Let $ABC$ be a triangle with $\angle A = 45^{\circ}$. Let $P$ be a point on the side $BC$ with $PB = 3$ and $PC = 5$. If $'O'$ is the circumcentre of the triangle $ABC$ then the length $OP$ is equal to

Using the property of the circumcentre, the length OP is given by the formula OP square equals R square minus the product of the two segments PB and PC, where R is the circumradius. The circumradius R is calculated as a divided by 2 sine A, yielding R equals BC divided by 2 sine 45 degrees, which is 8 divided by square root 2 or 4 square root 2. Substituting R square equals 32 and the product 3 times 5 equals 15 into the formula gives OP square equals 32 minus 15, so OP equals square root 17.

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