Geometry Problem 1521. 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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DBE = 70° (It will be explained on the sketch)
ReplyDeletehttps://photos.app.goo.gl/KzkMvaHgbfKy5cqt7
DeleteAgain one comment is deleted : https://photos.app.goo.gl/CxQ2Xvm1BvtF8zg19
DeleteLet <QBD = < QDB = p and let < OBE = < OEB = q
ReplyDeleteIn Triangle OQB,
(p+p-40) + (q + q - 40) = 140
So p + q = 220/2 = 110
Hence < DBE = p + q - 40 = 70
Sumith Peiris
Moratuwa
Sri Lanka
https://photos.app.goo.gl/kYPKJjPWmLn4TNQd7
ReplyDeleteNote that BOE and BQD are isosceles triangles
See sketch for angles d, o, q and e
We have d= 90-½.q
And e=90-½. o
So d+e= 180-½(q+o)
But q+o= 180- ∡(ABC)= 140
So d+e= 110
And∡(DBE)= 180-(d+e)= 70