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Solution of right triangle problem in junior high school mathematics
Answer: over d, DF is perpendicular to EC.

By s (ABC) =18 =1/2ac * be =1/2 * AC * BC * sinc.

s(DEC)= 2 = 1/2 * EC * DF = 1/2 * EC * DC * sinC

Divide the two formulas to get: (AC/DC)*(BC/EC)=9.

Namely: (1/cosc) 2 = 9.

Germany: COSC = 1/3

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