since tanx = 3/4, it implies;
Let AB = 4 units.
Let AD = 3 units.
tana = 4/3
a = atan(4/3)°
a is angle ADB.
b = 180-a
b = 90+90-atan(4/3)
b = (90+atan(3/4))°
b is angle BDC.
It implies;
The required angle, y is;
y = angle BCD = angle DCB since BD = CD.
Therefore;
y = ½(180-b)
y = ½(180-(90+atan(3/4))
y = ½(90+90-90-atan(3/4))
y = ½(90-atan(3/4))
y = ½atan(4/3)°
y = 26.5650511771°
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