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IGCSE MATHS Doubts

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princess12:
thanx
can any 1 explain me oct 2007 q 6,7
urgently fast
best of luck

Ghost Of Highbury:
@princess - the whole question?? Q6 and Q7 ?? any specific subpart??

Adzel:
o/n 2007 p4 q3 (explaination please)

Thanks in advance...

princess12:
the whole question a@di please
 ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ???

Ghost Of Highbury:
ans 6) ai) 1cm:200cm

               1cm:2m

              therefore AB = 26m/2m = 13 cm

                           BD = 30m/2m = 15 cm

                           Angle A shud be 80 degree.

aii) Measure them with ur protractor.

aiii) Bisect the angle DAB and extend it till it intersects the trapezium.

aiv) Draw the perpendicular bisector of AD.

av) check the attached diagram.

bi) Sine rule. (Sin D)/26 = (Sin 80)/30

Sin D/26 = 0.032826925

sin D = 26*0.032826925 = 0.8535

D = sin-1 0.8535 = 58.6 degree ~ 59 degree.

bii) Angle BDC = 100-58.6 = 41.4

     (BC)^2 = 182 + 302 - 2*18*30*cos 41.4 = 1224 - 810.11 = 414

     BC = = 20.3m

biii) area of triangle ADB + BDC = 0.5 * 26 *30sin'41.4' + 0.5 *18* 30sin'41.4' = 436

will answer the 7th question soon

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