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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 835.
Result: 150Solution: Let BC=AC=1 a:alfa, b:betaFrom sinus theorem in BDCDC=sina/sin(a+b)Sinus theorem in ADC using DC we getsina*sin(2a+3b)=sin2a*sin(a+b) since sin2a=2*sina*cosa we get2*sin(a+b)*cosa=sin(2a+b)+sinb=sin(2a+3b)sinb=sin(2a+3b)-sin(2a+b)=2*sinb*cos(2a+2b)=sinbeliminating sinb(b can't be zero) we getcos(2a+2b)=1/2 which means 2a+2b=60, a+b=30So x=180-30=150 degrees.
And using elementary geometry?
http://img38.imageshack.us/img38/3597/problem835.pngMake drawing per attached sketch1. Draw altitude CF and bisector AF of angle DACMake angle ACE= beta2. Due to symmetric properties , CDE and AEDB are isoscelesAnd BE, AD and CF are concurrent at point G3. In triangle AGC , GE is an angle bisector => angle BGF=angle AGF=angle AGE=angle CGE=60In triangle ABC we have 60+ 4alpha+4.beta=180 => alpha+ beta=30In triangle BDC x=180-alpha-beta= 150
Minor typo correction.Line 3 of my solution should read as "1. Draw altitude CF of triangle ABC and bisector AE of angle DAC "Peter Tran
Fie P punctul de pe segmentul AD astfel incat m(: PC este mediatoarea segmetului AB si bisectoarea unghiului =>m(In ∆BPC m(6α+6β=180 =>α+β=30 de unde in ∆BDC x=180-(α+β)=150
Let CX be the bisector of < ACB, X being on AD. Now thro' congruence or symmetry < AXC = < BXC. But D is the in centre of Tr. ACX, so XD bisects < BXC. It thus follows that < BXD = 60. So 2 alpha + 2 beta + 120 = 180 in Tr. ADC, implying that alpha + beta = 30.So from Tr. BDC x = 150Sumith PeirisMoratuwaSri Lanka