ABC is a triangle in which: m\(\angle A = 90^\circ\), AC = 15 cm and AB = 20 cm. Prove that:
\(\cos C \cos B - \sin C \sin B = \text{zero}\)
First, find BC using Pythagoras theorem:
\[ BC = \sqrt{AB^2 + AC^2} = 25 \text{ cm} \]
Now evaluate the expression:
\[ \cos C \cos B - \sin C \sin B \]
\[ = \left(\frac{15}{25}\right) \left(\frac{20}{25}\right) - \left(\frac{20}{25}\right) \left(\frac{15}{25}\right) \]
\[ = 0 \]
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