Question 1198890
**1. Find the Height of the Equilateral Triangle Base (Altitude)**

* In an equilateral triangle, the altitude bisects the base.
* Let the altitude of the equilateral triangle be 'h'.
* Using the Pythagorean theorem in one of the right-angled triangles formed by the altitude:
   (h)² + (4)² = (8)² 
   h² = 64 - 16 = 48
   h = √48 = 4√3 cm

**2. Find the Height of the Pyramid (VX)**

* Consider the right triangle VAX, where VX is the height of the pyramid.
* Using the Pythagorean theorem:
   (VX)² + (AX)² = (VA)²
   VX² + (4√3)² = (13)² 
   VX² + 48 = 169
   VX² = 121
   VX = 11 cm

**3. Find AX**

* From the previous step, we found that AX = 4√3 cm.

**4. Find the Angle VAB**

* Consider the right triangle VAX.
* Let the angle VAB be θ.
* cos(θ) = AX / VA 
* cos(θ) = (4√3) / 13
* θ = arccos((4√3) / 13) 
* θ ≈ 49.11 degrees

**Therefore:**

* **AX = 4√3 cm**
* **VX = 11 cm**
* **Angle VAB ≈ 49.11 degrees**