D311.76%20miles%20%2Fhour)
Make sure to not use the useless numbers on the slant. Also remember to divide by two at the end of calculating.
I’m not sure i’m pretty sure it’s 28