In an isosceles triangle ABC, the base is AC = 4 √7, and the side is 8. Find the sine of the outer angle at the vertex A.

Let’s build the height ВD of the triangle ABC.

Since the ABC triangle is isosceles, the height BD is the same as the median of this triangle, and then BD bisects the base of the AC in half.

АD = СD = АС / 2 = 4 * √7 / 2 = 2 * √7 cm.

By the Pythagorean theorem, in a right-angled triangle ABD, we determine the length of the leg BD.

BD ^ 2 = AB ^ 2 – AD ^ 2 = 64 – 28 = 36.

ВD = 6 cm.

In a right-angled triangle ABD, through the hypotenuse AB and leg BD, we determine the sine of the angle BAD.

SinBAD = BD / AB = 6/8 = 0.75.

Since the sines of adjacent angles are equal, then SinDAB = SinBAD = 0.75.

Answer: The sine of the outer angle is 0.75.