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