WAEC - Mathematics (2005 - No. 10)

In the diagram, PR is a diameter, ∠PRQ = (3x-8)^o and ∠RPQ = (2y-7)^o. s x in terms of y

\(x=\frac{75-2y}{3}\)

\(x=\frac{105-3y}{2}\)

\(x=\frac{105-2y}{3}\)

\(x=\frac{75-3y}{2}\)

Explanation

180 = ∠RPQ + ∠PRQ + ∠PQR Since PQR = 90 (theorem: angle in a semi circle)
180 = ∠RPQ + ∠PRQ + 90 => 180^o = (3x-8)^o+(2y-7)^o+90^o; 90+8+7 = 3x+2y =>\(\frac{105-2y}{3}=x\)

WAEC - Mathematics (2005 - No. 10)

Explanation

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