Q03 Test 28 English

Test No. 28 – Complete Solutions

111

Convert 45°45′45″ into decimal degrees.

🔑 Key Concept: To convert DMS (Degrees, Minutes, Seconds) to decimal degrees: \[ \text{Decimal Degrees} = \text{Degrees} + \frac{\text{Minutes}}{60} + \frac{\text{Seconds}}{3600} \] because \(1' = \dfrac{1}{60}^\circ\) and \(1'' = \dfrac{1}{3600}^\circ\).

Given: \( 45^\circ\ 45'\ 45'' \)

Step 1 – Write the conversion formula: \[ \text{Decimal Degrees} = 45 + \frac{45}{60} + \frac{45}{3600} \]

Step 2 – Calculate each term: \[ \frac{45}{60} = 0.75^\circ \] \[ \frac{45}{3600} = 0.0125^\circ \]

Step 3 – Add all terms: \[ 45 + 0.75 + 0.0125 = 45.7625^\circ \]

\( \boxed{45^\circ\ 45'\ 45'' = 45.7625^\circ} \)

112

Convert radians to degrees:  (i) \(\dfrac{7\pi}{6}\) rad    (ii) \(\dfrac{5\pi}{3}\) rad

🔑 Key Concept: To convert radians to degrees, multiply by \(\dfrac{180^\circ}{\pi}\): \[ \theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180^\circ}{\pi} \] When the angle contains \(\pi\), it cancels out giving an exact result.

(i) Convert \(\dfrac{7\pi}{6}\) rad to degrees

Step 1 – Apply the formula: \[ \frac{7\pi}{6} \times \frac{180^\circ}{\pi} \]

Step 2 – Cancel \(\pi\) and simplify: \[ = \frac{7 \times 180^\circ}{6} = \frac{1260^\circ}{6} = 210^\circ \]

\( \boxed{\dfrac{7\pi}{6}\ \text{rad} = 210^\circ} \)

(ii) Convert \(\dfrac{5\pi}{3}\) rad to degrees

Step 1 – Apply the formula: \[ \frac{5\pi}{3} \times \frac{180^\circ}{\pi} \]

Step 2 – Cancel \(\pi\) and simplify: \[ = \frac{5 \times 180^\circ}{3} = \frac{900^\circ}{3} = 300^\circ \]

\( \boxed{\dfrac{5\pi}{3}\ \text{rad} = 300^\circ} \)

113

Find the distance between \(C(-4,\,-2)\) and \(D(0,\,9)\).

🔑 Key Concept – Distance Formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Square the differences of both coordinates, add them, then take the square root.

Given: \( C(x_1,\, y_1) = (-4,\, -2) \)   and   \( D(x_2,\, y_2) = (0,\, 9) \)

Step 1 – Apply the distance formula: \[ d = \sqrt{(0 - (-4))^2 + (9 - (-2))^2} \]

Step 2 – Compute each difference and square it: \[ d = \sqrt{(4)^2 + (11)^2} \] \[ d = \sqrt{16 + 121} \] \[ d = \sqrt{137} \]

Step 3 – Evaluate: \[ d = \sqrt{137} \approx 11.70\ \text{units} \]

\( \boxed{d = \sqrt{137} \approx 11.70\ \text{units}} \)

114

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

🔑 Key Concept – Ratio of Areas of Similar Figures: For two similar figures 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 (pentagon SRQOP):   \(l_1 = PQ = 35\ \text{cm},\quad A_1 = ?\)
Figure 2 (pentagon WXYZ):   \(l_2 = XY = 25\ \text{cm},\quad A_2 = 98\ \text{cm}^2\)

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

Step 2 – Substitute known values: \[ \frac{A_1}{98} = \left(\frac{35}{25}\right)^2 \]

Step 3 – Simplify the ratio: \[ \frac{35}{25} = \frac{7}{5} = 1.4 \]

Step 4 – Square the ratio: \[ \frac{A_1}{98} = \left(\frac{7}{5}\right)^2 = \frac{49}{25} = 1.96 \]

Step 5 – Solve for \(A_1\): \[ A_1 = 1.96 \times 98 = 192.08\ \text{cm}^2 \]

\( \boxed{A_1 = \dfrac{49}{25} \times 98 = \dfrac{4802}{25} = 192.08\ \text{cm}^2} \)