tag:blogger.com,1999:blog-6933544261975483399.post8038124002762694445..comments2024-03-19T00:02:30.728-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1152: Triangle, Quadrilateral, Angle, 10, 20, 30 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-6933544261975483399.post-68112514491586145422017-10-20T22:35:16.949-07:002017-10-20T22:35:16.949-07:00Please note all above solutions never use given da...Please note all above solutions never use given data m(ACB)= 10 degrees. in fact m(ACB) can have value not equal 10 degrees.<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-80792262983184516362015-10-29T00:28:23.387-07:002015-10-29T00:28:23.387-07:00Please elaborate, Sumith, and if possible please w...Please elaborate, Sumith, and if possible please write to me at: ajitathle@gmail.com.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89815250440949074802015-10-26T19:14:47.685-07:002015-10-26T19:14:47.685-07:00It is but it needs to be proved Ajit
I'm sure...It is but it needs to be proved Ajit<br /><br />I'm sure u know the proofSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-81364866053792625832015-10-21T00:07:11.666-07:002015-10-21T00:07:11.666-07:00See triangle ABC and triangle ABD, AB is common, a...See triangle ABC and triangle ABD, AB is common, and ang.ADB=2*ang.ACB.<br />So, A, B, and C are cocyclic (circum angle/central angle).<br /><br />---> x=ang..BDC/2 = 15 deg.Emilnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65869580790769355492015-10-12T23:11:41.334-07:002015-10-12T23:11:41.334-07:00Triangle DAC: <(CAD) + <(ACD) = 180 - 50 =...Triangle DAC: <(CAD) + <(ACD) = 180 - 50 = 130. And <(BAD) +<(ACD)= 145.<br />So x=<(BAD) + <(ACD) - <(CAD) - <(ACD) =145 -130= 15.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1069859466389212592015-10-12T21:22:15.637-07:002015-10-12T21:22:15.637-07:00We observe that /_BCA=10° while /_ADB =20° & s...We observe that /_BCA=10° while /_ADB =20° & simultaneously /_ADC=155° =(Larger /_CDA)/2.<br />Hence would it be correct to say that the circumcircle of Tr. ABC will have point D as its centre and DA(=DB=DC) as its radius? If that be correct then /_BAC = x = /_BDC/2 =30/2=15°.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5060535871544307782015-10-10T22:35:54.277-07:002015-10-10T22:35:54.277-07:00Would it be correct to say that since /_ABC=155°=L...Would it be correct to say that since /_ABC=155°=Larger /_ADC)/2=310°)/2 and /_ADB=2*/_BCA, a circle drawn with centre at D & radius = DA will pass through A, B & C? This would imply /_x=30°/2=15°.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-29071996228286087702015-10-10T07:59:02.539-07:002015-10-10T07:59:02.539-07:00http://s3.postimg.org/ewu8vbicj/pro_1152.png
Draw...http://s3.postimg.org/ewu8vbicj/pro_1152.png<br /><br />Draw Av//CD<br />We have ∠ (DAv)=180-50=130<br />And ∠ (DAv)= ∠ (DAC)+ ∠ (CAx)= ∠ (BAD)-x+∠ (ACD)=145-x<br />So x=15<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com