In an isosceles triangle ABC (AB = BC) The height AH is drawn. Find the angle BAC if the angle BAC = 2, Angle HAB.

Let the value of the angle HAB = X0, then, by condition, the angle HAC = 2 * HAB = 2 * X0.

Then the angle BAC = X + 2 * X = 3 * X0.

The ABC triangle is isosceles with the AC base, then the BCA angle = BAC = 3 * X0.

AH – height, then triangle ABH is rectangular, in which the angle ABH = (90 – X) 0.

The sum of the interior angles of a triangle is 180, then:

180 = 3 * X + 3 * X + 90 – X.

5 * X = 90.

X = 90/5 = 180.

Then the angle BAC = 3 * 18 = 54.

Answer: The BAC angle is 54.