Evaluate, without using a calculator: \( \sum_{A=35^{\circ}}^{55^{\circ}} \cos ^{2} A \)

To evaluate \( \sum_{A=35^{\circ}}^{55^{\circ}} \cos^{2} A \), we can first recognize that this is a sum of cosines squared from 35 degrees to 55 degrees. The angles involved are: - \( 35^\circ, 36^\circ, 37^\circ, 38^\circ, 39^\circ, 40^\circ, 41^\circ, 42^\circ, 43^\circ, 44^\circ, 45^\circ, 46^\circ, 47^\circ, 48^\circ, 49^\circ, 50^\circ, 51^\circ, 52^\circ, 53^\circ, 54^\circ, 55^\circ \) This gives us a total of \( 55 - 35 + 1 = 21 \) terms. Using the identity \( \cos^2 A = \frac{1 + \cos(2A)}{2} \), we can rewrite the sum: \[ \sum_{A=35^{\circ}}^{55^{\circ}} \cos^{2} A = \sum_{A=35^{\circ}}^{55^{\circ}} \frac{1 + \cos(2A)}{2} = \frac{1}{2} \sum_{A=35^{\circ}}^{55^{\circ}} 1 + \frac{1}{2} \sum_{A=35^{\circ}}^{55^{\circ}} \cos(2A) \] First, we compute the first part: \[ \sum_{A=35^{\circ}}^{55^{\circ}} 1 = 21 \] So, \[ \frac{1}{2} \sum_{A=35^{\circ}}^{55^{\circ}} 1 = \frac{1}{2} \cdot 21 = \frac{21}{2} \] Now, we compute the second part involving \( \cos(2A) \): \[ \sum_{A=35^{\circ}}^{55^{\circ}} \cos(2A) = \cos(70^\circ) + \cos(72^\circ) + \cos(74^\circ) + \ldots + \cos(110^\circ) \] Notice that from \( A = 35^\circ \) to \( A = 55^\circ \), the angles \( 2A \) run from \( 70^\circ \) to \( 110^\circ \), which consists of even angles. The sum of cosines of angles equally spaced from \( 70^\circ \) to \( 110^\circ \) can be handled using the formula for the sum of cosines. We can use symmetry about \( 90^\circ \) and the fact that \( \cos(90^\circ + x) = -\sin(x) \) helps pair terms neatly around this center. We calculate these terms and notice: \[ \cos 70^\circ = \sin 20^\circ \quad \text{and} \quad \cos 110^\circ = -\sin 20^\circ \] This balancing suggests that these sums will yield values that are either symmetric or sum into zero. More precisely, since the number of terms is even (21 terms total), the average sum will be decreasing from \( 70^\circ \) to \( 110^\circ \). Numerically computed in a sum, we find after simplifying \( \sum_{A=35}^{55}\cos(2A) \approx 0 \) due to the symmetry and cancellation in the cosine terms when accurately computed. Therefore, we can sum both parts: \[ \sum_{A=35^{\circ}}^{55^{\circ}} \cos^{2} A = \frac{21}{2} + 0 \cdot \frac{1}{2} = \frac{21}{2} = 10.5 \] Thus, the final result is: \[ \boxed{10.5} \]