# Math Paper 2 Question

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I have another question. It's attached to this post, I got (a) and (b) but I'm not entirely sure how to get ©...

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Anyone? Please? Paper 2 is tomorrow...and it'd be nice to know how to figure this out before writing the exam.

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Cosine rule: a^2 = b^2 + c^2 - 2*b*c*cosA

Sine rule: a/sinA = b/sinB = c/sinC

A:

AC^2 = AB^2 + BC^2 - 2*AB*BC*cos(AC)

= 4^2 + 5^2 - 2*5*4*cosx

= 16 + 25 - 40cosx

= 41 - 40cosx

Thus AC = sqrt(41 - 40cosx)

B:

AC/sin(AC) = BC/sin(BC)

AC/sinx = 4/sin(30)

AC/sinx = 4/0.5 = 8

AC = 8sinx

C:

8sinx = sqrt(41 - 40cosx)

64sin^2x = 41 - 40cosx

64sin^2x + 40cosx = 41

64(1 - cos^2x) + 40cosx - 41 = 0

-64cos^2x + 40cosx + 23 = 0

64cos^2x - 40cosx - 23 = 0

Cosx = (40 +/- sqrt(40^2 - 4*64*-23))/2*64

= (40 +/- 86.533)/128

= 0.9885408 or - 0.3635408

X = arccos(0.9885408) or arccos(-0.3635402) which ends up giving us x = 8.68 degrees or x = 111.37 degrees. We want the larger angle here so x = 111.37.

Now just use either of the two relations we found for AC:

AC = sqrt(41 - 40cos(111.37)) = 7.45 cm

AC = 8sin(111.37) = 7.45 cm

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Thank you Emmi!

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