1. In the given figure, O is the centre of the circle and ∠ACB = 30°. Then ∠AOB = ? (a) 90° (b) 60° (c) 30° (d) 15° 6. The value of (x - 1)(x + 2) / (x^2 + (x + 2)^2) is equal to?... 1. In the given figure, O is the centre of the circle and ∠ACB = 30°. Then ∠AOB = ? (a) 90° (b) 60° (c) 30° (d) 15° 6. The value of (x - 1)(x + 2) / (x^2 + (x + 2)^2) is equal to? (a) x^2 + x + 2 (b) x^2 + 2 (c) x^2 + 1/x - 2 (d) 1/x 7. If (4, 19) is a solution of the equation y = αx + 3, then is α? (a) √4 (b) 6 (c) 3 (d) 5 8. What is the class mark of the class interval 90 - 120 ? (a) 90 (b) 105 (c) 115 (d) 120 9. In the adjoining figure, BC = AD, CA || AB and BD || AB. The rule by which ∆ ABC ≡ ∆ BAD is (a) ASA (b) RHS (c) SSS (d) SAS 10. If AB || HF and DE || FG, then the measure of ∠LDF is ? (a) 90° (b) 80° (c) 100° (d) 108° 11. If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then the greater of the two angles is: (a) 54° (b) 136° (c) 120° (d) 108° Read more

Steps to Solve

1. Finding the angle AOB
   Given that $ \angle ACB = 30^\circ $ and that $ O $ is the center of the circle, we can use the inscribed angle theorem. The inscribed angle ($ \angle ACB $) is half of the central angle ($ \angle AOB $).
   So, we have:
   $$ \angle AOB = 2 \times \angle ACB = 2 \times 30^\circ = 60^\circ $$

2. Determining the value of the expression
   We need to evaluate the expression $ (x - \frac{1}{2})(x + \frac{1}{2}) $.
   Using the difference of squares formula, this simplifies to:
   $$ (x - \frac{1}{2})(x + \frac{1}{2}) = x^2 - \left(\frac{1}{2}\right)^2 = x^2 - \frac{1}{4} $$ This indicates the need for simplification based on the given options.

3. Solving for the class mark
   The class mark of an interval is the midpoint of that interval. For the interval $ 90 - 120 $, the class mark is calculated as:
   $$ \text{Class mark} = \frac{90 + 120}{2} = \frac{210}{2} = 105 $$

4. Applying the given rule
   The triangle congruence rule states that if two triangles share a side and the corresponding angles are equal (Angle-Side-Angle, ASA), then the triangles are congruent. In this case, since $ \triangle ABC \cong \triangle BAD $, it validates the congruence of the triangles.

5. Finding the measure of the angle LDFE
   Given that $ AB \parallel CD $ and we can identify that $ \angle ABH = 80^\circ $. Because alternate interior angles are equal, $ \angle DEB = \angle ABH = 80^\circ $. Since $ \angle LDFE $ is supplementary to $ \angle DEB $ in a straight line:
   $$ \angle LDFE + \angle DEB = 180^\circ $$
   Hence,
   $$ \angle LDFE = 180^\circ - 80^\circ = 100^\circ $$

6. Finding the greater of the two angles
   For the angles $ 54^\circ $ and $ 136^\circ $, since they are in the ratio $ 2:3 $, we need to find the total angle sum first before calculating.
   Assuming the angles are $ 2x $ and $ 3x $ we can set up the equation:
   $$ 2x + 3x = 180^\circ $$
   Solving this gives:
   $$ 5x = 180^\circ \implies x = 36^\circ $$
   Then, the angles will be $ 72^\circ $ and $ 108^\circ $ respectively. The greater angle is $ 108^\circ $.