<PRE><B>Question 1186313</B>
Here&#39;s how to solve this problem:

**a. Find angle C (the angle that AC makes with BC):**

1.  **Area of BCED:** We know the area of trapezoid BCED is 50,977 sq.m.  The formula for the area of a trapezoid is (1/2) * (base1 + base2) * height.  In our case, the bases are BC and DE, and the height is the perpendicular distance between them.  However, we don&#39;t know the height yet.

2.  **Triangles ADE and ABC are similar:** Since DE is parallel to BC, triangles ADE and ABC are similar. This means their corresponding angles are equal, and the ratio of their corresponding sides is constant.

3.  **Ratio of corresponding sides:** DE/BC = 150/400 = 3/8.  Let this ratio be k.

4.  **Ratio of areas:** The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides.  Area(ADE)/Area(ABC) = k² = (3/8)² = 9/64

5.  **Area of ABC:** Let Area(ABC) = X. Then Area(ADE) = (9/64)X. Also, Area(BCED) = Area(ABC) - Area(ADE) = X - (9/64)X = (55/64)X. We know that the area of BCED is 50977. So (55/64)X = 50977, from which X = 59200.

6.  **Area of ADE:** Area(ADE) = (9/64) * 59200 = 8325

7.  **Area of ABC (using another approach):** Area(BCED) = Area(ABC) - Area(ADE). 50977 = X - (9/64)X. So, X = 59200.

8.  **Area of ABC (using sine formula):** Area(ABC) = (1/2) * BC * AB * sin(B) = (1/2) * BC * AC * sin(C). We know BC = 400 and B = 50°.  We can express AB and AC in terms of BC using the Law of Sines:
    AB/sin(C) = BC/sin(A) and AC/sin(B) = BC/sin(A)

9. **Angle A:** Since the sum of the angles in a triangle is 180, we have A + B + C = 180, or A + 50 + C = 180. So A = 130 - C.

10. **Area of ABC (substituting):** 59200 = (1/2) * 400 * AC * sin(C). So AC = 296/sin(C)

11. **Law of Sines:** AC/sin(50) = 400/sin(130-C). 296/sin(C) * sin(50) = 400/sin(130-C). 0.5662/sin(C) = 400/(sin130cosC - cos130sinC). 0.5662(sin130cosC - cos130sinC) = 400sinC. 0.5662(0.766cosC + 0.6428sinC) = 400sinC. 0.4349cosC + 0.364sinC = 400sinC. 0.4349cosC = 399.636sinC. tanC = 0.4349/399.636. C = 0.0623 degrees. It seems that this is wrong.

12. **Let&#39;s take a different approach**: Area of ABC = (1/2) * BC * h where h is perpendicular from A to BC. 59200 = (1/2)*400*h. h = 296. h = ABsinB = ACsinC. AB/AC = sinC/sinB. AB = ACsinC/sin50. Area of ABC = (1/2)BC*ACsinC = (1/2)*400*ACsinC = 59200. 200ACsinC = 59200. ACsinC = 296. Area of ADE = (1/2)*DE*h1 = (1/2)*150*h1 = 8325. h1 = 111. h1/h = 150/400. h1 = (150/400)*h = 0.375*296 = 111.

13. **Use the given area of BCED**: Area(BCED) = (1/2)(BC+DE)h2. 50977 = (1/2)(400+150)h2. h2 = 185.37

14. h2 is the height of the trapezoid which is not the height of the triangle.

**b. Find the area of ABC:**

We have already found the area of ABC = 59200 sq.m.

**c. Find the area of ADE:**

We have already found the area of ADE = 8325 sq.m.

**a. Angle C**

Area(ABC) = (1/2) * BC * AC * sin(C)
59200 = (1/2) * 400 * AC * sin(C)
296 = AC * sin(C)

We also know that Area(ADE) = (1/2) * DE * AE * sin(A) and since triangles are similar, the angles are the same.

Since we have two unknowns (AC and C), we need another equation.

We know that the ratio of the sides is 150/400 = 3/8. So AE/AC = AD/AB = DE/BC = 3/8.
AD/AB = 3/8, so AB = (8/3)AD. Also AE/AC = 3/8, so AC = (8/3)AE.

Area(ABC) = (1/2) * BC * AB * sin(B)
59200 = (1/2) * 400 * AB * sin(50)
AB = 59200/(200sin50) = 387.44m

AD = (3/8)AB = 145.29m
AE = (3/8)AC
DE/BC = AE/AC = AD/AB = 3/8

Area(ADE) = (1/2) * AD * AE * sin(A)

AC/sin50 = 400/sinA
AC = 400sin50/sinA

59200 = (1/2)*400*ACsinC
296 = ACsinC

ACsinC = 296
AC = 296/sinC

296/sinC * sin50 = 400/sinA
sinA = (400sinCsin50)/296

A = 180 - 50 - C
sin(130-C) = (400sinCsin50)/296

sin130cosC - cos130sinC = 2.28sinC
0.766cosC + 0.6428sinC = 2.28sinC
0.766cosC = 1.6372sinC
tanC = 0.766/1.6372 = 0.467
C = 25.02 degrees.

**Final Answers:**

a. Angle C ≈ 25.02 degrees
b. Area of ABC = 59200 sq.m.
c. Area of ADE = 8325 sq.m.
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