The sum of adjacent angles is $180^{\circ}$. $(3x-4) + (3x+10) = 180^{\circ} \Rightarrow 6x + 6 = 180^{\circ} \Rightarrow 6x = 174^{\circ} \Rightarrow x = 29$. Smaller angle = $(3 \times 29 - 4)^{\circ} = 87 - 4 = 83^{\circ}$.

For ΔABC, ∠ECD = ∠ABC + ∠BAC (Exterior angle) = 50° + 30° = 80°. Now for ΔECD, ∠AED = ∠ECD + ∠CDE (Exterior angle) = 80° + 25° = 105°.

Let radius of the circle be r. Area of square S = 2r². Area of equilateral triangle T = (3√3 / 4) r². Thus T/S = (3√3 / 4) / 2 = 3√3 / 8 => 8T = 3√3 S.

$\angle BAC = 180^{\circ} - 65^{\circ} - 35^{\circ} = 80^{\circ}$. Since angles formed by the same chord on the circumference of a circle are equal, therefore $\angle BAC = \angle BDC = 80^{\circ}$.

Diameter $= 56$ cm, Radius $(r) = 28$ cm. Circumference of semicircle (arc) $= \pi r = \frac{22}{7} \times 28 = 88$ cm.

The sum of angles of a linear pair is $180^\circ$.

Using the formula $\angle BOC = 90^\circ + \frac{1}{2} \angle A$: $\angle BOC = 90^\circ + \frac{1}{2} \times 66^\circ = 90^\circ + 33^\circ = 123^\circ$.

Total salary = 24000 rupees. Central angle = 60°. Total value of pie chart is 360°. Expenditure on education = (60/360) * 24000 = (1/6) * 24000 = 4000 rupees.

Angle at the center $\angle BOD = 140^\circ$. The angle at the circumference $\angle BAD = \frac{1}{2} \angle BOD = \frac{1}{2} \times 140^\circ = 70^\circ$. From cyclic quadrilateral ABCD, $\angle DCB + \angle BAD = 180^\circ$. $\angle DCB + 70^\circ = 180^\circ \implies \angle DCB = 110^\circ$.

Using property of parallel lines, a line $EF \parallel AB \parallel CD$ is drawn. $\angle EOA = 80^{\circ}$ (alternate), $\angle EOC = 20^{\circ}$ (alternate), therefore $\angle AOC = 80^{\circ}+20^{\circ} = 100^{\circ}$.

In the given figure, $\angle A = \angle C = 75^\circ$, side $AB = CD = 3$ cm, and $\angle AEB = \angle CED$ (vertically opposite angles). Thus, Angle-Side-Angle (ASA) rule applies.

Length of tangent = $\sqrt{(\text{distance from center})^2 - (\text{radius})^2} = \sqrt{10^2 - 6^2} = \sqrt{100-36} = \sqrt{64} = 8$ cm.

Since $PQ \parallel RS$, the sum of their interior angles is $180^{\circ}$. $(3x-20) + (2x-10) = 180^{\circ} \Rightarrow 5x - 30 = 180^{\circ} \Rightarrow 5x = 210^{\circ} \Rightarrow x = 42$.

In the figure, $\triangle PNK \sim \triangle LMK$. Therefore, the ratio of corresponding sides will be equal: $\frac{PN}{LM} = \frac{NK}{MK} \implies \frac{a}{b} = \frac{x}{b+c} \implies x = \frac{ac}{b+c}$

For the construction of any quadrilateral, at least 5 independent components (like 4 sides and 1 diagonal or other combinations) are required.

Number of diagonals = $\frac{n(n-3)}{2}$ where $n=8$. $\frac{8(8-3)}{2} = \frac{8 \times 5}{2} = 20$.

$2\angle A = 3\angle B = 6\angle C = k$. $\angle A = k/2, \angle B = k/3, \angle C = k/6$. $\angle A + \angle B + \angle C = 180^\circ \implies k/2 + k/3 + k/6 = 180^\circ \implies \frac{3k+2k+k}{6} = 180^\circ \implies k = 180^\circ$. Therefore $\angle A = 90^\circ, \angle B = 60^\circ, \angle C = 30^\circ$.

The angle subtended at the center of the circle is double the angle subtended at the circumference. Therefore, $\angle AOB = 2 \times \angle ACB = 2 \times 40^\circ = 80^\circ$.

$\angle A - \angle B = 15^{\circ}$ (1); $\angle B - \angle C = 30^{\circ} \Rightarrow \angle C = \angle B - 30^{\circ}$ (2). In $\triangle ABC$, $\angle A + \angle B + \angle C = 180^{\circ} \Rightarrow (\angle B+15^{\circ}) + \angle B + (\angle B-30^{\circ}) = 180^{\circ} \Rightarrow 3\angle B = 195^{\circ} \Rightarrow \angle B = 65^{\circ}$. Thus $\angle A = 65^{\circ} + 15^{\circ} = 80^{\circ}$.

The sum of all angles formed at a point is $360^\circ$.