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Latitude of P = 15ºN; Latitude of Q = 65ºN P and Q both lies in northern hemisphere Latitudinal difference = 65º - 15º = 50º

PQ = circumference of the base of the cone = 2\pir PQ = \(\frac{120}{360}\) \times 2\pir = \(\frac{120}{360}\) \times 2 \times \(\frac{22}{7}\) \times 21 = 44 cm ∴ 44 = 2\pir; 44 = 2 \times \(\frac{22}{7}\) \times r 44 = \(\frac{44}{7}\) r r = 7 cm OR r = \(\frac{R\theta}{360}\) = \(\frac{21 \times 120}{360}\) = 7cm.

Volume of tin = \pir\(^2\)h = \(\frac{22}{7}\) \times 7\(^2\) \times 5 = 770 cm\(^3\)

Since OA is ⊥ to OB ∴ ∠ AOB = 90º. Hence, 4x + 3x + 2x + 90º = 360º (sum of angles at a point) 9x + 90º = 360º 9x = 270º; x = 30º

If P and E are points on CD and AB, respectively ∠ CPE = 52^\circ (corresponding angles are equal) ∴ ∠ TPD = 52^\circ (vertically opposite angles are equal) ∴ 52^\circ + x^\circ + 35^\circ = 180^\circ (sum of angles in a triangle) ∴ x^\circ + 87^\circ = 180^\circ x = 93^\circ

∠BFC = 180^\circ - a^\circ (adjacent angles on a straight line) ∴ a^\circ + b^\circ + c^\circ + ∠BFC = 360^\circ (quadrilateral) a^\circ + b^\circ + c^\circ + (180^\circ - a^\circ) = 360^\circ b^\circ + c^\circ = 360^\circ - (180^\circ - a^\circ) - a^\circ b^\circ + c^\circ = 360^\circ - 180^\circ + a^\circ - a^\circ b^\circ + c^\circ = 180^\circ

To calculate the interior angle of a regular polygon, we use the formula: \(\text{Interior Angle} = \frac{(n - 2) \times 180^\circ}{n}\) where \( n \) is the number of sides. For a 5-sided polygon (pentagon): \(\text{Interior Angle} = \frac{(5 - 2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ\) Thus, the interior angle of a regular pentagon is \( 108^\circ \).

The triangle XYZ is a right-angled triangle at Y, with opposite side to \theta (YZ) = 11 cm and hypotenuse (XZ) = 13 cm. The length of XY = \(\sqrt{(13^2 - 11^2)}\) = \(\sqrt{(169 - 121)}\) = \(\sqrt{48}\) cm (or 4\(\sqrt{3}\) cm in simplified form). Cot \theta = \(\frac{\text{adjacent}}{\text{opposite}}\) = \(\frac{\text{XY}}{\text{YZ}}\) = \(\frac{\sqrt{48}}{11}\)

\(\frac{360^\circ}{\text{n}}\) must be a positive integer \(\frac{360^\circ}{72^\circ}\) = 5

∠BAC = 90^\circ ∠ACD = 90^\circ if AE and CE are the bisectors of ∠BAC and ∠ACD respectively ∠BAE = 45^\circ; ∠DCE = 45^\circ ∴ ∠EAC = ∠ECA = 45^\circ But ∠EAC + ∠ACE + ∠AEC = 180^\circ (angle in a △) ∴ 45^\circ + 45^\circ + ∠AEC = 180^\circ ∠AEC = 180^\circ - 90^\circ ∠AEC = 90^\circ

∴ Triangle ABC is isosceles. :∠ACB = ∠ABC ∠ACD = 180^\circ - ∠ACB (angle on a straight line) Let ∠ACB = ∠ABC = x .: 2x + 50 = 180^\circ (sum of angles in a triangle) 2x = 130^\circ x = 65^\circ .: ∠ACB = 65^\circ ==> ∠ACD = 180^\circ - 65^\circ = 115^\circ

From the figure, sin 30^\circ = \(\frac{\text{y}}{10}\) ∴ 10 \times sin 30^\circ = y y = (10 \times \(\frac{1}{2}\)) = 5 cm

In classical constructions with compass and straightedge, 60^\circ (equilateral), 90^\circ (square), and 135^\circ (90^\circ + 45^\circ) are constructible. 145^\circ is not constructible in general (not a multiple resulting from bisections of constructible angles).

Locus theorem states that: The locus of a point that moves so that it is equidistant from two given fixed points is the perpendicular bisector of the straight line joining them.

a. EXPLANATION FOR CONSTRUCTION i. Measure AB = 7.5 cm with a compass from a ruler. ii. Make an arc with this compass, placing the needle at point B iv. Construct an angle of 105^\circ at B v. Join the angle until it touches the previous arc made with a length of 8.1 cm vi. Call this point of intersection C, then join CA vii. Draw a line BQ to any length viii. Divide BQ into 5 equal parts with a compass ix. Join the fifth division to C and draw a line parallel to this to join the third division to line BC, and call the point where they meet D. x. At D, construct angle 90^\circ, the point where the line of construction of 90^\circ meets AC, call it "P" xi. Measure BP b. c. d |BP| = 5.5cm.