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Question: Let ABC be a triangle with ĐA = 45<sup>0</sup>. Let P be a point on side BC with PB = 3 & PC = 5. If...

Let ABC be a triangle with ĐA = 450. Let P be a point on side BC with PB = 3 & PC = 5. If O is circumcentre of triangle ABC then length OP is

A

18\sqrt { 18 }

B

17\sqrt { 17 }

C

19\sqrt { 19 }

D

15\sqrt { 15 }

Answer

17\sqrt { 17 }

Explanation

Solution

Let OP = x

let circumradius = R

\ BCsin⁡45∘\frac { B C } { \sin 45 ^ { \circ } } = 2R ̃ R = 424 \sqrt { 2 }

Now using power of a point.

PB. PC = PM. PN

15 = (R–x) (R + x) ̃ x=17x = \sqrt { 17 }