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http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_w05_qp_4.pdf

 

Question no 6

 Nov 19, 2016
 #1
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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

 Nov 19, 2016

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