+502 http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_w05_qp_4.pdf
Question no 6
+37155 Here is what I come up with
A 4 planes of symmetry
B 1/3 (6)(5)(3)
C.......waiting for printout
D Using Pythagorean theorem BD^2 = 6^2 + 5^2 BD=7.81 1/2 of BD =BF=3.905 cm
TAN(FBP) = 3/3.905 FBP = 37.53 degrees
E Pythag Theorem again PB^2 = 3.905^2 + 3^2 PB= 4.924 cm
C FM = 1/2 (6cm) = 3cm so TAN (PMF) = 3/3 = 1 so angle (PMF) = ARCTAN(1) = 45 degrees
