The answer is 100.

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Lets assume the mid point of AC is D. Then BHDC is cyclic quad. And hence MD.MC = MH.MB Lets assume AM = X, Then MC = 2X, MD = 3X/2. - X = X/2 We THEN have MD.MC = X² = AM^2 or AM² = MD.MC = MH.MB Thus MA is tangential to the circum-circle of Tr. AHB and thus /

ABH = /HAM which makes /_AHB=100°.