(a) Given that (sin x = 0.6, 0° leq x leq 90°), evaluate (2cos x + 3sin x), leaving your answer in the form (frac{m}{n}), where m and n are integers.
(b)
In the diagram, a semi-circle WXYZ with centre O is inscribed in an isosceles triangle ABC. If /AC/ = /BC/, |OC| = 30 cm and < ACB = 130°, calculate, correct to one decimal place, the (i) radius of the circle ; (ii) area oc the shaded portion. [Take (pi = frac{22}{7})].
Explanation
Given (sin x = 0.6 = frac{3}{5})
(therefore Opp = 3; Hyp = 5)
Using Pythagoras theorem,
(5^{2} = 3^{2} + Adj^{2})
(Adj = sqrt{5^{2} – 3^{2}})
(Adj = sqrt{25 – 9} = sqrt{16} = 4)
(therefore cos x = frac{4}{5} = 0.8)
(2cos x + 3sin x = 2(frac{4}{5}) + 3(frac{3}{5}))
= (frac{8}{5} + frac{9}{5})
= (frac{17}{5}).
(b)
Join O to X and O to Y.
Considering either (Delta OXC) or (Delta OYC),
(sin 65 = frac{r}{30})
(r = 30 sin 65)
(r = 30 times 0.9063)
(r = 27.189 cm approxeq 27.2 cm) (to one decimal place)
(b) Area of shaded portion = Area of triangle ABC – Area of semi-circle WXYZ.
= (frac{1}{2} bh – frac{1}{2} pi r^{2})
= ((frac{1}{2} times 2c times 30) – (frac{1}{2} times frac{22}{7} times (27.189)^{2}))
From (Delta COA, tan 65 = frac{c}{30})
(implies c = 30 tan 65 approxeq 64.34 cm)
(therefore) Area of shaded portion = (30(64.34) – frac{8131.659}{7})
= (1930.056 – 1161.666)
= (768.39 cm^{2})