In an isosceles triangle ABC, the angle C = 104 degrees AM is the height of the triangle. Find the angle MAB.

△ ABC – isosceles with base AB and obtuse <C = 104 °.

The angles at the base are (180 ° – 104 °) / 2 = 38 °.

The height AM drawn to side BC lies outside the triangle.

Consider △ AMC – rectangular (<M = 90 °).

<MAB + <ABM = 90 ° (acute rectangular corners △ -ka).

<ABM = 38 ° – common for △ ABC and △ AMC.

<MAB = 90 ° – <ABM = 90 ° – 38 ° = 52 °.

Answer: <MAB = 52 °.



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