Challenging Geometry Problem 1611. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
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Thanks for the hint Antonio
ReplyDeleteThe perpendicular from C to AD extended meets at E. CE extended meets AB extended at F
< FBC = 73 = < ACF = < AFC
So FC = CB = CD = DF, hence Triangle CDF is equilateral
Hence < BCD = 60 - 34 = 26 and
x = 39 - 26 = 13
Sumith Peiris
Moratuwa
Sri Lanka
Trigonometry Solution
DeleteIn Triangle ABC
b/a = sin 73 / sin 34 .......(1)
In Triangle ADC
b/a = sin (x+17) / sin 17 .....(2)
From (1) & (2)
sin (x + 7) / sin 17 = sin 73 / sin 34 = cos 17 / (2 sin 17. cos 17) = 1/ (2 sin 17)
So sin (x + 17) = 1/2 and hence
x + 17 = 30 and x = 13
Sumith Peiris
Moratuwa
Sri Lanka
Another Geometry Solution
ReplyDeleteExtend AB to P such that AP = AC = b. Let AD extended meet CP at Q
Now < AQC = 90 and
CQ = QP = a/2 (since CBP is isosceles)
In Right Triangle CDQ, CQ = CD / 2 = a/2 so it is a 30-60-90 Triangle
Hence <DCQ = 60 and
Therefore x = 73 - 60 = 13
Sumith Peiris
Moratuwa
Sri Lanka