Let's analyze each option to determine if the triangle must have an angle of at least $90^{\circ}$. We can use the Pythagorean theorem's logic to check the relationship between the sides. Let $a, b$, and $c$ be the lengths of the sides, with $c$ being the longest side.
- If $\(a^2+b^2=c^2\)$, the triangle is a right triangle, and the angle opposite side $c$ is exactly $\(90^{\circ}\)$.
- If $\(a^2+b^2>c^2\)$, the triangle is an acute triangle, and all angles are less than $\(90^{\circ}\)$.
- If $\(a^2+b^2<c^2\)$, the triangle is an obtuse triangle, and the angle opposite side $c$ is greater than $\(90^{\circ}\)$.
A. The angle included between sides whose lengths are 3 and 4

The lengths of the two sides are given as 3 and 4, but we do not know the length of the third side. The third side could be any length between $\(4-3=1\)$ and $\(4+3=7\)$ (due to the triangle inequality theorem).
- If the third side were 5 , we would have a right triangle $\(\left(3^2+4^2=5^2\right)\)$, and one angle would be exactly $\(90^{\circ}\)$.
- If the third side were less than 5 , the triangle would be acute, and no angle would be at least $\(90^{\circ}\)$.
- If the third side were greater than 5 , the triangle would be obtuse, and one angle would be greater than $\(90^{\circ}\)$.

Since we cannot know the length of the third side, the angle is not guaranteed to be at least $\(90^{\circ}\)$. This option is incorrect.

B. The triangle $A B C$, in which the sides are of lengths 6,8 and 11

The lengths of the sides are 6,8 , and 11 . The longest side is 11 . Let's check the relationship with the Pythagorean theorem.
- $\(a=6, b=8, c=11\)$
- $\(6^2+8^2\)$ vs. $\(11^2\)$
- $\(36+64\)$ vs. 121
- 100 vs. 121

Since $100<121$, we have $\(a^2+b^2<c^2\)$. This means the angle opposite the longest side (11) is greater than $\(90^{\circ}\)$. Therefore, an angle in this triangle must be at least $\(90^{\circ}\)$. This option is correct.
C. Triangle $A B C$, in which the sides are of lengths 7, 24 and 21.

The lengths of the sides are 7,24 , and 21 . The longest side is 24 . Let's check the relationship.
- $a=7, b=21, c=24$
- $\(7^2+21^2\)$ vs. $\(24^2\)$
- $\(49+441\)$ vs. 576
- 490 vs. 576

Since $490<576$, we have $\(a^2+b^2<c^2\)$. This means the angle opposite the longest side (24) is greater than $90^{\circ}$. Therefore, an angle in this triangle must be at least $90^{\circ}$. This option is correct.

D. Isosceles triangle $\(A B C\)$ in which two same sides are 10 cms each

The lengths of two sides are 10 and 10 . The third side could have any length $x$ such that $10- 10<x<10+10$, so $0<x<20$.
- If the third side is small (e.g., 1 ), the triangle is acute.
- If the third side is large (e.g., 18 ), we can check the relationship: $\(10^2+10^2\)$ vs. $\(18^2\)$.
- $100+100$ vs. 324
- 200 vs. 324
- Since $\(200<324\)$, the angle opposite the side of length 18 would be obtuse.

However, the third side could also be small enough that all angles are acute. For example, if the third side is 10 , it's an equilateral triangle, and all angles are $\(60^{\circ}\)$. Since the angle is not guaranteed to be at least $\(90^{\circ}\)$, this option is incorrect.

The only options where the triangle must have an angle of at least $\(90^{\circ}\)$ are $\(\mathbf{B}\)$ and $\(\mathbf{C}\)$.