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a + c + d = 180o[sum of angles in triangle] x = a[vertical opp. ∠s] y + x + d = 180o[sum of ∠s in a \Delta] y + a + d = 180o y + a + d = a + c + d y = c but b + y = 180o [sum of ∠s on a straight line ] b + c = 180o; Ans = A = 180o

The formula for the exterior angle of a regular polygon = \(\frac{360}{n}\). Among the options, only 72^\circ is divisible by 360^\circ to give an integer.

ACB = ABC [Base angles of isoceles triangle] ACB + ABC + 50o = 180o[sum of angles in a triangle] ACB + ABC = 180o - 50o = 130o ACB 130o/2 = 65o ACD + ACB = 180o[sum of ∠s on a straight line] ACD + 65o = 180o ACD = 180o - 65o = 115o

Note: sin 20^\circ = cos 70^\circ hence \(\frac{\sin 20}{\cos 70} = 1\) Also, cos 65^\circ = sin 25^\circ \(\frac{\cos 65}{\sin 25} = 1\) \(\therefore \frac{\sin 20}{\cos 70} + \frac{\cos 65}{\sin 25}\) = 1 + 1 = 2

\(\sin 30^\circ = \frac{y}{10}\) \(y = 10 \sin 30^\circ = 10 \times 0.5 = 5 cm\)

x\(^2\) = 12\(^2\) + 5\(^2\) x\(^2\) = 144 + 25 x\(^2\) = 169 x = \(\sqrt{169}\) = 13m.

The mode = 8 with the highest frequency of 4 times.

Mean of 20 observations = 4 All the observations = 20 x 4 = 80 Largest observed value = 23 Remaining observation = 80 - 23 = 57 Mean of observation = 57/19 = 3

Arranging the numbers in ascending order, we have 2.35, 2.50, 2.64, 2.72, 2.91. The median = 2.64

at least 10 => ≥ 10 6/36 = 1/6

2r, 6w, 5b balls Total balls = 2 + 6 + 5 = 13 balls p(black) = 5/13

\(3310_{5} - 1442_{5} = 1313_{5}\) You can either do a direct subtraction (which is easier), or convert both numbers in base 10, subtract and convert back to base 5 (if you're not sure).

3.1415926 \(\approxeq\) 3.14159 (to 5 d.p)

Error = (16.8cm - 15cm = 1.8cm\) \(\text{% error} = \frac{1.8}{15} \times 100% = \frac{180}{15}\) = \(12.00%\)

Present salary = N24,000. Increment after first year = \(\frac{10}{100} \times N24,000 = N2,400\) Salary at the beginning of second year = N(24,000 + 2,400) = N26,400. Increment after second year = \(\frac{10}{100} \times N26,400 = N2,640\) Salary at the beginning of third year = N(26,400 + 2,640) = N29,040. \(\therefore\) His salary at the beginning of the third year = N29,040.