Strategy:
1. Consider $\triangle A B D$. We know two sides are $A B=5$ and $A D=5$. Let $\angle A=\theta$. We can find the length of BD using the Law of Cosines.
2. Consider $\triangle B C D$. We know its sides are $B C=6$ and $C D=1$. The third side is BD , which connects the two triangles.
3. Apply the Triangle Inequality Theorem to $\(\triangle B C D\)$ to find the possible range of BD .
4. Use this range of BD to find the possible range of $\(\cos A\)$ (and thus $\(\angle A\)$ ).

Step 1: Apply the Law of Cosines to $\(\triangle A B D\)$
In $\(\triangle A B D\)$, let $\(\mathrm{BD}=x\)$.
By the Law of Cosines:

$$
\(\begin{aligned}
& x^2=A B^2+A D^2-2(A B)(A D) \cos A \\
& x^2=5^2+5^2-2(5)(5) \cos A \\
& x^2=25+25-50 \cos A \\
& x^2=50-50 \cos A \text { (Equation 1) }
\end{aligned}\)
$$


Step 2: Apply the Triangle Inequality Theorem to $\triangle B C D$
For $\triangle B C D$ to exist, the sum of any two sides must be greater than the third side.
Let the sides be $\mathrm{BC}=6, \mathrm{CD}=1$, and $\mathrm{BD}=x$.
- $\(B C+C D>B D \Longrightarrow 6+1>x \Longrightarrow 7>x\)$
- $\(B C+B D>C D \Longrightarrow 6+x>1 \Longrightarrow x>-5\)$ (always true since x is a length)
- $\(C D+B D>B C \Longrightarrow 1+x>6 \Longrightarrow x>5\)$

Combining these inequalities, we get the possible range for $x$ :

$$
\(5<x<7\)
$$

Step 3: Relate the range of $x$ to $\cos A$ using Equation 1
We have $x^2=50-50 \cos A$.
Substitute the bounds for $x$ :
- If $\(x=5: 5^2=50-50 \cos A \Longrightarrow 25=50-50 \cos A \Longrightarrow 50 \cos A=25 \Longrightarrow$ $\cos A=0.5 \Longrightarrow A=60^{\circ}\)$.
- If $\(x=7: 7^2=50-50 \cos A \Longrightarrow 49=50-50 \cos A \Longrightarrow 50 \cos A=1 \Longrightarrow$ $\cos A=0.02 \Longrightarrow A \approx 88.85^{\circ}\)$.

Since $\(5<x<7\)$, it means $\(25<x^2<49\)$.
Substitute this into $\(x^2=50-50 \cos A\)$ :

$$
\(25<50-50 \cos A<49\)
$$


Subtract 50 from all parts:

$$
\(\begin{aligned}
& 25-50<-50 \cos A<49-50 \\
& -25<-50 \cos A<-1
\end{aligned}\)
$$


Divide all parts by -50 . Remember to reverse the inequality signs when dividing by a negative number:

$$
\(\begin{aligned}
& \frac{-25}{-50}>\cos A>\frac{-1}{-50} \\
& 0.5>\cos A>0.02 \\
& \text { So, } 0.02<\cos A<0.5
\end{aligned}\)
$$


Now, we need to find the angles $A$ for which this is true.
We know that as A increases from $\(0^{\circ}$ to $180^{\circ}, \cos A\)$ decreases.
- If $\(\cos A=0.5\)$, then $\(A=60^{\circ}\)$.
- If $\(\cos A=0.02\)$, then $\(A \approx 88.85^{\circ}\)$.

Since $0.02<\cos A<0.5$, the possible range for angle A is:

$$
\(60^{\circ}<A<88.85^{\circ}\)
$$


Step 4: Check the given options
(A) 45 (Not in range)
(B) 60 (Not in range - $A$ must be greater than 60)
(C) 75 (Is in the range $\(60^{\circ}<A<88.85^{\circ}\)$ )
(D) 90 (Not in range)
(E) 120 (Not in range)

Therefore, $\(75^{\circ}\)$ is the only possible value for angle A among the given options.
The final answer is 75 .