A plane is drawn through the apex of the cone that intersects the base along a chord that contracts an arc of 90 degrees

A plane is drawn through the apex of the cone that intersects the base along a chord that contracts an arc of 90 degrees. find the S side of the surface of the cone if the generatrix is 10 and the angle in the section at the vertex is 60.

The area of ​​the lateral surface of the cone Sside = π * r * l, where r is the radius of the circumference of the base of the cone, l is the length of the generatrix of the cone, equal to 10.
Sside = π * r * l = 3.14 * 10 * r = 31.4r.
Find the radius of the circle.
According to the condition, the chord contracts the arc at 90 degrees, which means that the angle between the radii is 90. That is, we get a triangle OAB, where O is the center of the base circle, AB is the chord, and OA = OB = r. The plane drawn through the vertex is the triangle ACB, where C is the vertex, AC = AB are generators, AB is the chord. The angle between AC and AB is 60.
Let’s find the value AB:
AB = (AC ^ 2 + BC ^ 2 – AC * BC * cos60) ^ (1/2) = (200 – 100 * 1/2) ^ (1/2) = √150.
Now from the triangle OAB we find the radius of the circle. Since the BAW is rectangular, then by the Pythagorean theorem we write:
r ^ 2 + r ^ 2 = AB ^ 2; r ^ 2 = 150/2 = 75. We get r = √75 = 5√3.
Side = 31.4r = 31.4 * 5√3 = 157√3.