Q03 Test 32 English

Test No. 32 – Complete Solutions

127

Convert to decimal degrees:  (i) 65°32′15″    (ii) 42°18′45″

🔑 Key Concept – DMS to Decimal Degrees: \[ \text{Decimal Degrees} = \text{Degrees} + \frac{\text{Minutes}}{60} + \frac{\text{Seconds}}{3600} \] because \(1' = \dfrac{1}{60}°\) and \(1'' = \dfrac{1}{3600}°\).

(i) Convert 65°32′15″ to decimal degrees

Step 1 – Apply the formula: \[ 65 + \frac{32}{60} + \frac{15}{3600} \]

Step 2 – Calculate each term: \[ \frac{32}{60} = 0.5\overline{3}° \qquad \frac{15}{3600} = 0.004166...° \]

Step 3 – Add all terms: \[ 65 + 0.5333 + 0.0042 = 65.5375° \]

\( \boxed{65°\ 32'\ 15'' = 65.5375°} \)

(ii) Convert 42°18′45″ to decimal degrees

Step 1 – Apply the formula: \[ 42 + \frac{18}{60} + \frac{45}{3600} \]

Step 2 – Calculate each term: \[ \frac{18}{60} = 0.3° \qquad \frac{45}{3600} = 0.0125° \]

Step 3 – Add all terms: \[ 42 + 0.3 + 0.0125 = 42.3125° \]

\( \boxed{42°\ 18'\ 45'' = 42.3125°} \)

128

Convert to radians:  (i) 36°    (ii) 22.5°

🔑 Key Concept – Degrees to Radians: \[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \] Simplify the resulting fraction to its lowest terms for the exact answer.

(i) Convert 36° to radians

Step 1 – Apply the formula: \[ 36 \times \frac{\pi}{180} = \frac{36\pi}{180} \]

Step 2 – Simplify by dividing by 36: \[ \frac{36\pi}{180} = \frac{\pi}{5} \]

\( \boxed{36° = \dfrac{\pi}{5}\ \text{rad} \approx 0.6283\ \text{rad}} \)

(ii) Convert 22.5° to radians

Step 1 – Apply the formula: \[ 22.5 \times \frac{\pi}{180} = \frac{22.5\pi}{180} \]

Step 2 – Simplify (multiply numerator and denominator by 2 first): \[ \frac{22.5\pi}{180} = \frac{45\pi}{360} = \frac{\pi}{8} \]

\( \boxed{22.5° = \dfrac{\pi}{8}\ \text{rad} \approx 0.3927\ \text{rad}} \)

129

Find the distance between \(L(0,\,3)\) and \(M(-2,\,-4)\).

🔑 Key Concept – Distance Formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Given: \(L(x_1, y_1) = (0, 3)\),  \(M(x_2, y_2) = (-2, -4)\)

Step 1 – Apply the distance formula: \[ d = \sqrt{(-2 - 0)^2 + (-4 - 3)^2} \]

Step 2 – Compute and square the differences: \[ d = \sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} \]

Step 3 – Evaluate: \[ d = \sqrt{53} \approx 7.28\ \text{units} \]

\( \boxed{d = \sqrt{53} \approx 7.28\ \text{units}} \)

130

Find the midpoint of the line joining \(A(8,\,3)\) and \(B(2,\,-1)\).

🔑 Key Concept – Midpoint Formula: \[ M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right) \] The midpoint is the average of the x-coordinates and the average of the y-coordinates.

Given: \(A(x_1, y_1) = (8, 3)\),  \(B(x_2, y_2) = (2, -1)\)

Step 1 – Apply the midpoint formula: \[ M = \left(\frac{8 + 2}{2},\ \frac{3 + (-1)}{2}\right) \]

Step 2 – Simplify each coordinate: \[ M = \left(\frac{10}{2},\ \frac{2}{2}\right) = (5,\ 1) \]

\( \boxed{M = (5,\ 1)} \)