Consider triangle ABD, let y be angle ABD, by Cosine rule, cosy=[3^2+x^2-(x+1)^2]/[2(3)(x)] = (4-x)/(3x) Consider triangle BCD, cos(pi/2 – y) = [4^2+x^2-(x+2)^2]/[2(4)(x)], i.e. siny=(3-x)/(2x) Hence, 1 = [(4-x)/(3x)]^2 + [(3-x)/(2x)]^2 36x^2 = 4(x^2 – 8x + 16) + 9(x^2-6x+9) 23x^2+86x-145=0 (23x-29)(x+5)=0