💥 Q 06: Prep 3 Geometry Question Bank T1

ABCD is an isosceles trapezoid in which:
AD // BC, AD = 4 cm, AB = 5 cm, and BC = 12 cm.
Prove that:
\[ \frac{5 \tan B \cos C}{\sin^2 C + \cos^2 B} = 3 \]

\[ \frac{5 \tan B \cos C}{\sin^2 C + \cos^2 B} = \frac{5 \times \frac{3}{4} \times \frac{4}{5}}{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = 3 \]

MR/Nasef.N