(a) Given that \(\sin x = 0.6, 0° \leq x \leq 90°\), evaluate \(2\cos x + 3\sin x\), leaving your answer in the form \(\frac{m}{n}\), where m and n are inte...
Question 1 Report
(a) Given that \(\sin x = 0.6, 0° \leq x \leq 90°\), evaluate \(2\cos x + 3\sin x\), 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 \(\pi = \frac{22}{7}\)].
Answer Details
(a) Evaluating \(2\cos x + 3\sin x\)
Given \(\sin x = 0.6 = \dfrac{3}{5}\) with \(0^\circ \le x \le 90^\circ\), x is acute, so \(\cos x\) is positive.
\[\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - (0.6)^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 = \frac{4}{5}\]
Then
\[2\cos x + 3\sin x = 2\left(\frac{4}{5}\right) + 3\left(\frac{3}{5}\right) = \frac{8}{5} + \frac{9}{5} = \frac{17}{5}\]
(b) Semi-circle inscribed in isosceles triangle ABC
From the diagram, the diameter WZ lies along the top side AB, the centre O is on AB, and the semicircular arc is tangent to the two equal sides. The apex C is below, with \(|OC| = 30\ \text{cm}\) measured along the axis of symmetry, and \(\angle ACB = 130^\circ\).
By symmetry (AC = BC), the line CO bisects angle C, so the half-angle at C is
\[\frac{130^\circ}{2} = 65^\circ\]
(i) Radius of the circle. The radius drawn to the point where the arc touches a side is perpendicular to that side. In the right-angled triangle formed by O, C and the tangent point, OC is the hypotenuse and the radius r is opposite the \(65^\circ\) angle:
\[r = |OC|\sin 65^\circ = 30 \times 0.9063 = 27.19\]
\[r \approx 27.2\ \text{cm}\]
(ii) Area of the shaded portion. The shaded region is the part of triangle ABC lying outside the semicircle:
\[\text{Shaded} = \text{Area of } \triangle ABC - \text{Area of semicircle}\]
The centre O lies on AB, so OC = 30 cm is the height of the triangle from C to AB. The half-base is
\[\frac{1}{2}|AB| = 30\tan 65^\circ = 30 \times 2.1445 = 64.34\ \text{cm}\]
\[|AB| = 128.67\ \text{cm}\]
\[\text{Area of } \triangle ABC = \frac{1}{2}\times |AB| \times \text{height} = \frac{1}{2}\times 128.67 \times 30 = 1930.1\ \text{cm}^2\]
Area of the semicircle, with \(\pi = \dfrac{22}{7}\) and \(r = 27.2\):
\[\frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7}\times (27.2)^2 = \frac{1}{2}\times \frac{22}{7}\times 739.84 = 1162.6\ \text{cm}^2\]
Therefore
\[\text{Shaded area} = 1930.1 - 1162.6 = 767.5\ \text{cm}^2\]
Download The App On Google Playstore
Everything you need to excel in your exams
