Q03 Test 34 English

Test No. 34 – Complete Solutions

135

Find arc length and area given \(r = \dfrac{4.8}{\pi}\) cm, \(\theta = \dfrac{5\pi}{6}\) rad.

🔑 Key Concept – Arc Length and Area of a Sector: \[ \ell = r\theta \qquad\qquad A = \frac{1}{2}r^2\theta \] \(\theta\) is already in radians — no conversion needed.

Given: \(r = \dfrac{4.8}{\pi}\ \text{cm}\),  \(\theta = \dfrac{5\pi}{6}\ \text{rad}\)

Part A – Arc Length \(\ell\)

Step 1 – Apply the formula: \[ \ell = r\theta = \frac{4.8}{\pi} \times \frac{5\pi}{6} \]

Step 2 – Cancel \(\pi\) and simplify: \[ \ell = \frac{4.8 \times 5}{6} = \frac{24}{6} = 4\ \text{cm} \]

\( \boxed{\ell = 4\ \text{cm}} \)

Part B – Area of Sector \(A\)

Step 1 – Apply the formula: \[ A = \frac{1}{2}r^2\theta = \frac{1}{2} \times \left(\frac{4.8}{\pi}\right)^2 \times \frac{5\pi}{6} \]

Step 2 – Expand \(r^2\): \[ A = \frac{1}{2} \times \frac{23.04}{\pi^2} \times \frac{5\pi}{6} \]

Step 3 – Cancel one \(\pi\) and simplify: \[ A = \frac{1}{2} \times \frac{23.04 \times 5}{6\pi} = \frac{115.2}{12\pi} = \frac{9.6}{\pi} \]

Step 4 – Evaluate: \[ A = \frac{9.6}{3.14159} \approx 3.056\ \text{cm}^2 \]

\( \boxed{A = \dfrac{9.6}{\pi} \approx 3.056\ \text{cm}^2} \)

136

Find the value of \(2\sin 60°\cos 60°\).

🔑 Key Concept – Standard Trigonometric Values & Double Angle Identity: \[ \sin 60° = \frac{\sqrt{3}}{2} \qquad \cos 60° = \frac{1}{2} \] Also recognise the double angle identity: \[ 2\sin\theta\cos\theta = \sin 2\theta \] So \(2\sin 60°\cos 60° = \sin 120°\).

Method 1 – Direct substitution: \[ 2\sin 60°\cos 60° = 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} \]

Step 2 – Simplify: \[ = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \]

Method 2 – Using double angle identity: \[ 2\sin 60°\cos 60° = \sin(2 \times 60°) = \sin 120° = \sin(180° - 60°) = \sin 60° = \frac{\sqrt{3}}{2} \]

\( \boxed{2\sin 60°\cos 60° = \dfrac{\sqrt{3}}{2} \approx 0.866} \)

137

Find the distance between \(Q(0,\,0)\) and \(P(-8,\,-7)\).

🔑 Key Concept – Distance Formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] When one point is the origin \((0, 0)\), this simplifies to: \[ d = \sqrt{x^2 + y^2} \]

Given: \(Q(x_1, y_1) = (0, 0)\),  \(P(x_2, y_2) = (-8, -7)\)

Step 1 – Apply the distance formula: \[ d = \sqrt{(-8-0)^2 + (-7-0)^2} \]

Step 2 – Compute and square: \[ d = \sqrt{(-8)^2 + (-7)^2} = \sqrt{64 + 49} = \sqrt{113} \]

Step 3 – Evaluate: \[ d = \sqrt{113} \approx 10.63\ \text{units} \]

\( \boxed{d = \sqrt{113} \approx 10.63\ \text{units}} \)

138

Find the equation of the line passing through \((2,\,1)\) and \((-4,\,6)\).

🔑 Key Concept – Equation of a Line Through Two Points: First find the slope \(m\): \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Then use the point-slope form: \[ y - y_1 = m(x - x_1) \]

Given: \((x_1, y_1) = (2, 1)\),  \((x_2, y_2) = (-4, 6)\)

Step 1 – Find the slope \(m\): \[ m = \frac{6 - 1}{-4 - 2} = \frac{5}{-6} = -\frac{5}{6} \]

Step 2 – Apply point-slope form using \((2, 1)\): \[ y - 1 = -\frac{5}{6}(x - 2) \]

Step 3 – Expand the right side: \[ y - 1 = -\frac{5}{6}x + \frac{10}{6} \] \[ y - 1 = -\frac{5}{6}x + \frac{5}{3} \]

Step 4 – Rearrange to standard form (multiply through by 6): \[ 6y - 6 = -5x + 10 \] \[ 5x + 6y - 16 = 0 \]

\( \boxed{5x + 6y - 16 = 0} \)