Q03 Test 37 English

Test No. 37 – Complete Solutions

147

Find the value of \(\tan\dfrac{\pi}{6} + \cot\dfrac{\pi}{6}\).

🔑 Key Concept – Standard Trig Values at \(\dfrac{\pi}{6}\) (30°): \[ \tan\frac{\pi}{6} = \tan 30° = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \] \[ \cot\frac{\pi}{6} = \cot 30° = \frac{\cos 30°}{\sin 30°} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \] Note: \(\cot\theta = \dfrac{1}{\tan\theta}\), so \(\cot 30° = \sqrt{3}\).

Step 1 – Substitute standard values: \[ \tan\frac{\pi}{6} + \cot\frac{\pi}{6} = \frac{1}{\sqrt{3}} + \sqrt{3} \]

Step 2 – Find common denominator: \[ = \frac{1}{\sqrt{3}} + \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}} = \frac{1}{\sqrt{3}} + \frac{3}{\sqrt{3}} = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]

Step 3 – Rationalise the denominator: \[ \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx \frac{4 \times 1.732}{3} \approx \frac{6.928}{3} \approx 2.309 \]

\( \boxed{\tan\dfrac{\pi}{6} + \cot\dfrac{\pi}{6} = \dfrac{4\sqrt{3}}{3} \approx 2.309} \)

148

Find slope and inclination of the line formed by points \((2,\,7)\) and \((3,\,-2)\).

🔑 Key Concept – Slope and Inclination: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] The inclination \(\alpha\) is the angle the line makes with the positive x-axis: \[ \alpha = \tan^{-1}(m) \qquad 0° \leq \alpha < 180° \] For a negative slope, the inclination is between 90° and 180°.

Given: \((x_1, y_1) = (2, 7)\),  \((x_2, y_2) = (3, -2)\)

Step 1 – Calculate the slope: \[ m = \frac{-2 - 7}{3 - 2} = \frac{-9}{1} = -9 \]

Step 2 – Find the inclination: \[ \alpha = \tan^{-1}(-9) \] Since \(m\) is negative, the line slopes downward. The reference angle is: \[ \tan^{-1}(9) \approx 83.66° \] Since slope is negative, the inclination is in the second quadrant: \[ \alpha = 180° - 83.66° \approx 96.34° \]

\( \boxed{m = -9 \qquad \alpha \approx 96.34°} \)

149

Find equation: Line passing through \(A(-6,\,5)\) with slope \(7\).

🔑 Key Concept – Point-Slope Form: When a point \((x_1, y_1)\) and slope \(m\) are known: \[ y - y_1 = m(x - x_1) \]

Given: Point \(A(x_1, y_1) = (-6, 5)\),  slope \(m = 7\)

Step 1 – Apply point-slope form: \[ y - 5 = 7\bigl(x - (-6)\bigr) \] \[ y - 5 = 7(x + 6) \]

Step 2 – Expand: \[ y - 5 = 7x + 42 \]

Step 3 – Rearrange to standard form: \[ y = 7x + 42 + 5 \] \[ y = 7x + 47 \] \[ 7x - y + 47 = 0 \]

\( \boxed{7x - y + 47 = 0} \)

150

Find the unknown quantities in the following figures (similar shapes).

🔑 Key Concept – Ratio of Areas of Similar Figures: \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \] Rearranging to find the unknown area: \[ A_1 = A_2 \times \left(\frac{l_1}{l_2}\right)^2 \]

Given: Shape 1: \(l_1 = 3.6\ \text{cm},\quad A_1 = ?\)
Shape 2: \(l_2 = 5.76\ \text{cm},\quad A_2 = 18\ \text{cm}^2\)

Step 1 – Write the formula: \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \]

Step 2 – Substitute known values: \[ \frac{A_1}{18} = \left(\frac{3.6}{5.76}\right)^2 \]

Step 3 – Simplify the ratio: \[ \frac{3.6}{5.76} = \frac{360}{576} = \frac{5}{8} = 0.625 \]

Step 4 – Square the ratio: \[ \left(\frac{5}{8}\right)^2 = \frac{25}{64} \]

Step 5 – Solve for \(A_1\): \[ A_1 = 18 \times \frac{25}{64} = \frac{450}{64} = \frac{225}{32} = 7.03125\ \text{cm}^2 \]

\( \boxed{A_1 = \dfrac{225}{32} \approx 7.03\ \text{cm}^2} \)