Precalculus help?

Question

After rezoning the austin area for the new maple ridge high school, the school board mapped out the new
school district zones. If the straight-line distance from SPHS to RRHS is 2.5mi and from RRHS to MRHS is 3.2miles, and
from MRHS to SPHS is 3.3miles, what is the area of the MRHS zone as it is mapped out by the triangle formed by
the three high schools? If the district profile shows that roughly 4300 students live in each square mile in the district,
what should the enrollment at MRHS be estimated at?

 

Thank you

Answer 1

I have outlined and summarized diagramming in the Video above. Please have a look there first. I am writing out the math details here to shorten the video. .The detailed algebra steps are shown below. Try the Algebra solution after observing the video.

sin (Θ) = 3.2/3.3* sin(Φ)

cos(Θ) = (2.5/3.2) - (3.2/3.3)*cos(Φ)

Identity sin²(Θ) + cos²(Θ) =1 and sin²(Φ) + cos²(Φ) = 1

sin²(Θ) = (3.2/3.3)²sin²(Φ)

cos²(Θ) = (2.5/3.3)² - (3.2/3.3)²*cos²(Φ)

sin²(Θ) + cos²(Θ) = .(0.94)*sin²(Φ) + (0.94)*cos²(Φ) + 0.571.- (2 * 0.969 * 0.751) * cos(Φ)

1 = 0.94 + 0.571 - 1.44 cos(Φ)

- (1-0.941-0.571) =.-1.44*cos(Φ)

cos(Φ) = 0.357

Φ = cos^-1(0.357)

Φ = 69.1

Then from substitution into the sine relationship,

Sin(Θ) = (3.2/3.3) *sin(69.1)

sin(Θ) = 0.94* sin( 69.1) = 0.878

Θ = 61.4

The projection, z, for the point to the perpendicular ito the base z=3.3*sin(61.4) = 2.81 mi

So area on the left side of the triangle is A_L = 2.81 mi * cos (61.4)* /2 = 1.38/2 mi²

A_L = 0.69 mi²

Thr Area of the right side is the area of the rectangle divided by 2. A_R = 2.81 mi * cos(69.1) /2 = 0.501 mi²

The total area inside of the triangle path between the schools = A_R + A_L = 0.69 mi² +0.5 mi² = 1.19 mi²

The number of students residing in this general area is N_s = 4300 students/ mi²

Estimated school students = 1,19 mi² * 4300 students/ mi² = 5117 students.