It is given the trapezoid OABC,where OA=a,OC=c and CB=2a .Point M is the middle of AB and point N dhe middle of BC.Find OX

Question

BC=2a pls i need an answer as fast as possible

Answer 1

^{It can be shown that the length OX also depends on the angle θ between OC and OA, which has not been provided. However, if OC and OA are vectors, as it would appear from the diagram, then the dot product of the vectors is accosθ so cosθ=a.c/ac, where the vectors are shown in bold type and their magnitudes in normal type. It will then be possible to show OX either as a scalar length or as a vector. The question doesn’t specify which is required. Before continuing with a full solution I need clarification please.}

^{The following approach uses geometry and trigonometry.}

It’s helpful to draw in CA and AN.

Let θ=CÔA, then ∠ANB=∠OCB=180°-θ (interior and corresponding angles of a parallelogram). AN=OC=c, CN=NB=OA=a, CB=CN+NB=2a. cosθ=-cos(180-θ).

AB²=AN²+NB²-2AN.NBcosANB=c²+a²+2cacosθ.

OB²=OC²+CB²-2OC.CBcosOĈB=c²+4a²+4cacosθ (Cosine Rule),

We know  OA=a, AB, OB (from above), so we can find ∠OBA, which we’ll call λ:

OA²=OB²+AB²-2OB.ABcosλ,

a²=4a²+c²+4accosθ+a²+c²+2accosθ-2√((4a²+c²+4accosθ)(a²+c²+2accosθ))cosλ,

cosλ=(2a²+c²+3accosθ)/√((a²+c²+2accosθ)(4a²+c²+4accosθ)).

We know NM=½CA, NB=a, MB=½AB, so we can find ∠NMB, which we’ll call ɸ.

NM=½CA=½√(a²+c²-2accosθ) (NM is a semi-diagonal in the parallelogram with 3 vertices A, N, B corresponding to congruent vertices O, C, N in parallelogram OANC),

MB=½AB=½√(a²+c²+2accosθ),

NB²=NM²+MB²-2NM.MBcosɸ,

a²=¼(a²+c²-2accosθ)+¼(a²+c²+2accosθ)-½√((a²+c²-2accosθ)(a²+c²+2accosθ))cosɸ,

a²=c²-√((a²+c²)²-4a²c²cos²θ)cosɸ, 

cosɸ=(c²-a²)/√((a²+c²)²-4a²c²cos²θ).

XB/sinɸ=MB/sin(180-(λ+ɸ))=MB/sin(λ+ɸ) (Sine Rule)

XB=MBsinɸ/sin(λ+ɸ).

OX=OB-XB. Since we know OB and XB, we can calculate OX.

Comments

Editing was a nightmare for this especially since I needed to produce a concise solution with minimum text, so I’ve skipped a couple of steps in the calculations to save space and the final equation for OX would be too complicated and difficult to write out in full, but by substitution you can find OX from its components. This is the trig/geometry solution. For a vector solution you will need to make appropriate substitutions.