Altitudes $\overline{ax}$ and $\overline{by}$ of acute triangle $abc$ intersect at $h$. if $\angle ahb = 132^\circ$, then what is $\angle acb$?

48 degrees

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See the attached image for a visual of the problem and answer.

Based on the diagram, we have angle AHB = 132 degrees (given) equal in measure to angle XHY since these two angles are vertical angles. 

Angles HYC and HXC are right angles due to the nature of AX and BY being altitudes. Recall that altitudes are segments that go from one vertex to the opposite side and they are perpendicular to the opposite side.

Focus on quadrilateral HXCY. So far, we know that...
Angle XHY = 132 degrees
Angle HYC = 90 degrees
Angle HXC = 90 degrees

The angle we want to find is angle ACB, which is the same as angle YCX. This angle is the missing angle of the quadrilateral HXCY.

For any quadrilateral, the four angles must add to 360 degrees. 

(angle XHY) + (angle HYC) + (angle HXC) + (angle YCX) = 360
(132) + (90) + (90) + (angle YCX) = 360
312 + (angle YCX) = 360
312 + (angle YCX) - 312 = 360 - 312
angle YCX = 48 degrees

Since angle ACB is the same as angle YCX, we can say
angle ACB = angle YCX = 48 degrees

So in summary,
angle ACB = 48 degrees

Iam guessing c. or d.

Ithink it is 7. 16