PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
Here we have a cyclic quadrilateral PQRS with PR being a diameter of the circle. Let the centre of this circle be ‘O’.
We are given that `angleQPR` and `angleSPR = 72°` . This is shown in fig (2).
So we see that,
\[\angle QPS = \angle QPR + \angle RPS\]
\[ = 67°+ 72° \]
\[ = 139°\]
In a cyclic quadrilateral it is known that the opposite angles are supplementary.
`angleQPS + angleQRS = 180°`
`angleQRS = 180° - angleQPS`
`= 180° - 139°`
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