Given a triangle ABC with sides: AB = 9; BC = 6 and AC = 5. Plane M passes through the AC side, making

Given a triangle ABC with sides: AB = 9; BC = 6 and AC = 5. Plane M passes through the AC side, making an angle of 45 ° with the plane of the triangle. Find the distance between plane M and vertex B.

From vertex B we lower the height BN and get two right-angled triangles ABN and CBN with a common side BN. We take СN as x and then АN = 5 + x.

By the Pythagorean theorem, we derive BN ^ 2 for two triangles:

BN² = AB² – AN² = 9² – (5 + x) ² = 81 – 25 – 10 * x – x² = 56 – 10 * x – x².

BN² = BC² – CN² = 36 – x².

Equating the results:

56 – 10 * x – x² = 36 – x².

– 10X – x² + x² = 36 – 56.

– 10 * x = – 20.

x = 2.

Substitute and find BN

BN² = 36 – 2² = 32.

BN = √ (32).

Now from the vertex B we draw a segment BL perpendicular to the plane M, this is the distance between the plane M and the vertex B.

Consider the triangle BNL, it is rectangular and isosceles, because ВL is perpendicular to NL and the angle ВNL is 45 by condition.

Again, by the Pythagorean theorem, we derive BN²:

BN² = BL² + NL² since BN = √ (32) and BL = NL.

√ (32²) = 2BL².

32 = 2BL².

BL² = 32/2.

BL = √ (16).

BL = 4.

Answer: the distance between plane M and vertex B is 4.