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