# Through the vertex of the right angle C of the isosceles right-angled triangle ABC, in which the hypotenuse AB = 8 cm

**Through the vertex of the right angle C of the isosceles right-angled triangle ABC, in which the hypotenuse AB = 8 cm, the perpendicular CP is drawn to the plane of the triangle, and point P is connected to the vertices A and B. The planes of triangles APB and ABC form an angle of 60 ° between themselves. Find: the length of the perpendicular CP; area ΔАРВ, angle between the straight line ААР and the plane ΔАВС.**

Consider a right-angled triangle ABC, in which the angle is ACB, and AC = BC. Determine the length of the legs CB and AC according to the Pythagorean theorem AB ^ 2 = AC ^ 2 + BC ^ 2 = 2 * AC ^ 2.

AC ^ 2 = AB ^ 2/2 = 64/2 = 32.

AC = 4 * √2 cm.

Consider a right-angled triangle ACD, whose angle ADC = 90. Let us define, according to the Pythagorean theorem, the leg SD, which is the height of the triangle ABC.

CD ^ 2 = AC ^ 2 – AD ^ 2 = 32 – 16 = 16.

CD = 4 cm.

Consider a right-angled triangle РDC, in which the angle РDC = 30, then РС = СD * tg60 = 4 * √3. In the PDC triangle, the angle DPC = 180 – 90 – 69 = 30.

The CD leg lies opposite the angle 30, and is equal to half the length of the PD hypotenuse, then PD = 2 * CD = 2 * 4 = 8 cm.

Determine the area of the triangle APD.

Sarv = AB * PD / 2 = 8 * 8/2 = 32 cm2.

Let’s define the APC angle.

tgАРС = РС / АС = 4 * √3 / 4 * √2 = √ (3/2).

Angle APC = arctan √ (3/2).

Answer: CP = 4 * √3, Sarv = 32 cm2, APC angle = arctan √ (3/2).