Question on: JAMB Mathematics - 2010

Let A denote the area of (\bigtriangleup)PQR, then A = (\frac{1}{2} bh)

Using Sin 60^o = (\frac{h}{q})

h = q sin 60^o

So A = (\frac{1}{2}b(q \sin 60^o))

12(\sqrt{3} = \frac{1}{2} \times 8 \times q \times \frac{\sqrt{3}}{3})

12(\sqrt{3}) - 2q(\sqrt{3})

q = (\frac{12}{2} = 6)cm