Q03 Test 42 English

Test No. 42 – Complete Solutions

168

Convert to radians in terms of \(\pi\):  (i) 255°    (ii) 75°45′

🔑 Key Concept – Degrees to Radians: \[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \] For DMS angles, first convert to decimal degrees, then apply the formula.

(i) Convert 255° to radians

Step 1 – Apply the formula: \[ 255 \times \frac{\pi}{180} = \frac{255\pi}{180} \]

Step 2 – Simplify by dividing by 15: \[ \frac{255\pi}{180} = \frac{17\pi}{12} \]

\( \boxed{255° = \dfrac{17\pi}{12}\ \text{rad}} \)

(ii) Convert 75°45′ to radians

Step 1 – Convert to decimal degrees: \[ 75°45' = 75 + \frac{45}{60} = 75 + 0.75 = 75.75° \]

Step 2 – Apply the formula: \[ 75.75 \times \frac{\pi}{180} = \frac{75.75\pi}{180} \]

Step 3 – Simplify (multiply by 4/4 to clear decimal): \[ \frac{75.75\pi}{180} = \frac{303\pi}{720} = \frac{101\pi}{240} \]

\( \boxed{75°45' = \dfrac{101\pi}{240}\ \text{rad} \approx 1.3221\ \text{rad}} \)

169

Find equation of line with x-intercept \(-3\) and y-intercept \(4\).

🔑 Key Concept – Intercept Form: When x-intercept \(= a\) and y-intercept \(= b\) are known: \[ \frac{x}{a} + \frac{y}{b} = 1 \] The x-intercept is where the line crosses the x-axis: point \((a, 0)\).
The y-intercept is where it crosses the y-axis: point \((0, b)\).

Given: x-intercept \(a = -3\) → point \((-3,\ 0)\)
y-intercept \(b = 4\) → point \((0,\ 4)\)

Step 1 – Apply intercept form: \[ \frac{x}{-3} + \frac{y}{4} = 1 \]

Step 2 – Multiply through by 12 (LCM of 3 and 4): \[ -4x + 3y = 12 \]

Step 3 – Rearrange to standard form: \[ 4x - 3y + 12 = 0 \]

\( \boxed{4x - 3y + 12 = 0} \)

170

Points \(A(-1,\,h)\), \(B(3,\,2)\) and \(C(7,\,3)\) are collinear. Find \(h\).

🔑 Key Concept – Condition for Collinearity: Three points are collinear (lie on the same line) when the slope between any two pairs of points is equal: \[ \text{slope of } AB = \text{slope of } BC \] Equivalently, the area of the triangle formed by three collinear points is zero: \[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0 \]

Given: \(A(-1,\ h)\),  \(B(3,\ 2)\),  \(C(7,\ 3)\)

Step 1 – Set slope AB = slope BC: \[ \text{slope } AB = \frac{2 - h}{3 - (-1)} = \frac{2 - h}{4} \] \[ \text{slope } BC = \frac{3 - 2}{7 - 3} = \frac{1}{4} \]

Step 2 – Equate the slopes: \[ \frac{2 - h}{4} = \frac{1}{4} \]

Step 3 – Solve for \(h\): \[ 2 - h = 1 \] \[ h = 2 - 1 = 1 \]

\( \boxed{h = 1} \)

171

Ratio of volumes of two regular tetrahedrons is \(27:8\). Find the ratio of their edge lengths.

🔑 Key Concept – Edge Lengths from Volume Ratio: For two similar solids with corresponding lengths \(l_1\) and \(l_2\): \[ \frac{V_1}{V_2} = \left(\frac{l_1}{l_2}\right)^3 \] Rearranging to find the ratio of lengths: \[ \frac{l_1}{l_2} = \sqrt[3]{\frac{V_1}{V_2}} \]

Given: \(\dfrac{V_1}{V_2} = \dfrac{27}{8}\)

Step 1 – Apply the formula: \[ \frac{l_1}{l_2} = \sqrt[3]{\frac{V_1}{V_2}} = \sqrt[3]{\frac{27}{8}} \]

Step 2 – Evaluate the cube root: \[ \frac{l_1}{l_2} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}} = \frac{3}{2} \]

\( \boxed{\dfrac{l_1}{l_2} = \dfrac{3}{2} \quad \text{i.e. } 3:2} \)