# AB-perpendicular AC and AD – inclined to the alpha plane. angle ACB = angle ADB = 30 degrees

**AB-perpendicular AC and AD – inclined to the alpha plane. angle ACB = angle ADB = 30 degrees CD = 2√2AB Find angle CAD**

Since, according to the condition, AB is perpendicular to the alpha plane, and the segments BC and BD lie in this plane, then the segment AB is perpendicular to the segments BC and BD.

Then triangles ABC and ABD are rectangular with acute angles AB = ADB = 30.

Triangles ABC and ABD are equal in leg and adjacent angle, since leg AB is common, angle BAC = BAD = 60, then AC = AD.

The AB leg lies opposite an angle of 30, then AC = AD = 2 * AB cm.

Since AC = AD, the ACD triangle is isosceles.

Let’s build a perpendicular AO to the side CD, then CO = DO = CD / 2 = 2 * √2 * AB / 2 = √2 * AB,

In a right-angled triangle AOC SinCAO = CO / AC = √2 * AB / 2 * AB = √2 / 2.

The angle CAO = 45, and since AO is the bisector of the angle ACD, the angle CD = 2 * CAO = 2 * 45 = 90.

Answer: The CAD angle is 90.