Problem:
In β³ABC,β A=100β,β B=50β,β C=30β,AH is an altitude, and BM is a median. Then β MHC=
Answer Choices:
A. 15β
B. 22.5β
C. 30β
D. 40β
E. 45β
Solution:
Since β C=30β and AHβ₯HC,β CAH=60β and, in fact, β³AHC is a 30-60-90 triangle. Hence, AH=21βAC=AM. Thus, β³AHM is equilateral, β AHM=60β, and β MHC=90ββ60β=30β.
OR
Since M is the midpoint of hypotenuse AC of right triangle AHC,MH and MC are radii of the circle circumscribing β³AHC. Therefore MH=MC and β MHC=β C=30β.