Question

Let ABC be a triangle with $\angle A={45}^{\circ }$. Let P be a point on side BC with $PB=3$ and $PC=5$.

If $O$ is circumcenter of triangle ABC, then length $OP$ is?

• A. $\sqrt{18}$
• B. $\sqrt{17}$
• C. $\sqrt{19}$
• D. $\sqrt{15}$

Answer 1

The correct option is B $\sqrt{17}$

In $△BOC$
$\angle BOC=2A={90}^{\circ }$
$\Rightarrow \angle OBC=\angle OCB={45}^{\circ }$
In $△OBP$
from sine rule: $\frac{OP}{\sin {45}^{\circ }}=\frac{BP}{\sin \angle BOP}$
$\Rightarrow \sin \angle BOP=\frac{3}{OP\sqrt{2}}$ ....(1)
In $△OCP$
from sine rule: $\frac{OP}{\sin {45}^{\circ }}=\frac{CP}{\sin \angle COP}$
$\Rightarrow \sin (\frac{\pi }{2}-\angle BOP)=\frac{5}{OP\sqrt{2}}$
$\Rightarrow \cos \angle BOP=\frac{5}{OP\sqrt{2}}$ ....(2)

Squaring and adding (1) and (2), we get $1=\frac{34}{2{OP}^2}$
$\Rightarrow OP=\sqrt{17}$
Ans: B