Q 03 Test 27 English

Test No. 27 – Complete Solutions

107

Convert 109.42° into degrees, minutes, and seconds.

🔑 Key Concept: Multiply the decimal part of degrees by 60 to get minutes. Multiply the decimal part of minutes by 60 to get seconds. \[ 1^\circ = 60' \qquad\qquad 1' = 60'' \]

Given: \( 109.42^\circ \)

Step 1 – Separate the whole degrees: Whole degrees \(= 109^\circ\),   Decimal part \(= 0.42^\circ\)

Step 2 – Convert decimal degrees to minutes: \[ 0.42 \times 60 = 25.2' \] Whole minutes \(= 25'\),   Decimal part \(= 0.2'\)

Step 3 – Convert decimal minutes to seconds: \[ 0.2 \times 60 = 12'' \]

\( \boxed{109.42^\circ = 109^\circ\ 25'\ 12''} \)

108

Convert degrees to radians:  (i) 15°    (ii) 15°15′

🔑 Key Concept: To convert degrees to radians, multiply by \(\dfrac{\pi}{180}\): \[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \] Because \( 180^\circ = \pi \) radians.

(i) Convert 15° to radians

Step 1 – Apply the formula: \[ 15^\circ \times \frac{\pi}{180} = \frac{15\pi}{180} \]

Step 2 – Simplify the fraction: \[ \frac{15\pi}{180} = \frac{\pi}{12} \]

\( \boxed{15^\circ = \dfrac{\pi}{12}\ \text{rad} \approx 0.2618\ \text{rad}} \)

(ii) Convert 15°15′ to radians

Step 1 – Convert 15°15′ to decimal degrees: \[ 15^\circ 15' = 15^\circ + \frac{15}{60}^\circ = 15^\circ + 0.25^\circ = 15.25^\circ \]

Step 2 – Apply the formula: \[ 15.25 \times \frac{\pi}{180} = \frac{15.25\pi}{180} \]

Step 3 – Simplify: \[ \frac{15.25\pi}{180} = \frac{61\pi}{720} \approx \frac{61 \times 3.14159}{720} \approx \frac{191.637}{720} \approx 0.2662\ \text{rad} \]

\( \boxed{15^\circ 15' = \dfrac{61\pi}{720}\ \text{rad} \approx 0.2662\ \text{rad}} \)

109

Find the midpoint of the line joining \(A(2,\,3)\) and \(B(8,\,7)\).

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

Given: \( A(x_1,\, y_1) = (2,\, 3) \)   and   \( B(x_2,\, y_2) = (8,\, 7) \)

Step 1 – Apply the midpoint formula: \[ M = \left(\frac{2 + 8}{2},\ \frac{3 + 7}{2}\right) \]

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

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

110

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

🔑 Key Concept – Ratio of Areas of Similar Triangles: For two similar triangles with corresponding sides \(l_1\) and \(l_2\) and areas \(A_1\) and \(A_2\): \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \] The ratio of areas equals the square of the ratio of corresponding sides.

Given: Figure 1:   \(l_1 = 10\ \text{cm},\quad A_1 = 40\ \text{cm}^2\)
Figure 2:   \(l_2 = 25\ \text{cm},\quad A_2 = ?\)

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

Step 2 – Substitute known values: \[ \frac{40}{A_2} = \left(\frac{10}{25}\right)^2 \]

Step 3 – Simplify the ratio and square it: \[ \frac{10}{25} = \frac{2}{5} = 0.4 \] \[ \frac{40}{A_2} = (0.4)^2 = 0.16 \]

Step 4 – Solve for \(A_2\): \[ A_2 = \frac{40}{0.16} = 250\ \text{cm}^2 \]

\( \boxed{A_2 = 250\ \text{cm}^2} \)