Q03 Test 35 English

Test No. 35 – Complete Solutions

139

Find the value of \(2\cos\dfrac{\pi}{6}\sin\dfrac{\pi}{6}\).

🔑 Key Concept – Standard Trig Values & Double Angle Identity: \[ \cos\frac{\pi}{6} = \cos 30° = \frac{\sqrt{3}}{2} \qquad \sin\frac{\pi}{6} = \sin 30° = \frac{1}{2} \] Recognise the double angle identity: \[ 2\sin\theta\cos\theta = \sin 2\theta \] So \(2\cos\dfrac{\pi}{6}\sin\dfrac{\pi}{6} = \sin\!\left(2 \times \dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{3}\).

Method 1 – Direct substitution: \[ 2\cos\frac{\pi}{6}\sin\frac{\pi}{6} = 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} \]

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

Method 2 – Using double angle identity: \[ 2\cos\frac{\pi}{6}\sin\frac{\pi}{6} = \sin\!\left(\frac{2\pi}{6}\right) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \]

\( \boxed{2\cos\dfrac{\pi}{6}\sin\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2} \approx 0.866} \)

140

Write the equation of the line cutting x-axis at \((2,\,0)\) and y-axis at \((0,\,-4)\).

🔑 Key Concept – Intercept Form of a Line: When x-intercept \(= a\) and y-intercept \(= b\) are known, use the intercept form: \[ \frac{x}{a} + \frac{y}{b} = 1 \] Alternatively use the two-point slope method or find slope then y-intercept.

Given: x-intercept: \((2, 0)\) so \(a = 2\)
y-intercept: \((0, -4)\) so \(b = -4\)

Step 1 – Apply intercept form: \[ \frac{x}{2} + \frac{y}{-4} = 1 \]

Step 2 – Multiply through by 4 (LCM of 2 and 4): \[ 2x - y = 4 \]

Step 3 – Write in standard form: \[ 2x - y - 4 = 0 \]

Check – Slope from two points \((2,0)\) and \((0,-4)\): \[ m = \frac{-4 - 0}{0 - 2} = \frac{-4}{-2} = 2 \qquad \checkmark \]

\( \boxed{2x - y - 4 = 0} \)

141

A bag of rice has height 60 cm and weight 50 kg. Find the weight of rice in a similar bag of height 90 cm.

🔑 Key Concept – Ratio of Volumes (Weights) of Similar Solids: For two similar 3D objects with corresponding lengths \(l_1\) and \(l_2\), their volumes (and hence weights) scale as the cube of the ratio of lengths: \[ \frac{W_1}{W_2} = \left(\frac{l_1}{l_2}\right)^3 \] This is because volume is a three-dimensional measure.

Given: Bag 1: height \(l_1 = 60\ \text{cm}\),  weight \(W_1 = 50\ \text{kg}\)
Bag 2: height \(l_2 = 90\ \text{cm}\),  weight \(W_2 = ?\)

Step 1 – Write the formula: \[ \frac{W_1}{W_2} = \left(\frac{l_1}{l_2}\right)^3 \]

Step 2 – Substitute known values: \[ \frac{50}{W_2} = \left(\frac{60}{90}\right)^3 \]

Step 3 – Simplify the ratio: \[ \frac{60}{90} = \frac{2}{3} \qquad \left(\frac{2}{3}\right)^3 = \frac{8}{27} \]

Step 4 – Solve for \(W_2\): \[ \frac{50}{W_2} = \frac{8}{27} \] \[ W_2 = \frac{50 \times 27}{8} = \frac{1350}{8} = 168.75\ \text{kg} \]

\( \boxed{W_2 = 168.75\ \text{kg}} \)

142

Find the equation of the horizontal line passing through \((-9,\,7)\).

🔑 Key Concept – Horizontal Line: A horizontal line has a slope of zero and every point on it has the same y-coordinate. Its equation is simply: \[ y = c \] where \(c\) is the y-coordinate of any point on the line.

Given: Point \((-9,\ 7)\) lies on the line.

Step 1 – Identify the y-coordinate: The y-coordinate of the given point is \(7\).

Step 2 – Write the equation: Since the line is horizontal, every point has \(y = 7\): \[ y = 7 \]

\( \boxed{y = 7} \)