Geometry Problem 1529. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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Trigonometry Solution

ReplyDeleteLet < BDC = y

In Triangle ABC, a/sin30 = b/sin50 ...............(1)

In Triangle BDC, a/sin.y = b/ sin100.............(2)

(1)/(2); sin.y/ (1/2) = sin100/sin50 = 2cos50

So sin.y = cos50 = sin40

Hence y = 40 = < BDC

Sumith Peiris

Moratuwa

Sri Lanka

Geometry Solution

ReplyDeleteLet O be the Circumcentre of Triangle ABC. C being obtuse, O is outside the Triangle. Drop a perpendicular OX to CO extended. Drop a perpendicular BY to AC extended.

< COB = 2 X 30 = 60 so Triangle BCO is equilateral. < AOC = 2 X < ABC = 100, < ACO = 40 since Triangle ACO is isosceles

Triangles AOX & BCY are congruent ASA and so AX = BY

Hence Triangles ACX & BDY are congruent (Hypotenuse, side, 90)

Therefore <BDC = ACX = 40

Sumith Peiris

Moratuwa

Sri Lanka

Similar to previous.

ReplyDeleteLet O be circumcenter of ABC.

BOC is equilateral.

OAC is a 100-40-40 triangle.

OAC and CBD has two equal sides and one equal angle.

This means either BDC = ACO or BDC = 180-ACO.

Second one violates triangle conditions.

So OAC is similar CBD. So angle BDC = 40.