Q 03 Test 29 English

Test No. 29 – Complete Solutions

115

Convert 94°27′54″ into decimal degrees.

🔑 Key Concept: To convert DMS 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: \( 94^\circ\ 27'\ 54'' \)

Step 1 – Write the conversion formula: \[ \text{Decimal Degrees} = 94 + \frac{27}{60} + \frac{54}{3600} \]

Step 2 – Calculate each term: \[ \frac{27}{60} = 0.45^\circ \] \[ \frac{54}{3600} = 0.015^\circ \]

Step 3 – Add all terms: \[ 94 + 0.45 + 0.015 = 94.465^\circ \]

\( \boxed{94^\circ\ 27'\ 54'' = 94.465^\circ} \)

116

Convert degrees to radians:  (i) 75°    (ii) 315°

🔑 Key Concept: To convert degrees to radians, multiply by \(\dfrac{\pi}{180}\): \[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \] Simplify the fraction as much as possible to get the exact result.

(i) Convert 75° to radians

Step 1 – Apply the formula: \[ 75 \times \frac{\pi}{180} = \frac{75\pi}{180} \]

Step 2 – Simplify by dividing numerator and denominator by 15: \[ \frac{75\pi}{180} = \frac{5\pi}{12} \]

\( \boxed{75^\circ = \dfrac{5\pi}{12}\ \text{rad} \approx 1.3090\ \text{rad}} \)

(ii) Convert 315° to radians

Step 1 – Apply the formula: \[ 315 \times \frac{\pi}{180} = \frac{315\pi}{180} \]

Step 2 – Simplify by dividing numerator and denominator by 45: \[ \frac{315\pi}{180} = \frac{7\pi}{4} \]

\( \boxed{315^\circ = \dfrac{7\pi}{4}\ \text{rad} \approx 5.4978\ \text{rad}} \)

117

Point \(P(-5,\,3)\) lies on a circle with centre \(C(-2,\,7)\). Find the radius of the circle.

🔑 Key Concept – Radius of a Circle: If a point \(P\) lies on a circle with centre \(C\), then the radius \(r\) is simply the distance from the centre \(C\) to the point \(P\): \[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] This is just the distance formula applied between the centre and the point on the circle.

Given: Centre \( C(x_1,\, y_1) = (-2,\, 7) \),   Point on circle \( P(x_2,\, y_2) = (-5,\, 3) \)

Step 1 – Apply the distance formula: \[ r = \sqrt{(-5-(-2))^2 + (3-7)^2} \]

Step 2 – Simplify inside the root: \[ r = \sqrt{(-5+2)^2 + (-4)^2} \] \[ r = \sqrt{(-3)^2 + (-4)^2} \] \[ r = \sqrt{9 + 16} \] \[ r = \sqrt{25} \]

Step 3 – Evaluate: \[ r = 5\ \text{units} \]

\( \boxed{r = 5\ \text{units}} \)

118

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

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

Given: Figure 1: \(\quad l_1 = ?\ ,\quad A_1 = 13.5\ \text{cm}^2\)
Figure 2: \(\quad l_2 = 3\ \text{cm},\quad A_2 = 24\ \text{cm}^2\)

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

Step 2 – Substitute known values: \[ \frac{13.5}{24} = \left(\frac{l_1}{3}\right)^2 \]

Step 3 – Simplify the left side: \[ \frac{13.5}{24} = \frac{135}{240} = \frac{9}{16} \]

Step 4 – Take the square root of both sides: \[ \frac{l_1}{3} = \sqrt{\frac{9}{16}} = \frac{3}{4} \]

Step 5 – Solve for \(l_1\): \[ l_1 = 3 \times \frac{3}{4} = \frac{9}{4} = 2.25\ \text{cm} \]

\( \boxed{l_1 = 2.25\ \text{cm}} \)