Geometry Problem 1535. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

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< BOF = 45 and since KO = OH, < OHK = 22.5.....(1)

ReplyDeleteFrom Right Triangle AFH ; < AFH = arctan (1/2).....(2)

Now consider Triangle FMH

< AMK = < FMH = 180 - 22.5 - arctan(1/2) = 157.5 - 26.6 = 130.9 approximately

Sumith Peiris

Moratuwa

Sri Lanka