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B = Bird ; H = Hunter. Bearing of the hunter from the bird = S 72^\circ W.

1 + sec2 30o = sec 30o = \(\frac{2}{\sqrt{3}}\) \(\frac{(2)^2}{3}\) = \(\frac{4}{3}\) 1 + sec2 30o = sec 30o = 1 + \(\frac{4}{3}\) = 2\(\frac{1}{3}\)

Area of a regular hexagon = 8 x 8 x sin 60o = 32 x \(\frac{\sqrt{3}}{2}\) Area = 16\(\sqrt{3}\) x 6 = 96 \(\sqrt{3}\)cm2

\(\frac{\cot (90 - \theta)}{sin^2\theta}\) \(\cot (90 - \theta) = \tan \theta\) \(\therefore \frac{\cot (90 - \theta)}{\sin^{2} \theta} = \frac{\tan \theta}{\sin^{2} \theta}\) \(\tan \theta = \frac{\sqrt{3}}{3}\) \(\sin \theta = \frac{1}{2} \implies \sin^{2} \theta = \frac{1}{4}\) \(\frac{\cot(90 - \theta)}{\sin^{2} \theta} = \frac{\sqrt{3}}{3}\div\frac{1}{4}\) = \(\frac{4}{\sqrt{3}}\)

Length of an area = \(\frac{\theta}{360}\) x 2\(\pi\)r Longitude difference = 147 + 153 = 30NoN distance between xy = \(\frac{\theta}{360}\) x 2\(\pi\)R cos60o = \(\frac{300}{360}\) x 4 x 104 x \(\frac{1}{2}\) = 1.667 x 104km(1667 km)

Total students = 120 grade = 18 Zo = \(\frac{18}{120}\) x \(\frac{360}{1}\) = 54o

By Pythagoras, KZ2 = 252 - 52 KZ2 = (25 + 15)(25 - 15) = 400 KZ = \(\sqrt{400}\) = 20 area of XYZ = \(\frac{1}{2} \times 28 \times 15\) = 210 sq. cm

< PRQ = \(\frac{1}{2}\) < POQ = 90o < PSR + < PQR = 180o < PQR = 180o - 145o = 35o \(\bigtriangleup\)PQR is a right angled triangle x = 90 - < PQR = 90o - 35o = 55o

180 - \(\phi\)o = \(\theta\)o + \(\phi\)o (Sum of opposite interior angle equal to its exterior angle) 180 = 2\(\phi\) + \(\theta\)o

Area of sector = \(\frac{120}{360} \times \pi \times (3)^2 = 3 \pi\) Area of triangle = \(\frac{1}{2} \times 3 \times 3 \times \sin 120^o\) = \(\frac{9}{2} \times \frac{\sqrt{3}}{2} = \frac{9\sqrt {3}}{4}\) Area of shaded portion = 3\(\pi - \frac{9\sqrt {3}}{4}\) = 3 (\(\pi - 3 \frac{\sqrt{3}}{4}\))

In the figure, angle x = 20o + 80o + 40o = 140o

\(\frac{\sin y}{3} = \frac{\sin 120^o}{5}\) sin 120o = sin 60o 5 sin y = 3 sin 60o sin y = \(\frac{3 \sin 60^o}{5}\) \(\frac{3 \times 0.866}{5}\) = \(\frac{2.598}{5}\) y = sin-1 0.5196 = 30o 18'

\(\frac{0.0001432}{1940000}\) = k x 10n where 1 \(\leq\) k \(\leq\) 10 and n is a whole number. Using four figure tables, the eqn. gives 7.38 x 10-11 k = 7.381, n = -11

Let the selling price(SP from P to Q be represented by x i.e. SP = x When SP = x at 10% profit CP = \(\frac{100}{100}\) + 10 of x = \(\frac{100}{110}\) of x when Q sells to R, SP = N209 at loss of 5% Q's cost price = Q's selling price CP = \(\frac{100}{95}\) x 209 = 220.00 x = 220 = \(\frac{2200}{11}\) = 200 = N200.00

Total cost of production = N120.00 Labour Cost = \(\frac{70}{120}\) x \(\frac{360^o}{1}\) = 210o