In △ ABC, ∠ A=100°, ∠ B=50°, ∠ C=30°, AH is an altitude, and BM is a median. Then ∠ MHC=

1989 AMC 12 Problem 7

Attempt Problem 7 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

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\triangle ABC,

\angle A=100^\circ,

\angle B=50^\circ,

\angle C=30^\circ,

AH

is an altitude, and

BM

is a median. Then

\angle MHC=

15^\circ

22.5^\circ

30^\circ

40^\circ

45^\circ

Compare the horizontal and vertical changes from

H

to midpoint

M

Both changes are half the corresponding legs of right triangle

AHC

Because

M

is the midpoint of

AC,

the horizontal change from

H

M

\frac{HC}{2},

while the vertical change is

\frac{AH}{2}.

Thus

\begin{aligned} \tan\angle MHC &=\frac{\frac{AH}{2}}{\frac{HC}{2}}\\ &=\frac{AH}{HC}\\ &=\tan\angle ACH. \end{aligned}

Hence

\angle MHC=\angle C=30^\circ.

Thus the correct answer is C.