I think It would probably be 1
ABC ,⊿PQR,select an angle ∠ACB =θ
LET θ’s (opp,adj,hypo)=(AB,BC,AC) and (PQ,QR,PR)
Then by similarity,
sinθ=AB/AC=PQ/PR ,cosθ=BC/AC=QR/PR ,tanθ=AB/BC=PQ/QR
the corresponding ratio of (adjacent side/hypotenuse),and hence
the cosine ratio must be the same
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i think it help
The second bar representing Paris is misplaced.
The bar representing Munich is a bit larger than needed.
The bar representing Oslo is misplaced.
