1. Problem 28: Find the length of the arc in a circle of radius 5 cm subtending an angle of 33°15' at the center. Step 1: Convert the angle to decimal degrees: 33°15' = 33 + 15/60 = 33.25°. Step 2: Use the arc length formula $$L = r \theta$$ where $$\theta$$ is in radians. Step 3: Convert degrees to radians: $$\theta = 33.25 \times \frac{\pi}{180} = \frac{133}{360} \pi$$ radians. Step 4: Calculate arc length: $$L = 5 \times \frac{133}{360} \pi = \frac{665}{360} \pi = \frac{665\pi}{360}$$ cm. Step 5: Simplify fraction: $$\frac{665}{360} = 1 + \frac{305}{360} = 1 + \frac{61}{72}$$. Step 6: So, $$L = \pi \times (1 + \frac{61}{72}) = \pi \times \frac{133}{72}$$ cm. Step 7: Numerically, $$\pi \approx 3.1416$$, so $$L \approx 3.1416 \times 1.8472 = 5.8$$ cm. Answer choices are in cm with fractions; closest is option 2) 3 (65/72) cm, but our calculation shows about 5.8 cm, so re-check. Re-check: Step 3 angle in radians: $$33.25 \times \frac{\pi}{180} = 0.5804$$ radians. Step 4: $$L = 5 \times 0.5804 = 2.902$$ cm. So the arc length is approximately 2.9 cm. Option 4) 2 (65/72) cm = 2 + 65/72 = 2.9027 cm, matches our result. Final answer: 4) 2 (65/72) cm. 2. Problem 29: A train moves on a circular curve of radius 1500 m at 66 km/h. Find the angle turned in 10 seconds. Step 1: Convert speed to m/s: $$66 \times \frac{1000}{3600} = 18.33$$ m/s. Step 2: Distance traveled in 10 s: $$18.33 \times 10 = 183.3$$ m. Step 3: Angle in radians: $$\theta = \frac{\text{arc length}}{r} = \frac{183.3}{1500} = 0.1222$$ radians. Step 4: Convert radians to degrees: $$0.1222 \times \frac{180}{\pi} = 7.0^\circ$$. Answer: 3) 7°. 3. Problem 30: Given $$\sin \theta = \frac{5}{7}$$, find $$\tan \theta$$ for $$0 < \theta < 90^\circ$$. Step 1: Use identity $$\sin^2 \theta + \cos^2 \theta = 1$$. Step 2: Calculate $$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{5}{7}\right)^2} = \sqrt{1 - \frac{25}{49}} = \sqrt{\frac{24}{49}} = \frac{2\sqrt{6}}{7}$$. Step 3: Calculate $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{5/7}{2\sqrt{6}/7} = \frac{5}{2\sqrt{6}}$$. Answer: 3) 5/(2√6). 4. Problem 31: Calculate $$\sin^2 30^\circ + \sin^2 45^\circ + \sin^2 60^\circ + \tan 45^\circ$$. Step 1: $$\sin 30^\circ = \frac{1}{2} \Rightarrow \sin^2 30^\circ = \frac{1}{4}$$. Step 2: $$\sin 45^\circ = \frac{\sqrt{2}}{2} \Rightarrow \sin^2 45^\circ = \frac{1}{2}$$. Step 3: $$\sin 60^\circ = \frac{\sqrt{3}}{2} \Rightarrow \sin^2 60^\circ = \frac{3}{4}$$. Step 4: $$\tan 45^\circ = 1$$. Step 5: Sum: $$\frac{1}{4} + \frac{1}{2} + \frac{3}{4} + 1 = \frac{1}{4} + \frac{1}{2} + \frac{3}{4} + 1 = \frac{1+2+3}{4} + 1 = \frac{6}{4} + 1 = \frac{3}{2} + 1 = \frac{5}{2}$$. Answer: 3) 5/2. 5. Problem 32: If $$A + B + C = 180^\circ$$, find $$\tan \frac{A + B}{2}$$. Step 1: Since $$A + B + C = 180^\circ$$, then $$A + B = 180^\circ - C$$. Step 2: $$\tan \frac{A + B}{2} = \tan \frac{180^\circ - C}{2} = \tan (90^\circ - \frac{C}{2})$$. Step 3: Use identity $$\tan (90^\circ - x) = \cot x$$. Step 4: So, $$\tan \frac{A + B}{2} = \cot \frac{C}{2}$$. Answer: 3) cot(C/2). 6. Problem 33: Find the value of $$\frac{D}{90}$$ given options. Since no further context is given, the problem is ambiguous. Answer: None of the options can be confirmed without more information. Final JSON: