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Angular distance = 40o + 20o = 60o using θ/360 x 2πr for distance between p and Q = 60/360 x 2πRcos4o = 1/3πRcos4o

by using trial and error Y = sinx since \(\pi\) in radian = 180º

Sin\(\theta\) = -0.9063; \(\theta\) = sin-1(0.9063) \(\theta\) = -64.99 = -65^\circ \(\theta\) = 180^\circ - (-65); 180 + 65 = 245^\circ

x/25 = tan 30o x = 25/√3 = 25√3/3

x/25 = tan 30o x = 39 x 1/√3 = 13√3m

From the figure, < ATO = 90^\circ < AOT = 180^\circ - 145^\circ = 35^\circ \(\therefore\) < TAO = 180^\circ - (90^\circ + 35^\circ) = 55^\circ (sum of angle in a triangle)

Since X and Y are mutually exclusive, Pr(X ∩ Y) = 0

No expanded working has been attached for this question yet. The verified answer is Option A.

Sum of ∠s in a pentagon = (n - 2)180 = 540^\circ x^\circ + 2x^\circ + x^\circ + 60^\circ + x^\circ + 10^\circ + x^\circ - 10^\circ = 540^\circ 6x^\circ + 60^\circ = 540^\circ; x = 80^\circ

Since < URP = 25^\circ, then < XPU = 4 x 25^\circ = 100^\circ \(\therefore\) < TPQ = 180^\circ - 100^\circ = 80^\circ \(\therefore\) < PQT = 180^\circ - (80^\circ + 25^\circ) = 75^\circ < SQR = 75^\circ - 25^\circ = 50^\circ (exterior angle = 2 opp interior angles) \(\therefore\) < USQ = 180^\circ - 50^\circ = 130^\circ

Area = 1/2 x 16 x 16sin60o = 64√3cm2

Area of \(\Delta\)DCF = 24cm2 Area of Quad= 2 x 24 = 48cm2

Given (-2, 0) and (0, -4) Gradient = \(\frac{-4 - 0}{0 - (-2)}\) = \(\frac{-4}{2}\) = -2

\(\frac{x}{4}\) = \(\frac{\tan 60}{1}\) x = 4 tan60 = 4\(\sqrt{3}\) BD = 2x = 8\(\sqrt{3}\)

The sum of the exterior angles of a regular polygon = 360^\circ. \(\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360^\circ\) \(5x + 50 = 360^\circ \implies 5x = 360^\circ - 50^\circ = 310^\circ\) \(x = \frac{310^\circ}{5} = 62^\circ\)