Given a regular triangular prism ABCA1B1C1. AB = 4 cm, AA1 = 2 cm. Find the angle between planes AB1C and ABC.

Since ABCA1B1C1 is a regular triangular prism, ABC and A1B1C1 are equal right triangles, with sides AB = BC = AC = A1B1 = B1C1 = A1C1 – 4 cm. The side faces ABB1A1 = ACC1A1 = BCC1B1 are rectangles in which the width (AA1 = BB1 = CC1) is 2 cm, and the length is 4 cm.
1. Plane AB1C is an isosceles triangle ACB1 (since its sides B1A and B1 C are diagonals of equal rectangles).
Let us find the length В1С according to the Pythagorean theorem from the triangle CBB1:
В1С = √ (ВС ^ 2 + BB1 ^ 2);
В1С = √ (4 ^ 2 + 2 ^ 2) = √ (16 + 4) = √20 = 2√5 (cm).
2. Since AB1C is an isosceles triangle, then B1H is both the height and the median, then AH = CH = AC / 2. By the Pythagorean theorem, we find the length B1H from the right-angled triangle B1HC:
B1H = √ (B1C ^ 2 – (AC / 2) ^ 2);
B1H = √ ((2√5) ^ 2 – (4/2) ^ 2) = √ (4 * 5 – 2 ^ 2) = √ (20 – 4) = √16 = 4 (cm).
3. Consider a right-angled triangle В1ВН. B1B = 2 cm – leg, BH = 2√3 cm – leg, B1H = 4 cm – hypotenuse (since it lies opposite the right angle).
The angle between the planes ABC and AB1C is the angle between the height В1Н and ВН, that is, the angle В1НВ.
The sine of an angle in a right triangle is the ratio of the opposite leg to the hypotenuse. The sine of the angle В1НВ between the planes AB1C and ABC will be equal to:
sinB1HB = B1B / B1H;
sinB1HB = 2/4 = 1/2.
Angle B1HB = 30 degrees.
Answer: angle B1HB = 30 degrees.