P2 maths

I ihave been doing a past paper and i have done all the questions apart from this one any help would be greatly apreciated.

The diagram shows the rectangular cross-section PQRS of a letter rack. A rectangular envelope ABCD rests in the vertical plane PQRS inside the letter rack. QR is horizontal. QR= 30cm AD = 27 cm CD = 18cm. the bottom edge, bc, of the envelope, makes an angle x with the base QR of teh rack.

(a) Prove that 9cos x + 6sin x =10.

(b) Express 9cosx + 6sinx in the form Rcos(x- alpha), where R>0 and 0<alpha<90 giving the values of R and alpha to 2 decimal places.

(c) Hence or otherwise find x giving your answer to the nearest thenth of a degree.

Thanks in advance

Ooooh, I remember doing that question! Can't remember if it was in my real exam or a practice paper, but I remember doing it. Unfortunately I have no maths skills left whatsoever so I can't help you with it I'm afraid :redface:

Reply 2

JamesF

1

Part i)
Try to find qr. By symmetry we have qb = ds.
So using some trig, find br and ds. Add these and you get qr which is 30.

Reply 3

Ralfskini

10

droid

I ihave been doing a past paper and i have done all the questions apart from this one any help would be greatly apreciated.

The diagram shows the rectangular cross-section PQRS of a letter rack. A rectangular envelope ABCD rests in the vertical plane PQRS inside the letter rack. QR is horizontal. QR= 30cm AD = 27 cm CD = 18cm. the bottom edge, bc, of the envelope, makes an angle x with the base QR of teh rack.

(a) Prove that 9cos x + 6sin x =10.

(b) Express 9cosx + 6sinx in the form Rcos(x- alpha), where R>0 and 0<alpha<90 giving the values of R and alpha to 2 decimal places.

(c) Hence or otherwise find x giving your answer to the nearest thenth of a degree.

Thanks in advance

a) qb+br=30, br=27cosx and qb=18sinx
so 27cosx+18sinx=30, 9cosx+6sinx=10.

part b) and c) are much easier.

P2 maths

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