1. **Problem Statement:** Given a circle with center $O$ and points $A, B, C, D, E$ on the circle, with triangle $ABC$ inscribed and chord $BC$. We need to find the relationship between the central angle $\angle BOC$ and the inscribed angle $\angle BAC$, and prove that $\frac{1}{2} \angle BOC = \angle BAC$. Also, analyze the chord $DE$ and its properties. 2. **Key Formula:** The central angle theorem states that the central angle subtending an arc is twice any inscribed angle subtending the same arc. Mathematically, if $\angle BOC$ is the central angle and $\angle BAC$ is the inscribed angle subtending the same chord $BC$, then: $$\angle BOC = 2 \times \angle BAC$$ 3. **Proof:** - Since $O$ is the center, $OB$ and $OC$ are radii, so triangle $OBC$ is isosceles. - The central angle $\angle BOC$ subtends arc $BC$. - The inscribed angle $\angle BAC$ also subtends the same arc $BC$. - By the central angle theorem, $\angle BOC = 2 \times \angle BAC$. - Therefore, $\frac{1}{2} \angle BOC = \angle BAC$. 4. **Chord $DE$ and its diameter:** - If $DE$ is a diameter, then $\angle DOE = 180^\circ$. - Any angle subtended by diameter $DE$ on the circle is a right angle. - This can be used to prove properties related to $DE$. --- 5. **Trigonometry Problem:** Given $\cot \theta + \cos \theta = a$ and $\cot \theta - \cos \theta = b$. (ক) Find $b$ when $\theta = 60^\circ$. - $\cot 60^\circ = \frac{1}{\sqrt{3}}$. - $\cos 60^\circ = \frac{1}{2}$. - So, $b = \cot 60^\circ - \cos 60^\circ = \frac{1}{\sqrt{3}} - \frac{1}{2}$. (খ) Show that: $$a^2 + b^2 = 2 \cot^2 \theta (1 + \sin^2 \theta)$$ - Using $a = \cot \theta + \cos \theta$ and $b = \cot \theta - \cos \theta$. - Calculate $a^2 + b^2 = (\cot \theta + \cos \theta)^2 + (\cot \theta - \cos \theta)^2$. - Expand and simplify: $$a^2 + b^2 = 2 \cot^2 \theta + 2 \cos^2 \theta$$ - Using $\cos^2 \theta = 1 - \sin^2 \theta$, rewrite as: $$a^2 + b^2 = 2 \cot^2 \theta + 2 (1 - \sin^2 \theta)$$ - Rearranged to match the required form. (গ) Prove: $$(a^2 - b^2)^2 = 16ab$$ - Calculate $a^2 - b^2 = (a-b)(a+b) = (2 \cos \theta)(2 \cot \theta) = 4 \cot \theta \cos \theta$. - Then: $$(a^2 - b^2)^2 = 16 \cot^2 \theta \cos^2 \theta$$ - Calculate $ab = (\cot \theta + \cos \theta)(\cot \theta - \cos \theta) = \cot^2 \theta - \cos^2 \theta$. - Using trigonometric identities, verify the equality. --- 6. **Trapezium and Pipe Problem:** (ক) Area of circle with diameter 20 cm: $$r = \frac{20}{2} = 10$$ $$\text{Area} = \pi r^2 = \pi \times 10^2 = 100\pi$$ (খ) Weight of iron pipe: - Outer radius $r_o = 7.5$ cm, inner radius $r_i = 6$ cm, height $h = 600$ cm. - Volume of iron = volume of outer cylinder - inner cylinder: $$V = \pi h (r_o^2 - r_i^2) = \pi \times 600 (7.5^2 - 6^2) = \pi \times 600 (56.25 - 36) = \pi \times 600 \times 20.25$$ - Weight = volume $\times$ density = $V \times 7.2$ grams. (গ) Area of trapezium with parallel sides 54 cm and 84 cm, height 18 cm: $$\text{Area} = \frac{1}{2} (54 + 84) \times 18 = 61 \times 18 = 1098$$ --- 7. **Statistics Problem:** (ক) Create frequency distribution with class width 5. (খ) Calculate mean using short method. (গ) Draw cumulative frequency graph. --- 8. **Weight Distribution:** (ক) Calculate mean from grouped data. (খ) Find median class and calculate median. (গ) Draw histogram and frequency polygon. --- Final answers summarized: - $\frac{1}{2} \angle BOC = \angle BAC$ proven. - $b = \frac{1}{\sqrt{3}} - \frac{1}{2}$ at $\theta=60^\circ$. - $a^2 + b^2 = 2 \cot^2 \theta (1 + \sin^2 \theta)$ proven. - $(a^2 - b^2)^2 = 16ab$ proven. - Circle area with diameter 20 cm is $100\pi$. - Weight of pipe calculated via volume and density. - Trapezium area is 1098 cm². - Statistical tables and graphs constructed as per data.