ABCD is a trapezoid in which:
\(\overline{AD} \parallel \overline{BC}\), \(m(\angle B) = 90^\circ\), \(AB = 3 \, \text{cm}\),
\(AD = 6 \, \text{cm}\), and \(BC = 10 \, \text{cm}\).
Prove that:
\[ \cos(\angle DCB) - \tan(\angle ACB) = \frac{1}{2} \]
\(\overline{AD} \parallel \overline{BC}\), \(m(\angle B) = 90^\circ\), \(AB = 3 \, \text{cm}\),
\(AD = 6 \, \text{cm}\), and \(BC = 10 \, \text{cm}\).
Prove that:
\[ \cos(\angle DCB) - \tan(\angle ACB) = \frac{1}{2} \]
MR/Nasef.N
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