Question

Let ABC be a triangle with $\angle$A = ${45}^0$. Let P be a point on the side BC with PB = 3 and PC = 5. If 'O' is the circumcenter of the triangle ABC then the length of OP is equal to -

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

Answer 1

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

In $△BOC$

$\angle BOC=2A={90}^{\circ }$

In $△OBP$

$\frac{OP}{\sin {45}^{\circ }}=\frac{BP}{\sin \angle BOP}⟹\sin \angle BOP=\frac{3}{OP\sqrt{2}}\cdots \left(1\right)$

In $△OCP$

$\frac{OP}{\sin {45}^{\circ }}=\frac{CP}{\sin \angle COP}⟹\sin (\frac{\pi }{2}-\angle BOP)=\frac{5}{OP\sqrt{2}}$

$⟹\cos \angle BOP=\frac{5}{OP\sqrt{2}}\cdots \left(2\right)$

$(1{)}^2+(2{)}^2⟹1=\frac{34}{2O{P}^2}⟹OP=\sqrt{17}$