In the tetrahedron dabc, the edge cd is perpendicular to the plane abc AC = BC = 10 cm

In the tetrahedron dabc, the edge cd is perpendicular to the plane abc AC = BC = 10 cm, AB = 16 cm, CD = 6 cm. Find the linear angle of the dihedral ung CABD.

AC = BC = 10cm, AB = 16cm, CD = 6cm.

Let the vertex of the linear angle be point K, then:

∠CABD = tg CD / CK, where CD = 6cm.

Find CK from triangle ABC, where point K divides side AB in half and angle ACB in half, which means it divides triangle ABC into two identical triangles ACK and СВK. Then the CК is a leg in the CВK triangle.

СK ^ 2 = CB ^ 2 – (ВK) ^ 2,

CK ^ 2 = CB ^ 2 – (AB / 2) ^ 2,

CK ^ 2 = 10 ^ 2 – (16/2) ^ 2,

CK ^ 2 = 100 – (8) ^ 2,

СK ^ 2 = 100 – 64,

СK ^ 2 = 36,

СK = 6cm.

∠CABD = tg 6/6,

∠CABD = tg 1,

∠CABD = n / 4 = 45 °.